Force

(i)
We know that,

Moment of force = Force × Perpendicular distance

We can see from the given figure that vector C has least perpendicular distance from point O.

Hence, vector C will have least moment about O.

(ii) We know that,

Moment of force = Force × Perpendicular distance

We can see from the given figure that vector A has greatest perpendicular distance from point O

Hence, vector A will have greatest moment about O.

(iii)(a) Clockwise moments are produced by vectors A and B.

Reason : If turning effect on the body is clockwise then moment of force is called clockwise moment and it is taken as a negative value.

(b) Anticlockwise moment is produced by vector C.

Reason : If turning effect on the body is anticlockwise then moment of force is called anticlockwise moment and it is taken as a positive value.

(iv) Sum of torques due to vectors A, B and C = Resultant torque about point O

$-(4 \times 0.9) - (4 \times 0.8) + (4 \times 0.6)Nm \\[0.5em] = -3.6 - 3.2 + 2.4 Nm \\[0.5em] = -4.4 Nm$

So, the resultant torque is 4.4Nm (clockwise direction).

(a) As two forces of same magnitude are acting on a body at the same point and they are in opposite direction so the resultant force will be zero.

F – F = 0

(b) When two forces of same magnitude act on a body at two different points at a separation r and in opposite direction then the moment of force will be Fr

Moment of forces = F × r

Two forces of magnitude F act at point A and point B. The body rotates in anticlockwise direction.

Two equal and opposite parallel forces, not acting along the same line form a couple.

A couple is always needed to produced a rotation.

Two examples in our daily life where we use couple are:

1. Turning a water tap.
2. Tightening the cap of a bottle.

At point A and point B two forces act which rotate the bar in anticlockwise direction.

The perpendicular distance between the two forces called AB is the couple arm.

Moment of force F at A = F x OA (anticlockwise)

Moment of force F at B = F x OB (anticlockwise)

Total moment of couple = F x OA + F x OB

F x (OA + OB) = F x AB

= F x d

Hence proved : Moment of force = Force x Couple arm

When we hang a metre rule horizontally from a fixed support with the help of a strong thread at point O as shown.

We use two slotted weights W₁ and W₂

We hang two spring balances on either side of the thread. The metre rule may tilt to one side.

We adjust the two spring balance distances from the support by keeping one at A and the other at B so that the scale again becomes horizontal.

Let W₁ at a distance OA = l₁ be the weight suspended on the right side of thread from the spring balance at A, while W₂ at a distance OB = l₂ be the weight suspended on the left side of thread from the spring balance at B.

The weight W₁ tends to turn the scale clockwise and the weight W₂ tends to turn the scale anticlockwise.

Clockwise moment = W₁ × l₁

Anticlockwise moment = W₂ × l₂

In equilibrium, when the scale is horizontal, it is found that

Clockwise moment = Anticlockwise moment

So, W₁l₁ = W₂l₂

Hence, the principle of moments is proved.

linear motion

Reason — Linear motion refers to the motion of an object in a straight line, such as when a stationary rigid body moves in response to a force applied in the direction of the force.

rotational motion

Reason — If a body is pivoted at a point and the 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 force and the motion of body is called the rotational motion.

both, the force and its perpendicular distance from the axis

Reason — The moment of a force (also known as torque) about an axis is calculated by multiplying the magnitude of the force by the perpendicular distance from the axis to the line of action of the force. So, it depends on both the magnitude of the force and its perpendicular distance from the axis.

maximum

Reason — The torque produced by a force about an axis depends not only on the magnitude of the force but also on the perpendicular distance from the axis of rotation to the line of action of the force. Increasing this distance (having it at a maximum) maximizes the torque produced by the force.

positive, negative

Reason — The moment of force for anticlockwise moment is taken as positive and for clockwise moment is taken as negative.

both the direction of force and the point of application of force

Reason — The direction of rotation of a pivoted body depends on both the direction of the applied force and the point at which the force is applied relative to the pivot point. Changing either or both of these factors can alter the direction of rotation of the body.

two equal and opposite parallel forces

Reason — Two equal and opposite parallel forces, not acting along the same line form a couple.

