Rolling - NEET Physics Questions
Question 11: 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 12: 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 13: 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 $E_t / E = 5:7$.