Get notes, summary, questions and answers, MCQs, extras, competency-based questions and PDFs of Force: ICSE Class 10 Physics (Concise/Selina Workbook). However, the notes should only be treated as references, and changes should be made according to the needs of the students.
Summary
A force can make an object move. This movement can be in a straight line, which is called translational motion. It can also be a turning movement around a fixed point, known as rotational motion. The turning effect of a force is called the moment of force, or torque. The size of this turning effect depends on two things: the magnitude of the force and the perpendicular distance from the fixed point, or pivot, to the line of the force. If you apply a force farther from the pivot, you need less force to create the same turning effect. This is why door handles are placed far from the hinges. Moments can be clockwise or anticlockwise.
Sometimes, two equal and opposite forces act on an object to make it rotate. This pair of forces is called a couple. A couple creates rotation without any linear movement. Turning a steering wheel with both hands is an example of applying a couple.
An object is in equilibrium when the forces acting on it cause no change in its motion. For an object to be in equilibrium, the total force must be zero, preventing any change in linear motion. Also, the total turning effect must be zero. This leads to the principle of moments, which states that for an object to be balanced, the sum of all clockwise moments about a pivot must equal the sum of all anticlockwise moments.
Every object has a point called the centre of gravity. It is the point where the entire weight of the object can be considered to act. An object can be balanced perfectly if it is supported at its centre of gravity. For an object to be stable, its centre of gravity should be as low as possible.
When an object moves in a circle at a constant speed, it is in uniform circular motion. Although its speed is constant, its direction of motion is always changing. This change in direction means it is accelerating. This acceleration is caused by a force that constantly pulls the object toward the center of the circle, called the centripetal force. If you whirl a stone on a string, the tension in the string provides the centripetal force. An observer moving with the object might feel an outward push. This is called the centrifugal force. It is not a real force but a “pretend” one that appears due to the rotation.
Workbook solutions
Exercise 1 (A)
MCQ
1. The motion of a stationary rigid body in a straight path when a force is applied in the direction of force is called :
(a) rotational motion
(b) linear motion
(c) angular motion
(d) zigzag motion
Answer: (b) linear motion
2. The motion of a door when a force is applied on its handle is:
(a) rotational motion
(b) linear motion
(c) angular motion
(d) zig-zag motion
Answer: (a) rotational motion
3. The moment of a force about a given axis depends on
(a) only the magnitude of force
(b) only the perpendicular distance of force from the axis
(c) neither the force nor the perpendicular distance of force from the axis
(d) both the force and its perpendicular distance from the axis.
Answer: (d) both the force and its perpendicular distance from the axis.
4. For producing the maximum turning effect on a body by a given force, the perpendicular distance of the line of action of force from the axis of rotation should be:
(a) minimum
(b) it does not matter
(c) maximum
(d) zero
Answer: (c) maximum
5. The moment of force for anticlockwise moment is taken as __________ and for clockwise moment is taken as __________.
(a) positive, negative
(b) negative, positive
(c) positive, positive
(d) negative, negative
Answer: (a) positive, negative
6. The direction of rotation of a pivoted body can be changed by changing:
(a) the direction of force
(b) the point of application of force
(c) both the direction of force and the point of application of force
(d) the magnitude of force.
Answer: (c) both the direction of force and the point of application of force
7. A couple is formed when __________ are not acting along the same line.
(a) two unequal and parallel forces
(b) two equal and parallel forces
(c) two equal and opposite parallel forces
(d) two unequal and opposite parallel forces
Answer: (c) two equal and opposite parallel forces
8. As per the principle of moments in equilibrium :
(a) Sum of anticlockwise moments is greater than sum of clockwise moments
(b) Sum of anticlockwise moments is equal to sum of clockwise moments
(c) Sum of anticlockwise moments is less than sum of clockwise moments
(d) none of the above
Answer: (b) Sum of anticlockwise moments is equal to sum of clockwise moments
9. A body is acted upon by two unequal forces in opposite direction, but not along the same line. The effect is that:
(a) the body will only have rotational motion
(b) the body will only have translational motion
(c) the body will have neither rotational nor translational motion
(d) the body will have both rotational as well as translational motion
Answer: (d) the body will have both rotational as well as translational motion
10. Out of the following, which one is an example of dynamic equilibrium?
(1) Moon revolving around the earth.
(2) A pebble fixed at the end of a string whirling in a circular path.
(3) An aeroplane moving at a constant height.
(4) A beam balance balanced in a horizontal position.
(a) (3) and (4)
(b) (1), (2) and (3)
(c) (1) and (2)
(d) (1) and (3)
Answer: (b) (1), (2) and (3)
Very Short Questions
1. State the condition when on applying a force, a body has: (a) translational motion, (b) rotational motion.
Answer: (a) When a force acts on a stationary rigid body which is free to move, the body starts moving in a straight path in the direction of the applied force. This is called linear or translational motion.
(b) When a body is pivoted at a point i.e. not free to move and a force is applied on the body at a suitable point, it rotates the body about the axis passing through the pivoted point. This is the turning effect of the force and the motion of the body is called rotational motion.
2. State whether the moment of force is a scalar or vector quantity?
Answer: The moment of force is a vector quantity.
3. Write the expression for the moment of force about a given axis.
Answer: The expression for the moment of force about the axis passing through the point O is:
Moment of force = Force × Perpendicular distance of force from point O
or, Moment of force = F × OP
4. State one way to reduce the moment of a given force about a given axis of rotation.
Answer: One way to reduce the moment of a given force about a given axis of rotation is to decrease the perpendicular distance of the line of action of the applied force from the axis of rotation. For example, to open or shut a door, if the force is applied at a point Q (near the hinge R), much greater force is required to open the door compared to applying force at the handle P which is at the maximum distance from the hinges.
5. State one way to obtain a greater moment of a force about a given axis of rotation.
Answer: One way to obtain a greater moment of a force about a given axis of rotation is to increase the perpendicular distance of the line of action of the applied force from the axis of rotation. For example, a spanner, used to tighten or loosen a nut, has a long handle to produce a large moment of force by a small force applied normally at the end of its handle.
6. Complete the following sentences :
(i) The S.I. unit of moment of force is ……………….
(ii) In equilibrium, algebraic sum of moments of all forces about the point of rotation is ……………….
(iii) In a beam balance when the beam is balanced in a horizontal position, it is in ………………. equilibrium.
(iv) The moon revolving around the earth is in ………………. equilibrium.
Answer: (i) The S.I. unit of moment of force is newton × metre (or N m).
