Rolling - NEET Physics Questions
Question 11: easy

Assertion (A): When the disc rolls without slipping, friction is required because condition of pure rolling is velocity of point of contact is zero.


Reason (R): The force of friction in the case of a disc rolling without slipping down an inclined plane is zero.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: For pure rolling, the point of contact velocity is zero, and friction provides the necessary torque.


Reason (R) is false: For a disc rolling without slipping down an inclined plane, friction is present and acts up the incline to provide the torque for rotation. Thus, (A) is true, (R) is false.

Question 12: easy

Assertion (A): A body is rolling without slipping on a surface. There must be frictional force to start such a motion.


Reason (R): In rolling without slipping, work done against the frictional force is zero on rolling body.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: Friction provides the necessary torque to initiate the angular acceleration required for rolling.


Reason (R) is true: In pure rolling, the point of contact is instantaneously at rest, so the work done by static friction is zero. Both statements are true, but R does not explain A.

Question 13: easy

Assertion (A): A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling).


Reason (R): For pure rolling, work done against frictional force is zero.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Assertion (A) is true; friction provides the torque for rolling. Without friction, the wheel slips. Reason (R) is true; in pure rolling, the contact point is stationary, so static friction does no work. However, (R) does not explain (A).

Question 14: moderate

A solid cylinder of mass $3\text{ kg}$ is rolling on a horizontal surface with velocity $4\text{ m s}^{-1}$. It collides with a horizontal spring of force constant $200\text{ Nm}^{-1}$. The maximum compression produced in the spring will be:

(2012 Pre)

1. $0.5\text{ m}$
2. $0.6\text{ m}$
3. $0.7\text{ m}$
4. $0.2\text{ m}$
View Answer

By mechanical energy conservation, kinetic energy converts to spring potential energy: $\frac{3}{4}mv^2 = \frac{1}{2}kx^2$. Substituting the values gives $x = 0.6\text{ m}$.

Question 15: moderate

A solid sphere of radius R is placed in smooth horizontal surface. A horizontal force F is applied, at height ‘h’ from the lowest point. For the maximum acceleration of centre of mass, which is correct:

(2002)

1. h = R
2. h = 2R
3. h = 0
4. No relation between h and R
View Answer

Acceleration of the centre of mass is given by $a = frac{F}{m} + frac{tau}{I}R_{text{eff}}$. For a smooth surface with no friction, force torque about centre is $tau = F(h-R)$. Maximizing acceleration depends on applying force at the top point where $h = 2R$ to maximize translational effect without opposing torque constraints, or simply using Newton's second law where $a = F/m$ is independent of $h$ unless specified with rotation, but for rolling/sliding conditions $h=2R$ yields specific torque relations.

Question 16: easy

A disc is rolling the velocity of its centre of mass is $v_{\text{cm}}$ then which one will be correct:

(2001)

1. The velocity of highest point is $2 v_{\text{cm}}$ and point of contact is zero
2. The velocity of highest point is $v_{\text{cm}}$ and point of contact is $v_{\text{cm}}$
3. The velocity of highest point is $2 v_{\text{cm}}$ and point of contact is $v_{\text{cm}}$
4. The velocity of highest point is $2 v_{\text{cm}}$ and point of contact is $2 v_{\text{cm}}$
View Answer

For pure rolling, the velocity of the topmost point is $$v*{\text{cm}} + \omega R = 2v_{\text{cm}}$$ and the point of contact is $$v_{\text{cm}} - \omega R = 0$$.

Question 17: easy

If a sphere is rolling, the ratio of the translational energy to total kinetic energy is given by:

(1991)

1. $7 : 10$
2. $2 : 5$
3. $10 : 7$
4. $5 : 7$
View Answer

Translational kinetic energy is $E_t = \frac{1}{2}mv^2$ and rotational kinetic energy is $E_r = \frac{1}{2}I\omega^2 = \frac{1}{5}mv^2$. Total energy $E = \frac{7}{5}mv^2$, so the ratio $E_t / E = 5:7$.