Sum of anticlockwise moments is equal to sum of clockwise moments

Reason — According to the principle of moments in equilibrium :

sum of anticlockwise moments = sum of clockwise moments.

the body will have rotational as well as translational motion

Reason — When a body is acted upon by two unequal forces in opposite directions, but not in the same line, then the effect is that the body will have rotational as well as translational motion.

(1), (2) and (3)

Reason — Dynamic equilibrium refers to a state where two or more opposing processes occur at the same rate, resulting in a stable overall condition.

1. Moon revolving around the earth is an example of dynamic equilibrium as the gravitational force pulling the moon towards the Earth is balanced by the centrifugal force of the moon's motion, resulting in a stable orbit.

2. A pebble fixed at the end of a string whirling in a circular path is also an example of dynamic equilibrium as the tension in the string pulling the pebble towards the center is balanced by the centrifugal force of the pebble's motion, keeping it in a stable circular path.

3. An aeroplane moving at a constant height is also an example of dynamic equilibrium. It achieves dynamic equilibrium when upward lift on it balances its weight downwards.

4. A beam balance balanced in a horizontal position is not an example of dynamic equilibrium because the balance may remain still, it's a static equilibrium where opposing forces (gravity pulling down on each side of the balance) are already balanced without any ongoing processes happening.

As we know,

Moment of Force = Force × Perpendicular distance from the point O

Moment of force = F × r

Substituting values in the above formula we get,

10 = 20 x r

⇒ r = $\dfrac{10}{20}$ = 0.5 m

Hence, distance of the point O from the line of action of the force is 0.5 m.

It is given that,

Required least force = 10N

r = 25cm = 0.25m

We know, Moment of force = F x r

Substituting the values of F and r we get,

Moment of force = 10 x 0.25 = 2.5 Nm

Hence, moment of force needed to turn the nut = 2.5 Nm

In first case,

mass = 50 kg

Initial length = R = 1m

F₁ = mg = 50 x 9.8 = 490 N

Torque 1 = R x F₁ = 1 x 490 = 490 Nm

Now in second case,

New length = r = R - 20% x R = R - 0.2 R = 0.8 R = 0.8 x 1 = 0.8m

Let required force = F₂

So,

Torque 2 = r x F₂ = 0.8 x F₂

According to the question,

Torque 1 = Torque 2

490 = 0.8 x F₂

F₂ = $\dfrac {49}{0.8}$ = 612.5N

Here, F₂-F₁ = 612.5 - 490 = 122.5 N

∴ Force required to produce the same torque will increase by 122.5 N

In first case,

F₁ = 100 N

r₁ = 0.5 m

Torque 1 = r₁ x F₁ = 0.5 x 100 = 50 Nm

∴ Torque exerted on the bolt = 50 Nm

In second case,

F₂ = F₁ + 20% x F₁ = F₁ + 0.2 x F₁ = 1.2 F₁ = 1.2 x 100 = 120 N

r₂ = 0.4 m

Torque 2 = r₂ x F₂ = 0.4 x 120 = 48 Nm

Here, Torque 1 > Torque 2

Torque 1 - Torque 2 = 50 Nm - 48 Nm = 2 Nm

∴ The torque will decrease by 2 Nm

We know from the question,

F = 2 N
Diameter = 2m
So, Radius or OB = 1m

(i) Moment of force at O = F x r

Substituting the values of F and r we get,

$2 \times 1 = 2Nm$ (ii) Moment of force at A = F x r

Substituting the values of F and r we get,

$2 \times 2 = 4Nm$

Given,

OA = 2m
OB = 4m
F1 = 5N
F2 = 3N

As we know,

Moment of force = F x r

Substituting the values of F and r

(i) Moment of force F1 about O

$5 \times 2 = 10Nm$

Therefore, moment of force F1 about O is 10Nm (anticlockwise)