(ii) In equilibrium, algebraic sum of moments of all forces, acting on the body, about the axis of rotation is zero.
(iii) In a beam balance when the beam is balanced in horizontal position, it is in static equilibrium.
(iv) The moon revolving around the earth is in dynamic equilibrium.
Short Questions
1. Define moment of force and state its S.I. unit.
Answer: The moment of a force (or torque) is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the axis of rotation. The S.I. unit of moment of force is newton × metre, which is abbreviated as N m.
2. State two factors affecting the turning effect of a force.
Answer: The turning effect of a force on a body depends on the following two factors:
(1) the magnitude of the force applied, and
(2) the perpendicular distance of the line of action of the force from the axis of rotation (or pivoted point).
3. When does a body rotate ? State one way to change the direction of rotation of the body. Give a suitable example to explain your answer.
Answer: When a body is pivoted at a point (i.e., not free to move) and a force is applied on the body at a suitable point, it rotates the body about the axis passing through the pivoted point; the body rotates due to the moment of force (or torque) about the pivoted point.
The direction of rotation of a body can be changed by changing the point of application of force. For example, as shown in the figure, for turning a steering wheel, the sense of rotation of the wheel can be changed by changing the point of application of force without changing the direction of force. In fig. (a), when force F is applied at point A of the wheel, the wheel rotates anticlockwise; while in fig. (b), the wheel rotates clockwise when the same force F is applied in the same direction at point B of the wheel.
4. What do you understand by the clockwise and anticlockwise moment of force ? When is it taken positive?
Answer: Conventionally, if the effect on the body is to turn it anticlockwise, the moment of force is called anticlockwise moment and it is taken positive. While if the effect on the body is to turn it clockwise, the moment of force is called clockwise moment and it is taken negative.
5. Why is it easier to open a door by applying the force at the free end of it ?
Answer: It is easier to open a door by applying the force at its free end because the handle is provided near the free end of the door so that a smaller force being at a larger perpendicular distance from the hinges produces the maximum moment of force that is required to open or shut the door.
6. The stone of a hand flour grinder is provided with a handle near its rim. Give reason.
Answer: The upper movable circular stone of a hand flour grinder is provided with a handle near its rim (i.e., at the maximum distance from the centre) so that it can easily be rotated about the iron pivot at its centre by applying a smaller force at the handle.
7. It is easier to turn the steering wheel of a large diameter than that of a small diameter. Give reason.
Answer: For a steering wheel of a large diameter, the force is applied at a larger perpendicular distance from the axis of rotation. Since a larger perpendicular distance means less force is needed to produce the same turning effect, it is easier to turn a large diameter steering wheel compared to a small diameter one.
8. A spanner (or wrench) has a long handle. Why ?
Answer: A spanner, used to tighten or loosen a nut, has a long handle to produce a large moment of force by a small force applied normally at the end of its handle.
9. A jack screw is provided with a long arm. Explain why?
Answer: A jack screw used to lift a heavy load such as a vehicle, has a long arm so that less effort is needed to rotate it so as to raise or lower the load table.
10. The adjacent diagram (Fig. 1.26) shows a heavy roller, with its axle at O, which is to be raised on a pavement XY. If there is friction between the roller and pavement, show by an arrow on the diagram the point of application and the direction of force to be applied. If pivoted at O, now will it go up?
Answer: To raise the roller on the pavement XY, a force should be applied at the rim of the roller in a direction that creates an anticlockwise moment about the edge of the pavement XY, generally upwards and forwards. The point of application would be on the rim.
If the roller is pivoted at its axle O, applying a force tangentially will cause it to rotate about O.
11. Define moment of couple. Write its S.I. unit.
Answer: The moment of a couple is equal to either force multiplied by the perpendicular distance between the two forces, also known as the couple arm.
The S.I. unit of moment of force is newton × metre, which is abbreviated as N m.
12. What do you mean by equilibrium of a body?
Answer: A body preserving its state, whether static or dynamic, in the presence of two or more forces is said to be in equilibrium. When a number of forces acting on a body produce no change in its state of rest or of linear or rotational motion, the body is said to be in a state of equilibrium.
13. State the condition when a body is in (i) static, (ii) dynamic equilibrium. Give one example each of static and dynamic equilibrium.
Answer: (i) When a body remains in a state of rest under the influence of several forces, the body is in static equilibrium. For example, if a body lying on a table top is pulled by a force F to its left and by an equal force F’ to its right along the same line, the body does not move. The applied forces are equal and opposite and along the same line, so they balance each other. Hence, the body remains at rest, in static equilibrium.
(ii) When a body remains in the same state of motion, either translational or rotational, under the influence of several forces, the body is said to be in dynamic equilibrium. For example, a rain drop reaches the earth’s surface with a constant velocity. The weight of the falling drop is balanced by the sum of the buoyant force and the force due to friction of air. Thus, the net force on the drop is zero, so it falls down with a constant velocity.
14. State two conditions for a body, acted upon by several forces, to be in equilibrium.
Answer: The following two conditions must be satisfied for a body to be in equilibrium:
(1) The resultant of all the forces acting on the body should be zero.
(2) The algebraic sum of moments of all the forces acting on the body about the point of rotation should be zero. This means the sum of the anticlockwise moments about the axis of rotation must be equal to the sum of the clockwise moments about the same axis.
5. State the principle of moments. Name one device based on it.
Answer: According to the principle of moments, in equilibrium, the sum of the anticlockwise moments is equal to the sum of the clockwise moments.
A physical balance, or beam balance, is a device that works on the principle of moments.
Long Questions
1. A, B and C are three forces each of magnitude 4 N acting in the plane of paper as shown in Fig. 1.27. Point O lies in the same plane.
(i) Which force has the least moment about O? Give reason.
Answer: Force C has the least moment about O because force C is nearest to O.
(ii) Which force has the greatest moment about O? Give reason.
Answer: Force A has the greatest moment about O because force A is farthest from O.
(iii) Name the forces producing (a) clockwise, (b) anticlockwise moments.
Answer: (a) Forces A and B produce clockwise moments. (b) Force C produces an anticlockwise moment.
(iv) What is the resultant torque about the point O ?
Answer: The resultant torque about the point O is 4.4 Nm (clockwise).
2. A body is acted upon by two forces each of magnitude F, but in opposite directions. State the effect of the forces if
(a) both forces act at the same point of the body.
Answer: If both forces act at the same point of the body, the resultant force is 0, the moment of forces is 0, and there is no motion.
