Rolling

Practice Questions:

Question 1: 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 $x = 0.6\text{ m}$.

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Question 2: 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.

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Question 3: 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 $$v_{\text{cm}} - \omega R = 0$$.

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Question 4: 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 $E_t / E = 5:7$.

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