(ii) Moment of force F2 about O

$3 \times 4 = 12Nm$

Therefore, moment of force F2 about O is 12Nm (clockwise)

(iii) Total moment of two forces about midpoint is

$12 - 10 = 2Nm$

Therefore, total moment of two forces about O is 2Nm (clockwise)

Given,
Ab = 4m
OA = 2m
OB = 2m
Force at A = 10N
Force at B = 10N

As we know ,

Moment of force = F x r

Substituting the values of F and r

Moment of force about o at point A

$10 \times 2 = 20Nm \text{ clockwise}$

Moment of force about o at point B

$10 \times 2 = 20Nm \text{ clockwise}$

Total moment of forces about the centre O

$20 + 20 = 40Nm \\[0.5em]$

Total moment of force about the pivot O is 40 Nm (clockwise)

Given,
Force at A = 10N
Force at B = 10N
Distance between A and B = 50cm = 0.5m

(i) Resultant moment of two forces at point A is

$10 \times 0.5 = 5Nm \text{ clockwise} \\[0.5em]$

Therefore, resultant moment of two forces at point A is 5 Nm (clockwise)

(ii) Resultant moment of two forces at point B is

$10 \times 0.5 = 5Nm \text{ clockwise} \\[0.5em]$

Therefore, resultant moment of two forces at point B is 5 Nm (clockwise)

(iii) Perpendicular distance of point O from either of the forces F = 10N is 0.25 m

Moment of force F at point A about O

$10 \times 0.25 =2.5Nm \text{ clockwise} \\[0.5em]$

Moment of force F at point B about O

$10 \times 0.25 =2.5Nm \text{ clockwise} \\[0.5em]$

Resultant moment of two forces about o is

$2.5 + 2.5 = 5Nm \text{ clockwise} \\[0.5em]$

Therefore, Resultant moment of two forces about o is 5Nm (clockwise)

Moment of couple = either force × couple arm

= 6 x 0.5 = 3 Nm

Let us assume that a 50 gf weight produces an anticlockwise moment about the middle point ( 50 cm ).

Now, if a weight of 100 gf produces a clockwise moment about the middle point and d cm be the distance from the middle.

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment.

$50 gf \times 50cm = 100gf \times d cm \\[0.5em] \Rightarrow d = \dfrac{50\times50}{100}\\[0.5em] \Rightarrow d = 25 cm$

Therefore, a weight of 100gf will be suspended at a distance of 25cm to keep the ruler balanced.

(i) Diagram of the arrangement is shown below:

(ii) As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment.

$W \times (58-50) = 20gf \times (100-58) \\[0.5em] \Rightarrow W = \dfrac{20\times42}{8} \\[0.5em] \Rightarrow W = 105 gf$

From the given figure we can see that,

Anticlockwise moment = 40 gf × 40 cm
Clockwise moment = 80 gf × d cm

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment.

$40gf \times 40cm = 80gf \times d \\[0.5em] \Rightarrow d = \dfrac{40 \times 40}{80}\\[0.5em] \Rightarrow d = 20 cm \\[0.5em]$

Therefore the bar will be in balanced position if weight of 80gf is placed at a point of 20cm to the right of O.

(i) We know that when anticlockwise moment is equal to clockwise moment then the rule will be in equilibrium.

Let us check,

(a) Anticlockwise moment = Fr

$40gf \times (50-10)cm = 40 \times 40 \\[0.5em] = 1600 gfcm$

(b) Clockwise moment = Fr

$20gf \times (90-50)cm = 20 \times 40 \\[0.5em] = 800gfcm$

So, the meter rule will not be in equilibrium. It will turn in anticlockwise direction.

(ii) To balance the rule in equilibrium 40gf should be kept on right side.

Let distance from middle = d cm

So clockwise moment becomes

$800 gf cm + (40gf \times d cm)$

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

$1600gfcm = 800gfcm + 40gf \times d cm$

Therefore,

$d = \dfrac{800}{40} = 20cm$

So, the additional weight should be placed at 70cm mark to bring the rule in equilibrium.