(b) the two forces act at two different points of the body at a separation r.
Answer: If the two forces act at two different points of the body at a separation r, the resultant force is 0, and the moment of forces is Fr. The forces tend to rotate the body about the mid-point between the two forces.
3. Draw a neat labelled diagram to show the direction of two forces acting on a body to produce rotation in it. Also mark the point O about which the rotation takes place.
Answer: A neat labelled diagram should show a bar AB pivoted at a point O. Two equal and opposite forces, each of magnitude F, are applied at the ends A and B, perpendicular to the bar. The distance AB is the couple arm (d). The forces together form a couple which rotates the bar about the point O.
4. What do you understand by the term couple ? State its effect on a body. Give two examples in our daily life where couple is applied to turn a body.
Answer: Two equal and opposite parallel forces, not acting along the same line, form a couple. A couple is always needed to produce a rotation. The two forces cannot produce translational motion as their resultant sum in any direction is zero, but each force is capable of producing a turning effect on the bar in the same direction. Thus, the two forces together form a couple which rotates the body.
Two examples in daily life where a couple is applied to turn a body are turning a water tap and turning the steering wheel of a truck.
5. Prove that Moment of couple = Force × couple arm.
Answer: Consider a bar AB which is pivoted at a point O. At the ends A and B, two equal and opposite forces, each of magnitude F, are applied. The perpendicular distance between the two forces is AB (= d), which is called the couple arm.
Moment of force F at end A = F × OA (anticlockwise)
Moment of force F at end B = F × OB (anticlockwise)
Total moment of couple (i.e., moment of both the forces) = F × OA + F × OB
= F × (OA + OB) = F × AB
Since AB = d (couple arm),
Total moment of couple = F × d (anticlockwise)
Thus, Moment of couple = Either force × perpendicular distance between the two forces (or couple arm).
6. Describe a simple experiment to verify the principle of moments, if you are supplied with a metre rule, a fulcrum and two springs with slotted weights.
Answer: To verify the principle of moments, suspend a metre rule horizontally from a fixed support by means of a strong thread at O (the fulcrum). Suspend two spring balances (or use springs with slotted weights) A and B on the metre rule on either side of the thread O. Suspend some slotted weights W₁ from spring balance A and W₂ from spring balance B. The metre rule may tilt to one side. Adjust either the slotted weights on the spring balances or the position of the spring balances on either side of the thread from O in such a way that the metre rule becomes horizontal again.
Let the weight suspended from the spring balance A on the right side of the thread be W₁ at a distance OA = l₁ from O.
Let the weight suspended from the spring balance B on the left side of the thread be W₂ at a distance OB = l₂ from O.
The weight W₁ tends to turn the metre rule clockwise, while the weight W₂ tends to turn the metre rule anticlockwise.
Clockwise moment of weight W₁ about the point O = W₁ × l₁.
Anticlockwise moment of weight W₂ about the point O = W₂ × l₂.
In equilibrium, when the metre rule is horizontal, it is found that W₁l₁ = W₂l₂.
i.e., clockwise moment = anticlockwise moment.
This verifies the principle of moments.
Numericals
1. The moment of a force of 20 N about a fixed point O is 10 N m. Calculate the distance of the point O from the line of action of the force.
Answer: Given,
Moment of force = 10 N m
Force, F = 20 N
Let the perpendicular distance of the point of application of force from the point O be r metre.
We know that,
Moment of force = force × distance
10 = 20 × r
or, r = 10 / 20
or, r = 0.5 m
Thus, the distance of the point O from the line of action of the force is 0.5 m.
2. A nut is opened by a wrench of length 25 cm. If the least force required is 10 N, find the moment of force needed to turn the nut.
Answer:
Given,
Force, F = 10 N
Length of wrench, r = 25 cm = 0.25 m
We know that,
Moment of force = F × r
Moment of force = 10 N × 0.25 m
Moment of force = 2.5 N m
Thus, the moment of force needed to turn the nut is 2.5 N m.
3. A 50 kg mass is being lifted by using a 1 metre long lever. If the length of the lever is reduced by 20%, how much more or less force is to be applied to achieve the same torque ?
Answer: In the first case:
Mass = 50 kg, so Force, F₁ = 50 kgf
Length of lever, r₁ = 1 m
Torque needed = F₁ × r₁ = 50 kgf × 1 m = 50 kgf m
In the second case:
The length of the lever is reduced by 20%.
New length of lever, r₂ = 1 m – (20% of 1 m) = 1 m – 0.2 m = 0.8 m
The torque required is the same, i.e., 50 kgf m.
Let the new force required be F₂.
Torque = F₂ × r₂
50 = F₂ × 0.8
or, F₂ = 50 / 0.8 = 62.5 kgf
Difference in force:
Increase in force = F₂ – F₁ = 62.5 kgf – 50 kgf = 12.5 kgf
The question asks for the answer in Newtons.
Using the relation 1 kgf = 9.8 N,
Increase in force = 12.5 × 9.8 N = 122.5 N
Thus, 122.5 N more force is to be applied.
4. A mechanic applies a force of 100 N at the end of a spanner to tighten a bolt. The length of the spanner is 0.5 metres. How much torque is exerted on the bolt ? If the mechanic increases the force applied by 20% but holds the spanner closer at 0.4 metres, would it result in an increase or decrease of torque ?
Answer: In the first case:
Force, F₁ = 100 N
Length of spanner, r₁ = 0.5 m
Torque exerted, τ₁ = F₁ × r₁ = 100 N × 0.5 m = 50 N m
In the second case:
The force is increased by 20%.
New force, F₂ = 100 N + (20% of 100 N) = 100 N + 20 N = 120 N
New length of spanner, r₂ = 0.4 m
New torque exerted, τ₂ = F₂ × r₂ = 120 N × 0.4 m = 48 N m
Comparison:
The initial torque was 50 N m and the new torque is 48 N m.
Decrease in torque = τ₁ – τ₂ = 50 N m – 48 N m = 2 N m
Thus, the torque exerted on the bolt is initially 50 N m, and it would result in a decrease of torque by 2 N m.
5. A wheel of diameter 2 m is shown in Fig. 1.28 with axle at O. A force F = 2 N is applied at B in the direction shown in figure. Calculate the moment of force about (i) the centre O, and (ii) the point A.
Answer: Given,
Diameter of the wheel = 2 m
Radius of the wheel = Diameter / 2 = 2 m / 2 = 1 m
Force, F = 2 N applied at B.
