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Rotational Motion

JEE Main level, single correct, medium

5 of 280

Question 5 of 280

A solid sphere ( $A$ ) of mass $5m$ and a spherical shell ( $B$ ) of mass $m$ , both having same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of $A$ and $B$ , they start rolling without slipping with an acceleration of $a_A$ and $a_B$ , respectively. The ratio of $a_A$ and $a_B$ is __________.

Show answer and solution

Answer: (A) $5 : 21$

For a body of mass

M

and radius

R

rolling without slipping under a force

F

applied tangentially at the highest point, the torque about the point of contact with the ground is

\tau = F(2R)

.

The moment of inertia about the point of contact is

I = I_{cm} + MR^2

.

Using

\tau = I\alpha

and the rolling without slipping condition

a = \alpha R

, we get:

F(2R) = (I_{cm} + MR^2) \dfrac{a}{R} \Rightarrow a = \dfrac{2FR^2}{I_{cm} + MR^2}

For the solid sphere

A

of mass

5m

I_{cm,A} = \dfrac{2}{5}(5m)R^2 = 2mR^2

a_A = \dfrac{2FR^2}{2mR^2 + 5mR^2} = \dfrac{2F}{7m}

For the spherical shell

B

of mass

m

I_{cm,B} = \dfrac{2}{3}mR^2

a_B = \dfrac{2FR^2}{\dfrac{2}{3}mR^2 + mR^2} = \dfrac{2F}{\dfrac{5}{3}m} = \dfrac{6F}{5m}

The ratio of their accelerations is:

\dfrac{a_A}{a_B} = \dfrac{\dfrac{2F}{7m}}{\dfrac{6F}{5m}} = \dfrac{2}{7} \times \dfrac{5}{6} = \dfrac{5}{21}

Answer:

5 : 21

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