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

$20 Kgf \times 2m = 40 Kgf \times d \\[0.5em] \Rightarrow d = \dfrac{20 \times 2}{40} \\[0.5em] \Rightarrow d = 1m$

So it is proved that horizontal position can be achieved if a person sits at a distance of 1m from the centre on the side opposite to the boy.

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

$100gf \times 40cm = W \times 60cm$

Weight on the longer pan

$W = \dfrac{(100 \times 40) }{60} \\[0.5em] W = 66.67gf$

So, a weight of 66.67gf kept on the pan of longer arm will balance the object of weight 100 gf kept on the other pan.

(i) The total anticlockwise moment about the centre o

$= 150gf \times 40cm \\ = 6000gf cm$

(ii) The total clockwise moment about the centre o

$= 250gf \times 20cm \\[0.5em] = 5000gf cm \\[0.5em]$

(iii) The difference of anticlockwise and clockwise moment

$= 6000 - 5000 \\[0.5em] = 1000gf cm \\[0.5em]$

(iv) As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

$150gf \times 40cm = 250gf \times 20cm + 100gf \times d \\[0.5em] 6000gf = 5000gf +100gf \times d \\[0.5em] d = \dfrac {1000}{100} \\[0.5em] d = 10cm$

So d = 10 cm on the right side of o

(i) Moment of force = Fr

Substituting the values of F and r we get,

$10gf \times 50cm = 500gfcm$

(ii) In order to make the rule horizontal by applying least force distance has to be max as moment of force will remain same.

So,

$F = \dfrac{500}{100} = 5gf$

Therefore, a force of 5 gf upwards at 100 cm can balance the rule.

(b) Suppose, if the mass of the meter rule be M.

Given, uniform meter rule, so weight (Mg) will act at 25 cm.

Mg produces anticlockwise moment about point o.

Now, in order to balance the 20g weight is tied at 50cm point and it acts in the clockwise direction.

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

$Mg(29 - 25) = 0.02 g (50-29) \\[0.5em] M = \dfrac{21\times 0.02}{4} \\[0.5em] = 0.105 kg \\[0.5em] = 105g$

The weight of half the meter rule is 105 g.

(c) The balancing point will shift towards the 25 cm mark.

(i) As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

$100g \times (50-40)cm = m \times (40-20)cm \\[0.5em] \Rightarrow 100 \times 10 = m \times 20 \\[0.5em] \Rightarrow m = 50g \\[0.5em]$

(ii) When the mass m is shifted to mark 10cm , it results in rule being shifted on the side of mass m in anticlockwise direction.

(iii) Anticlockwise moment is produced when a mass of m grams is moved towards the mark of 10cm.

$100gf \times (50-40)cm = 1000gfcm$

Therefore, the resultant moment will be

$1500gfcm - 1000 gfcm = 500gfcm (anticlockwise)$

(iv) As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

So,

$100gf \times (50-40)cm + 50gf \times d = 50gf \times (40-10)cm \\[0.5em] \Rightarrow 1000gfcm + 50gf \times d = 1500gfcm \\[0.5em] \Rightarrow 50gfd = 500gfcm \\[0.5em] \Rightarrow d = 10 cm$

Hence, we can balance 50gm at 50cm

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

R₁ + R₂ = W

Given, the system is in equilibrium so

$R_{1}\times\dfrac{l}{2} = R_{2}\times\dfrac{l}{2}$

Since, R₁ = R₂ and 2R₁ = W

Therefore, R₁ = W/2 = R₂

The moment of force is equal to the product of the magnitude of the force applied and the perpendicular distance of the line of action of the force from the axis of rotation(or pivoted point).

The S.I unit of moment of force is Newton × meter = Newton meter (Nm)

The following are the two factors on which moment of force about a point depends

(a) The magnitude of the force applied and,

(b) The distance of the line of action of the force from the axis of rotation(or the pivoted point).