(i) Moment of force about the centre O:
The perpendicular distance of the line of action of force F from the centre O is the radius of the wheel, OB.
Distance = OB = 1 m
Moment of force about O = F × OB = 2 N × 1 m = 2 N m
The force tends to rotate the wheel clockwise.
So, the moment of force is 2 N m (clockwise).
(ii) Moment of force about the point A:
The perpendicular distance of the line of action of force F from the point A is the diameter of the wheel, AB.
Distance = AB = 2 m
Moment of force about A = F × AB = 2 N × 2 m = 4 N m
The force tends to rotate the wheel clockwise about point A.
So, the moment of force is 4 N m (clockwise).
6. The diagram in Fig. 1.29 shows two forces F₁ = 5 N and F₂ = 3 N acting at points A and B respectively of a rod pivoted at a point O, such that OA = 2 m and OB = 4 m. Calculate : (i) the moment of force F₁ about O. (ii) the moment of force F₂ about O. (iii) total moment of the two forces about O.
Answer: Given,
F₁ = 5 N, F₂ = 3 N
OA = 2 m, OB = 4 m
(i) Moment of force F₁ about O:
The force F₁ tends to rotate the rod anticlockwise.
Moment of F₁ about O = F₁ × OA = 5 N × 2 m = 10 N m
The moment is 10 N m (anticlockwise).
(ii) Moment of force F₂ about O:
The force F₂ tends to rotate the rod clockwise.
Moment of F₂ about O = F₂ × OB = 3 N × 4 m = 12 N m
The moment is 12 N m (clockwise).
(iii) Total moment of the two forces about O:
By convention, anticlockwise moment is taken as positive and clockwise moment as negative.
Total moment = (Moment of F₁) + (Moment of F₂)
Total moment = (+10 N m) + (-12 N m) = -2 N m
The negative sign indicates that the resultant moment is clockwise.
So, the total moment is 2 N m (clockwise).
7. Two forces each of magnitude 10 N act vertically upwards and downwards respectively at the two ends A and B of a uniform rod of length 4 m which is pivoted at its mid point O as shown in Fig. 1.30. Determine the magnitude of the resultant moment of forces about the pivot O.
Answer: Given,
Force at A, F_A = 10 N (upwards)
Force at B, F_B = 10 N (downwards)
Length of rod AB = 4 m
The rod is pivoted at its midpoint O, so OA = OB = 4 m / 2 = 2 m.
Moment of force at A about O = F_A × OA = 10 N × 2 m = 20 N m
This force tends to rotate the rod clockwise.
Moment of force at B about O = F_B × OB = 10 N × 2 m = 20 N m
This force also tends to rotate the rod clockwise.
Since both moments are in the same direction (clockwise), the resultant moment is their sum.
Resultant moment = 20 N m + 20 N m = 40 N m (clockwise).
8. Fig. 1.31 shows two forces each of magnitude 10 N acting at points A and B at a separation of 50 cm, in opposite directions. Calculate the resultant moment of the two forces about the point (i) A, (ii) B and (iii) O situated exactly at the middle of the two forces.
Answer: Given,
Force at A, F_A = 10 N
Force at B, F_B = 10 N
Separation AB = 50 cm = 0.5 m
(i) Resultant moment about point A:
Moment due to force at A is zero since the distance is zero.
Moment due to force at B = F_B × AB = 10 N × 0.5 m = 5 N m. This force causes a clockwise rotation about A.
Total moment about A = 5 N m (clockwise).
(ii) Resultant moment about point B:
Moment due to force at B is zero.
Moment due to force at A = F_A × AB = 10 N × 0.5 m = 5 N m. This force also causes a clockwise rotation about B.
Total moment about B = 5 N m (clockwise).
(iii) Resultant moment about point O (midpoint):
Point O is at the middle, so OA = OB = 0.5 m / 2 = 0.25 m.
Moment due to force at A = F_A × OA = 10 N × 0.25 m = 2.5 N m (clockwise).
Moment due to force at B = F_B × OB = 10 N × 0.25 m = 2.5 N m (clockwise).
Total moment about O = 2.5 N m + 2.5 N m = 5 N m (clockwise).
9. A steering wheel of diameter 0.5 m is rotated anticlockwise by applying two forces each of magnitude 6 N. Draw a diagram to show the application of forces and calculate the moment of the forces applied.
Answer: The two forces form a couple. To rotate the wheel anticlockwise, two equal and opposite forces are applied tangentially at the ends of a diameter.
Given,
Magnitude of each force, F = 6 N
Diameter of the wheel (couple arm), d = 0.5 m
Moment of couple = Force × couple arm
Moment = F × d
Moment = 6 N × 0.5 m = 3 N m
The moment of the forces applied is 3 N m.
10. A uniform metre rule is pivoted at its mid-point. A weight of 50 gf is suspended at one end of it. Where should a weight of 100 gf be suspended to keep the rule horizontal ?
Answer: The pivot is at the mid-point, which is the 50 cm mark.
Let the 50 gf weight be suspended at the 100 cm end.
Distance of this weight from the pivot = 100 cm – 50 cm = 50 cm.
This creates a clockwise moment.
Clockwise moment = 50 gf × 50 cm = 2500 gf cm.
To balance the rule, an equal anticlockwise moment is required. This will be provided by the 100 gf weight. Let its distance from the pivot be x.
Anticlockwise moment = 100 gf × x.
By the principle of moments, in equilibrium:
Anticlockwise moment = Clockwise moment
100 × x = 2500
x = 2500 / 100 = 25 cm
This distance is from the pivot (50 cm mark) on the other side.
Position on the rule = 50 cm – 25 cm = 25 cm mark.
This position is 25 cm from the 0 cm end. The “other end” is the 0 cm end.
So, the weight should be suspended at a distance of 25 cm from the other end.
11. A uniform metre rule balances horizontally on a knife edge placed at the 58 cm mark when a weight of 20 gf is suspended from one end. (i) Draw a diagram of the arrangement. (ii) What is the weight of the rule ?
Answer: (i) The diagram would show a metre rule (0-100 cm) with a pivot at the 58 cm mark. The weight of the rule (W) acts at its centre (50 cm mark), creating an anticlockwise moment. To balance this, the 20 gf weight must be suspended at the 100 cm end to create a clockwise moment.
(ii) Let the weight of the rule be W. Since the rule is uniform, its weight acts at its midpoint, the 50 cm mark.
The pivot is at the 58 cm mark.
The weight of the rule W creates an anticlockwise moment.
Distance of W from the pivot = 58 cm – 50 cm = 8 cm.
Anticlockwise moment = W × 8.
The suspended 20 gf weight creates a clockwise moment. It is suspended at the end further from the pivot, which is the 100 cm mark.
Distance of 20 gf from the pivot = 100 cm – 58 cm = 42 cm.
Clockwise moment = 20 gf × 42 cm = 840 gf cm.
For the rule to be in equilibrium,
Anticlockwise moment = Clockwise moment
W × 8 = 840
W = 840 / 8 = 105 gf.
The weight of the rule is 105 gf.
12. The diagram below (Fig. 1.32) shows a uniform bar supported at the middle point O. A weight of 40 gf is placed at a distance 40 cm to the left of point O. How can you balance the bar with a weight of 80 gf ?
Answer: The weight of 40 gf at 40 cm to the left of O creates an anticlockwise moment.
Anticlockwise moment = 40 gf × 40 cm = 1600 gf cm.
To balance the bar, a clockwise moment of 1600 gf cm is needed. This will be provided by the 80 gf weight placed at a distance x to the right of O.
Clockwise moment = 80 gf × x.
In equilibrium,
Clockwise moment = Anticlockwise moment
80 × x = 1600
x = 1600 / 80 = 20 cm.
The bar can be balanced by placing the weight of 80 gf at a distance of 20 cm to the right of point O.
13. Fig. 1.33 shows a uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40 gf at the 10 cm mark and a weight of 20 gf at the 90 cm mark. (i) Is the metre rule in equilibrium ? If not, how will the rule turn ? (ii) How can the rule be brought to equilibrium by using an additional weight of 40 gf ?
Answer: The fulcrum is at the mid-point O (50 cm mark).
(i) Checking for equilibrium: Anticlockwise moment is produced by the 40 gf weight at the 10 cm mark.
Distance from fulcrum = 50 cm – 10 cm = 40 cm.
Anticlockwise moment = 40 gf × 40 cm = 1600 gf cm.
Clockwise moment is produced by the 20 gf weight at the 90 cm mark.
Distance from fulcrum = 90 cm – 50 cm = 40 cm.
Clockwise moment = 20 gf × 40 cm = 800 gf cm.
Since the anticlockwise moment (1600 gf cm) is not equal to the clockwise moment (800 gf cm), the rule is not in equilibrium. It will turn anticlockwise.
(ii) Bringing to equilibrium: The net anticlockwise moment is 1600 – 800 = 800 gf cm.
To balance the rule, an additional clockwise moment of 800 gf cm is needed. This can be provided by placing the additional 40 gf weight on the right side of the fulcrum at a distance x from it.
Additional clockwise moment = 40 gf × x.
40 × x = 800
x = 800 / 40 = 20 cm.
The position of this weight on the rule will be 50 cm (fulcrum) + 20 cm = 70 cm mark.
The rule can be brought to equilibrium by placing the additional weight of 40 gf at the 70 cm mark.
14. When a boy weighing 20 kgf sits at one end of a 4 m long see-saw, it gets depressed at this end. How can it be brought to the horizontal position by a man weighing 40 kgf.
Answer: The see-saw is 4 m long, so its pivot is at the center, 2 m from each end.
The boy (weight = 20 kgf) sits at one end, at a distance of 2 m from the pivot.
Moment due to the boy = 20 kgf × 2 m = 40 kgf m.
To bring the see-saw to a horizontal position, the man (weight = 40 kgf) must sit on the other side to produce an equal and opposite moment. Let the man’s distance from the pivot be x.
Moment due to the man = 40 kgf × x.
In equilibrium,
Moment due to man = Moment due to boy
40 × x = 40
x = 1 m.
The man should sit at a distance of 1 m from the centre on the side opposite to the boy.
15. A physical balance has its arms of length 60 cm and 40 cm. What weight kept on the pan of the longer arm will balance an object of weight 100 gf kept on the other pan ?
Answer: Let the weight on the longer arm (60 cm) be W.
The object of weight 100 gf is on the shorter arm (40 cm).
Moment on the longer arm = W × 60.
Moment on the shorter arm = 100 gf × 40 cm = 4000 gf cm.
For the balance to be in equilibrium, the moments must be equal.
W × 60 = 4000
W = 4000 / 60 = 400 / 6 = 66.67 gf.
A weight of 66.67 gf kept on the pan of the longer arm will balance the object.
16. The diagram in Fig. 1.34 shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf hang from the points A and B respectively of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate : (i) the total anticlockwise moment about O, (ii) the total clockwise moment about O, (iii) the difference of anticlockwise and clockwise moments, and (iv) the distance from O where a 100 gf weight should be placed to balance the metre rule.
Answer: The rule is pivoted at its centre O, so the weight of the rule produces no moment about O.
(i) Total anticlockwise moment about O:
This is caused by the 150 gf weight at distance OA = 40 cm.
Anticlockwise moment = 150 gf × 40 cm = 6000 gf cm.
(ii) Total clockwise moment about O:
This is caused by the 250 gf weight at distance OB = 20 cm.
Clockwise moment = 250 gf × 20 cm = 5000 gf cm.
(iii) Difference of moments:
Difference = Anticlockwise moment – Clockwise moment = 6000 gf cm – 5000 gf cm = 1000 gf cm.
(iv) Distance to balance the rule:
There is a net anticlockwise moment of 1000 gf cm. To balance this, an additional clockwise moment of 1000 gf cm is needed. This will be provided by placing a 100 gf weight at a distance x from O on the right side.
100 gf × x = 1000 gf cm
x = 1000 / 100 = 10 cm.
The 100 gf weight should be placed at a distance of 10 cm from O on the right side.
17. A uniform metre rule of weight 10 gf is pivoted at its 0 mark. (i) What moment of force depresses the rule ? (ii) How can it be made horizontal by applying least force?
Answer:
(i) Moment of force that depresses the rule:
The rule is uniform, so its weight of 10 gf acts at its centre, the 50 cm mark.
The pivot is at the 0 mark.
The distance of the weight from the pivot is 50 cm.
The moment that depresses the rule (clockwise moment) = 10 gf × 50 cm = 500 gf cm.
(ii) How to make it horizontal with least force:
To make the rule horizontal, an anticlockwise moment of 500 gf cm is needed.
Moment = Force × Perpendicular distance.
To apply the least force, the perpendicular distance must be maximum. The maximum distance from the pivot (0 mark) is the other end of the rule, at the 100 cm mark.
Let the least force be F.