If a body is pivoted at a point and the force is applied on the body at a suitable point, it rotates the body about the axis passing through the pivoted point.

The direction of rotation can be changed by changing the direction of force.

Figures given below shows the clockwise and anticlockwise movement in a circle pivoted at the centre by changing the point of application of force F from A to point B

If the effect on the body is to turn it clockwise, moment of force is called the clockwise moment and it is taken as negative, while if the effect on the body is to turn it anticlockwise, moment of force is called the anticlockwise moment and it is taken as positive.

It is easier to open the door by applying the force at the free end because when the perpendicular distance is large less force is required to turn the body.

The stone of hand flour grinder is provided with a handle near its rim so that by applying small force at the handle it can be rotated easily about the fixed point at its centre.

It is easier to turn the steering wheel of a large diameter in comparison to the wheel of a smaller diameter as less force is required in the first case due to the larger distance from the centre of the rim.

A spanner is provided with a long handle so as to create a larger turning moment with less force.

A jack screw which is used to lift a heavy vehicle has a long arm so that less force is needed to rotate it to raise or lower the vehicle.

Application of Force has to be done as shown in the diagram.

No, if pivoted at O it will not go up.

Moment of couple is product of either force and the perpendicular distance between the two forces.

The S.I unit of moment of couple is Nm

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 equilibrium.

(i) When a body remains in the state of rest under the influence of several forces, the body is said to be in static equilibrium.

Example — A toy lying on a floor is in static equilibrium.

(ii) When a body remains in the same state of motion (translation or rotational), under the influence of several forces, the body is said to be in dynamic equilibrium.

Example — An apple falling from a tree with a constant velocity.

The two conditions for a body to be in equilibrium are

1. The resultant of all the forces acting on a 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.

According to the principle of moments, if the algebraic sum of moments of all the forces, acting on the body, about the axis of rotation is zero, the body is in equilibrium. Therefore, as per the principle of moments,

sum of anticlockwise moments = sum of clockwise moments

For example, a physical balance works on the principle of moments.

(a) Translation motion is produced when a body is free to move.

(b) Rotational motion is produced when the body is fixed at a point.

The moment of force is a vector quantity.

The expression for the moment of force is given by:

Moment of force about a given axis = Force × perpendicular distance of force from the axis of rotation.

The moment of force can be reduced by decreasing the perpendicular distance of force from the axis of rotation.

We can obtain a greater moment of force by increasing the perpendicular distance of force from the axis of rotation.

(i) The S.I. unit of moment of force is Newton metre (Nm)

(ii) In equilibrium algebraic sum of moments of all forces about the point of rotation is zero.

(iii) In beam balance when the beam is balanced in a horizontal position, it is in static equilibrium.

(iv) The moon revolving around the earth is in dynamic equilibrium.

To determine centre of gravity for a triangular lamina, make three fine holes a, b, c near its edges. After that, suspend it from hole a along with a plumb line. Ensure that the lamina is freely suspended. When it comes to rest, draw straight line ad along the plumb line. Repeat the experiment by suspending the lamina through hole 'b' and then through hole 'c'. This gives us two straight lines, be and cf respectively. The three lines ad, be and cf intersect each other at a common point G. This common point G is the position of the centre of gravity of triangular lamina.

All of the above

Reason — With respect to centre of gravity :

• The position of centre of gravity depends on its shape i.e., on the distribution of mass.
• It is not necessary that the centre of gravity of a body should always be within the material of the body.
• By the concept of centre of gravity, a body of weight W can be considered as a point particle of weight W at its centre of gravity.

0

Reason — 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 zero. This is because the center of gravity is chosen in such a way that the torques (moments) due to the weight of the individual particles balance out, resulting in a net torque of zero.

$\dfrac{2\text{h}}{3}$

Reason — The centre of gravity of a hollow cone is at a height $\dfrac{\text{h}}{3}$ from the base.