F × 100 cm = 500 gf cm
F = 500 / 100 = 5 gf.
It can be made horizontal by applying a force of 5 gf upwards at the 100 cm mark.
18. A uniform half metre rule can be balanced at the 29.0 cm mark when a mass 20 g is hung from its one end. (a) Draw a diagram of the arrangement. (b) Find the mass of the half metre rule. (c) In which direction would the balancing point shift if 20 g mass is shifted inside from its one end?
Answer: (a) The diagram would show a half-metre rule (0-50 cm) with a pivot at 29.0 cm. The mass of the rule (M) acts at its centre (25 cm). The 20 g mass is hung at the 50 cm end to balance the rule.
(b) Mass of the half metre rule:
Let the mass of the rule be M. Its weight acts at the centre, 25 cm.
The pivot is at 29.0 cm.
The mass of the rule creates an anticlockwise moment.
Distance for M from pivot = 29.0 cm – 25 cm = 4.0 cm.
Anticlockwise moment = M × 4.0.
The 20 g mass is hung at one end. To balance, it must be at the 50 cm end.
Distance for 20 g from pivot = 50 cm – 29.0 cm = 21.0 cm.
Clockwise moment = 20 g × 21.0 cm = 420 g cm.
In equilibrium,
M × 4.0 = 420
M = 420 / 4.0 = 105 g.
(c) Direction of shift:
If the 20 g mass is shifted inside from its end (i.e., moved from 50 cm towards the pivot), its distance from the pivot decreases. This reduces the clockwise moment. The anticlockwise moment due to the rule’s mass will then be greater. To restore balance, the pivot must be moved to decrease the lever arm of the rule’s mass, which means moving the pivot towards the 25 cm mark.
19. A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm by suspending an unknown mass m at the mark 20 cm. (i) Find the value of m. (ii) To which side the rule will tilt if the mass m is moved to the mark 10 cm ? (iii) What is the resultant moment now ? (iv) How can it be balanced by another mass of 50 g ?
Answer: (i) Find the value of m:
The fulcrum is at 40 cm.
The mass of the rule (100 g) acts at its centre (50 cm), creating a clockwise moment.
Distance = 50 cm – 40 cm = 10 cm.
Clockwise moment = 100 g × 10 cm = 1000 g cm.
The unknown mass m at 20 cm creates an anticlockwise moment.
Distance = 40 cm – 20 cm = 20 cm.
Anticlockwise moment = m × 20.
In equilibrium, m × 20 = 1000, so m = 50 g.
(ii) To which side the rule will tilt:
The mass m (50 g) is moved to the 10 cm mark.
New distance from fulcrum = 40 cm – 10 cm = 30 cm.
New anticlockwise moment = 50 g × 30 cm = 1500 g cm.
Since the new anticlockwise moment (1500 g cm) is greater than the clockwise moment (1000 g cm), the rule will tilt on the side of mass m (anticlockwise).
(iii) Resultant moment now:
Resultant moment = 1500 g cm – 1000 g cm = 500 g cm (anticlockwise).
(iv) How can it be balanced:
To balance the net anticlockwise moment of 500 g cm, an additional clockwise moment of 500 g cm is needed. This can be provided by a 50 g mass placed at a distance x to the right of the fulcrum.
50 g × x = 500 g cm
x = 10 cm.
The position on the rule is 40 cm (fulcrum) + 10 cm = 50 cm mark.
It can be balanced by suspending the mass 50 g at the mark 50 cm.
20. In Fig. 1.35, a uniform bar of length l m is supported at its ends and loaded by a weight W kgf at its middle. In equilibrium, find the reactions R₁ and R₂ at the ends.
Answer: For the bar to be in equilibrium, two conditions must be met:
- The net force on the bar is zero (translational equilibrium).
- The net moment about any point is zero (rotational equilibrium).
From translational equilibrium:
The total upward force must equal the total downward force.
Upward forces = R₁ + R₂
Downward force = W
So, R₁ + R₂ = W —(i)
From rotational equilibrium:
Let’s take moments about the support point for R₁.
The weight W acts at the middle, at a distance of l/2 from the end. This creates a clockwise moment.
Clockwise moment = W × (l/2).
The reaction force R₂ acts at the other end, at a distance of l. This creates an anticlockwise moment.
Anticlockwise moment = R₂ × l.
For equilibrium,
Anticlockwise moment = Clockwise moment
R₂ × l = W × (l/2)
R₂ = W/2.
Now, substitute the value of R₂ in equation (i):
R₁ + (W/2) = W
R₁ = W – W/2
R₁ = W/2.
Thus, the reactions at the ends are R₁ = W/2 kgf and R₂ = W/2 kgf.
Exercise 1 (B)
MCQ
Here are the answers to the multiple-choice questions:
1. With respect to centre of gravity, which of the following statements are correct?
(a) The position of centre of gravity of a body depends on its shape.
(b) It is not necessary that the centre of gravity of a body should always be within the material of the body.
(c) A body of weight W can be considered as a point particle of weight W at its centre of gravity.
(d) All of the above
Answer: (d) All of the above
2. The centre of gravity of a body is the point about which the algebraic sum of moments of weights of all the particles constituting the body is :
(a) 1
(b) > 1
(c) < 1
(d) 0
Answer: (d) 0
3. The centre of gravity of a hollow cone of heigth h is at distance x from its vertex where the value of x is:
(a) h/3
(b) h/4
(c) 2h/3
(d) 3h/4
Answer: (c) 2h/3
4. Where should the centre of gravity of a body be located for stable equilibrium ?
(a) above the base
(b) near the geometric centre
(c) can be anywhere
(d) both (a) and (b)
Answer: (d) both (a) and (b)
5. Two objects have the same mass but different shapes. Which object would have a higher centre of gravity?
(a) the one with a larger base
(b) the one with a smaller base
(c) both would have the same centre of gravity
(d) it depends on the material of the obejct
Answer: (b) the one with a smaller base
Very Short Questions
1. Can the centre of gravity of a body be situated outside its material? Give an example.
Answer: Yes, the centre of gravity of a body can be situated outside its material. For example, the centre of gravity of a ring is at its centre where there is no material.
2. What is the position of the centre of gravity of a : (a) rectangular lamina (b) cylinder ?
Answer: The position of the centre of gravity is:
(a) for a rectangular lamina, at the point of intersection of its diagonals.
(b) for a cylinder, at the mid point on its axis.
3. At which point is the centre of gravity situated in : (a) a triangular lamina and (b) a circular lamina ?
Answer: The centre of gravity is situated:
(a) in a triangular lamina, at the point of intersection of its medians.