Hence, from the vertex the height is :

h - $\dfrac{\text{h}}{3}$ = $\dfrac{\text{3h-h}}{3}$ = $\dfrac{2\text{h}}{3}$

both (1) and (2)

Reason — For stable equilibrium, the center of gravity of a body should be located above the base of the body. Additionally, it should be as close as possible to the geometric center.

the one with a smaller base

Reason — An object with a larger base of support will have a low centre of gravity because the distribution of mass is spread out over a larger area, which tends to lower the center of gravity whereas that with a smaller base will have a higher centre of gravity because the mass is concentrated over a smaller area.

The centre of gravity of a body is defined as the point about which the algebraic sum of moments of weights of particles constituting the body is zero and the entire weight of the body is to act at this point.

The position of the centre of gravity of any body will depend on the type of shape it has i.e. (on the distribution of mass on the body) in it. It changes if the shape of the body is deformed.

For example, the centre of gravity of a wire is a its mid point. However, if the wire is turned in a way that it makes a circle then its centre of gravity will be at the centre of the circle.

The diagram is shown below:

When a heavy load is placed on the back, the center of gravity of the body shifts backward. To prevent falling backward and to keep the body’s center of gravity above the feet (the base of support), a person bends forward. This forward bending compensates for the backward shift, allowing the person to remain stable and balanced while carrying the load.

Precaution:
To ensure the ship sails smoothly, the heavier A-type containers should be loaded closer to the ship’s center of gravity (near the bottom and central part of the ship), while the lighter B-type containers should be loaded on top or towards the sides.

Reason:
A-type containers are 50% heavier than B-type containers. If these heavy containers are loaded improperly (e.g., on the upper decks or on one side), they will raise the ship’s center of gravity and make it unstable, increasing the risk of tilting, especially in rough seas.

The position of centre of gravity of the rim is marked by the letter G in the below diagram:

The position of centre of gravity G for the three pieces of card board are marked in the below diagrams:

Yes, the centre of gravity can be outside the material of the body.

In the case of a uniform ring the centre of gravity is in the centre of the ring and that point is outside the material of the ring.

Centre of gravity for the following figures are as follows :

(a) Rectangular lamina — at the point of intersection of the diagonals.

(b) Cylinder — at the mid point of the axis of cylinder.

The centre of gravity for the following figures are as follows :

(a) A triangular lamina — at the point of intersection of the medians.

(b) A circular lamina — at the centre of the circular lamina.

The centre of gravity of a uniform ring is situated at the centre of the ring.

(i) False

Reason — The position of the center of gravity of a body depends on its mass distribution. When a body is deformed, its mass distribution can change, leading to a potential shift in the position of its center of gravity.

(ii) True

When a body moves with a constant speed in a circular path, its motion is said to be in uniform circular motion.

Uniform circular motion is said to be accelerated as the velocity changes with continuous change in direction of motion.

The below diagram shows a particle moving in a circular path with a constant speed and its direction of velocity:

A uniform circular motion is accelerated as the speed is same however the direction changes continuously. So, it is said to be a uniform accelerated motion.

Centripetal force is required for this acceleration and its direction is towards the centre of the circular path.

(a) The pebble seems to move in a circular path when we are sitting outside the circular disc.

(b) The pebble seems to be stationary in front of us when we are standing at the centre of the disc.

(a) Velocity of stone is variable.

(b) Acceleration of stone is variable.

(c) The direction of acceleration of stone at any instance is to the centre of circular path.

(d) The frictional force provides the centripetal force.

(e) Centrifugal force acts on the hand and its direction is away from the centre.