(b) in a circular lamina, at its centre.
4. Where is the centre of gravity of a uniform ring situated ?
Answer: The centre of gravity of a uniform ring is situated at its centre.
5. State whether the following statements are true or false.
(i) ‘The position of the centre of gravity of a body remains unchanged even when the body is deformed.’
(ii) ‘The centre of gravity of a freely suspended body always lies vertically below the point of suspension’.
Answer:
(i) False.
(ii) True.
Short Answer Questions
1. Define the term ‘centre of gravity of a body’.
Answer: The centre of gravity (C.G) of a body is the point about which the algebraic sum of moments of weights of all the particles constituting the body is zero. The entire weight of the body can be considered to act at this point, howsoever the body is placed.
2. State a factor on which the position of the centre of gravity of a body depends. Explain your answer with an example.
Answer: The position of the centre of gravity of a body of given mass depends on its shape, i.e., on the distribution of mass (of particles) in it. It changes if the body is deformed. For example, the centre of gravity of a straight uniform wire is at the middle of its length. But if the same wire is bent into the form of a circle, its centre of gravity will then be at the centre of the circle.
3. A square card board is suspended by passing a pin through a narrow hole at its one corner. Draw a diagram to show its rest position. In the diagram, mark the point of suspension by the letter S and the centre of gravity by the letter G.
Answer: When a body is freely suspended from a point, it comes to rest in such a position that its centre of gravity lies vertically below the point of suspension. A diagram would show the square card board suspended from one corner, marked S. The centre of gravity, G, which is at the intersection of the diagonals of the square, would be shown vertically below S. A line would connect S and G, representing the plumb line’s direction when the card board is at rest.
4. Humans bend forward when they carry a heavy load on their back. Give reason.
Answer: Humans bend forward when they carry a heavy load on their back to align their centre of gravity, ensuring that the combined centre of gravity of the person and the load remains over their feet for stability.
5. You are assigned to look after a loading of a cargo ship with containers marked A and B. The A type containers are 50% more heavier than B type containers. What precaution would you take and why so that the ship sails smoothly even in rougher seas?
Answer: To prevent capsizing of ships, the centre of gravity is kept lower. Therefore, the heavier A type containers should be placed lower in the cargo ship, and the lighter B type containers can be placed higher. This ensures the ship’s stability, allowing it to sail smoothly even in rougher seas.
6. A uniform flat circular rim is balanced on a sharp vertical nail by supporting it at a point A, as shown in Fig. 1.41. Mark the position of the centre of gravity of the rim in the diagram by the letter G.
Answer: The centre of gravity of a uniform flat circular rim is at its geometric centre. In Fig. 1.41, the position of the centre of gravity G should be marked at the centre of the circular rim.
7. Fig. 1.42 shows three pieces of card board of uniform thickness cut into three different shapes. On each diagram draw two lines to indicate the position of the centre of gravity G.
Answer: To indicate the position of the centre of gravity G on each diagram in Fig. 1.42: For the first shape, which appears to be a rectangle or parallelogram, draw its two diagonals. The point where these two diagonals intersect is the centre of gravity G.
For the second shape, which appears to be a triangle, draw two of its medians (a line from a vertex to the midpoint of the opposite side). The point where these two medians intersect is the centre of gravity G.
For the third shape, which appears to be a circular disc, draw two of its diameters. The point where these two diameters intersect is the centre of gravity G.
Long Questions
1. Explain how will you determine experimentally the position of the centre of gravity for a triangular lamina (or a triangular piece of card board).
Answer: To determine experimentally the position of the centre of gravity for a triangular lamina (or a triangular piece of card board), you would first make three fine holes, say at points a, b, and c, near its edge. Then, suspend the given lamina along with a plumb line from the hole a, on a pin (or a nail) clamped horizontally on a retort stand. Check that the lamina is free to oscillate on the nail about the point of suspension. When the lamina has come to rest, draw a straight line, say ad, along the plumb line.
Repeat this procedure by suspending the lamina first through the hole b, and when it comes to rest, draw another straight line, say be, along the plumb line. Then, suspend the lamina through the hole c, and when it comes to rest, draw a third straight line, say cf, along the plumb line.
It will be noticed that the lines ad, be, and cf intersect each other at a common point G. This point G is the position of the centre of gravity of the triangular lamina. This method is generally used for an irregular lamina.
Exercise 1(C)
MCQ
1. Which of the following quantities remains constant in a uniform circular motion?
(a) Velocity
(b) Speed
(c) Acceleration
(d) Both velocity and speed
Answer: (b) Speed
2. The direction of motion in circular motion :
(a) is linear
(b) is along the tangent at that point of the circular path
(c) towards the centre
(d) none of these
Answer: (b) is along the tangent at that point of the circular path
3. The direction of centripetal force is always ;
(a) along the tangent at that point of the circular path
(b) towards the centre
(c) outwards from the centre
(d) none of the above
Answer: (b) towards the centre
4. Centrifugal force is :
(a) a real force
(b) the force of reaction of centripetal force
(c) a fictitious force
(d) directed towards the centre of the circular path :
Answer: (c) a fictitious force
5. The difference between centripetal and centrifugal force is :
(a) they both act in the same direction
(b) they both act in opposite directions
(c) they both have different magnitudes
(d) none of the above
Answer: (b) they both act in opposite directions
6. Which of the following is an example of uniform circular motion ?
(a) a car accelerating on a straight road
(b) a pendulum swinging back and forth
(c) a satellite orbiting the earth at a constant altitude
(d) a ball rolling down a hill
Answer: (c) a satellite orbiting the earth at a constant altitude
7. Assertion (A) : When a beam is in static equilibrium, the sum of clockwise moments is equal to the sum of anticlockwise moments.
Reason (R) : According to the principle of moments for a body in equilibrium, the sum of moments acting in one direction must be equal to the sum of moments in opposite direction.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (a) both A and R are true and R is the correct explanation of A
8. Assertion (A) : A heavier object placed at a far distance from the pivot point will have the same moment as a lighter object placed close to the pivot point.
Reason (R) : The moment of force is determined by the magnitude of the force along with its distance from the pivot point.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (c) assertion is false but reason is true
9. Assertion (A) : The centre of gravity of an irregularly- shaped object always lies at its geometric centre.
Reason (R) : The centre of gravity depends on the distribution of mass within an object.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (c) assertion is false but reason is true
Very Short Questions
1. Is it possible to have an accelerated motion with a constant speed? Name such type of motion.
Answer: Yes, it is possible to have an accelerated motion with a constant speed. Such type of motion is uniform circular motion.
2. Give an example of motion in which speed remains uniform, but the velocity changes.
Answer: An example of motion in which speed remains uniform, but the velocity changes is uniform circular motion.
3. Name the force required for circular motion. State its direction.
Answer: The force required for circular motion is centripetal force. Its direction is towards the centre of the circular path.
4. Is centrifugal force a real force?
Answer: No, centrifugal force is not a real force; it is a fictitious force.
5. State whether the following statements are true or false by writing T/F against them.
Answer: (a) Earth moves around Sun with a uniform velocity. F
(b) The motion of Moon around Earth in a circular path is an accelerated motion. T
(c) A uniform linear motion is unaccelerated, while a uniform circular motion is an accelerated motion. T
(d) In a uniform circular motion, the speed continuously changes because the direction of motion changes. F
(e) A boy experiences a centrifugal force on his hand when he rotates a piece of stone tied at one end of a string, holding the other end in the hand. F
Short Questions
1. Differentiate between uniform linear motion and uniform circular motion.
Answer: In uniform linear motion, the speed and velocity, both are constant and acceleration is zero i.e., uniform linear motion is an unaccelerated motion, while in a uniform circular motion the velocity is variable (although the speed is uniform), so it is an accelerated motion.
2. What is a centripetal force?
Answer: A force is needed to change the direction of motion of a particle (or to change the velocity of a particle) i.e., to produce acceleration. A particle moving in a circular path, continuously changes its direction of motion at each point of its path. This change in direction of motion can not be brought without a force. Thus, the motion in circular path is possible only under the influence of a force which is termed as the centripetal force. Centripetal force is the force acting on a body moving in a circular path, in a direction towards the centre of the circular path.
3. Explain the motion of a planet around the sun in an elliptical path.
Answer: A planet moves around the sun in an elliptical path for which the gravitational force of attraction on the planet by the sun provides the necessary centripetal force.
4. (a) How does a centripetal force differ from a centrifugal force with reference to the direction in which they act?
Answer: Centripetal force acts in a direction towards the centre of the circular path, while centrifugal force is in a direction opposite to the direction of centripetal force, i.e., away from the centre of the circular path.
(b) Is centrifugal force the force of reaction of the centripetal force?
Answer: No, the centrifugal force is not the force of reaction of the centripetal force because action and reaction do not act on the same body.
(c) Compare the magnitudes of centripetal and centrifugal force.
Answer: The magnitude of centrifugal force is the same as that of the centripetal force. Their magnitudes are 1:1.
5. State two differences between centripetal and centrifugal force.
Answer: Two differences between centripetal and centrifugal force are:
- Centripetal force is directed towards the centre of the circular path, while centrifugal force is directed away from the centre of the circular path.
- Centripetal force is a real force, while centrifugal force is a fictitious force.
Long Questions
1. Explain the meaning of uniform circular motion. Why is such motion said to be accelerated?
Answer: When a particle moves with a constant speed in a circular path, its motion is said to be uniform circular motion. In such a motion, a particle travels equal distances along the circular path in equal intervals of time, so the speed of particle is uniform, but the direction of motion of the particle changes at each point of the circular path. The continuous change in the direction of motion implies that the velocity of the particle is non-uniform (or variable) i.e., the motion is accelerated. Thus, the velocity of the particle in circular motion is variable or the circular motion is an accelerated motion even though the speed of the particle is uniform.
2. Draw a neat labelled diagram for a particle moving in a circular path with a constant speed. In your diagram show the direction of velocity at any instant.
Answer: The direction of velocity at any instant in a circular path: A particle moving in a circular path in a horizontal plane with uniform speed v in an anticlockwise direction travels each quarter of circle AB, BC, CD and DA in the same interval of time t = T/4 where T is the time taken by the particle to complete one round of the circular path. Thus, the speed of the particle is constant (or uniform), but the direction of motion of the particle is different at different points of the circular path. At any point, the direction of motion is along the tangent drawn at that point of the circular path. At the point A, the direction of motion of the particle is towards north; after completing quarter of circle, at the point B, the direction of motion of the particle is towards west; after completing half circle, at the point C, the direction of motion of the particle is towards south and after completing three-quarters of circle when the particle is at the point D, its direction of motion is towards east.
3. A uniform circular motion is an accelerated motion. Explain it. State whether the acceleration is uniform or variable? Name the force responsible to cause this acceleration. What is the direction of force at any instant? Draw a diagram in support of your answer.
Answer: In a uniform circular motion, the velocity of the particle is variable (although the speed is uniform), so it is an accelerated motion. The acceleration is variable (or non-uniform). The force responsible to cause this acceleration is the centripetal force. At each point of the circular path, this force is directed towards the centre of the circle.
4. A small pebble tied at one end of a string is placed near the periphery of a circular disc, at the centre of which the other end of the string is tied to a peg. The disc is rotating about an axis passing through its centre.
(a) What will be your observation when you are standing outside the disc? Explain.
Answer: When standing outside the disc, the person on the ground at M outside the merry-go-round observes that the pebble (ball) is moving in a circular path. This is because the tension T in the string provides the centripetal force needed for the circular motion.
(b) What will be your observation when you are standing at the centre of the disc? Explain.
Answer: When standing at the centre of the disc (on the platform of the merry-go-round at A), the person observes that the pebble (ball) is stationary placed just in front of him at P. As the merry-go-round rotates, the position of the person on the platform changes, and the pebble (ball) reaches at respective positions, as if it remains at rest always just in front of him. This observation is explained by considering the centrifugal force alongwith the force of tension in the string; the two forces are equal and opposite, so the net force on the pebble is zero.
5. A piece of stone tied at the end of a thread is whirled in a horizontal circle with uniform speed by hand. Answer the following questions :
(a) Is the velocity of stone uniform or variable?
Answer: The velocity of the stone is variable.
(b) Is the acceleration of stone uniform or variable?
Answer: The acceleration of the stone is variable (or non-uniform).
(c) What is the direction of acceleration of stone at any instant?
Answer: The direction of acceleration of the stone at any instant is towards the centre of the circular path.
(d) Which force provides the centripetal force required for circular motion?
Answer: The tension in the string provides the centripetal force required for circular motion.
(e) Name the force and its direction which acts on the hand.
Answer: The force which acts on the hand is the reaction of the tension in the string, and its direction is away from the centre of the circular path.
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