Speed

Reason — In uniform circular motion, where an object travels along a circular path at a constant speed, the direction of motion changes continuously, implying the presence of acceleration. Hence, velocity changes due to change in direction and speed remains constant.

is along the tangent at that point of the circular path

Reason — In circular motion, the direction of motion of an object is along the tangent at that point of the circular path where the object is located at any given moment.

towards the centre

Reason — The centripetal force is always directed towards the center of the circular path. It is responsible for keeping an object moving along the curved path by continuously changing its direction towards the center without altering its speed. This force acts perpendicular to the direction of motion, pointing inward, towards the center of the circular path.

a fictitious force

Reason — The centrifugal force is a fictitious force because it is assumed to exist only to explain observations made from the frame of reference of an observer who is also moving in a circular motion along with the object that is being observed.

they both act in opposite direction

Reason — Centripetal Force is a real force which acts towards the centre of the circle whereas centrifugal force is an assumed force which acts away from centre of the circle. Both the forces act in opposite directions.

a satellite orbiting the earth at a constant altitude

Reason —

• A satellite orbiting the Earth at a constant altitude experiences uniform circular motion because it maintains a constant speed while continuously moving along a circular path around the Earth.

• The car moving along a straight path and may be changing its speed, but it is not following a circular path.

• A pendulum exhibits periodic motion as it swings back and forth due to gravity but it is not circular motion.

• The ball is rolling along a sloped path, not a circular one.

Both A and R are true and R is the correct explanation of A.

Explanation

Assertion (A) is true because when the sum of the clockwise moments acting on the beam is equal to the sum of the anticlockwise moments, their resultant sum is zero and the beam has no rotational motion. Hence, it is in static equilibrium.

Reason (R) is true because it provides a logical explanation for Assertion (A), stating the principle of moments for a body in equilibrium, which aligns with the definition of static equilibrium.

assertion is false but reason is true.

Explanation

The moment of a force 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. Hence, a heavier object placed at a far distance from the pivot point will have a greater moment than a lighter object placed close to the pivot point. Therefore, Assertion (A) is false.

Reason (R) is true. The moment of force is determined by the magnitude of the force along with its distance from the pivot point.

assertion is false but reason is true.

Explanation

Assertion (A) is false. The center of gravity of an irregularly-shaped object does not always coincide with its geometric center. It depends on the distribution of mass within the object.

Reason (R) is true. The center of gravity does depend on the distribution of mass within an object.

The force acting on a body moving in a circular path, in a direction towards the centre of circular path is called Centripetal force.

The motion of a planet around the Sun in an elliptical path is an example of uniform circular motion. The gravitational force of attraction on the planet by the Sun provides the necessary centripetal force for this uniform circular motion as it is always directed towards the centre of the Sun.

(a) Both the forces act in opposite direction with reference to the direction.

(b) No, centrifugal force is not the force of reaction of the centripetal force.

(c) The magnitude of centripetal and centrifugal force is 1:1

Yes, an accelerated motion with constant speed is possible. Uniform circular motion is such type of motion.

Motion of artificial satellite is one of the cases where speed is same however velocity changes.

Centripetal force is required for circular motion. The direction of Centripetal force is always directed towards the centre of circle.

No the centrifugal force is a fictitious force.

(a) False
Reason — The Earth does not move around the Sun with uniform velocity. It follows an elliptical orbit, which means its speed varies as it travels around the Sun.

(b) True
Reason — The motion of Moon around Earth in a circular path is an accelerated motion because in a circular motion, the direction of velocity is continuously changing, which in turn causes change in acceleration.

(c) True
Reason — In uniform linear motion, an object moves in a straight line with a constant speed. As there is no change in speed or direction, hence, there is no acceleration. On the other hand, in uniform circular motion, an object moves in a circular path at a constant speed and the direction of the object's velocity continuously changes as it moves around the circle. This change in direction results in centripetal acceleration.

(d) False
Reason — In uniform circular motion, the speed of the object remains constant, but its velocity, which includes both speed and direction, changes continuously.

(e) False
Reason — When the boy rotates the stone tied to a string and holds the other end in his hand, the force experienced by his hand is centripetal, not centrifugal.
Centripetal force is the force directed toward the center of the circular path, required to keep an object moving in a circular motion. Centrifugal force, on the other hand, is a fictitious force that appears to act outward from the center of rotation in a rotating reference frame.