Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

Cylinder Q reaches the ground with larger angular speed.

Both cylinders P and Q reach the ground at the same time.

Cylinders P has larger linear acceleration than cylinder Q.

Both cylinders reach the ground with same translational kinetic energy.

Rotational Motion PYQ — JEE Advanced 2012 · Shift 2 · Q49 with Solution

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Correct answer: D

Rolling down an incline: acceleration formula

For a rigid body rolling without slipping down an incline of angle $\theta$ ,

$a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$

where:

$m$ = mass,
$R$ = radius,
$I$ = moment of inertia about its axis.

Compare moments of inertia of P and Q

Both cylinders have the same mass and same radius.

Cylinder P has most mass near the surface $\Rightarrow I_P$ is larger.
Cylinder Q has most mass near the axis $\Rightarrow I_Q$ is smaller.

Hence,

$I_P > I_Q$

Therefore,

$a_P = \frac{g\sin\theta}{1 + \frac{I_P}{mR^2}} < \frac{g\sin\theta}{1 + \frac{I_Q}{mR^2}} = a_Q$

So cylinder Q has larger linear acceleration and reaches first.

Thus:

A is false
B is false

Translational kinetic energy at the bottom

Since both start from the same height, total loss of potential energy is the same:

$mgh = K_{\text{trans}} + K_{\text{rot}}$

For rolling without slipping,

$K_{\text{trans}} = \frac{1}{2}mv^2, \qquad K_{\text{rot}} = \frac{1}{2}I\omega^2$

and

$v = \omega R$

So,

= \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right)$$ Thus, $$v^2 = \frac{2gh}{1 + \frac{I}{mR^2}}$$ Hence smaller $I$ gives larger $v$. So $v_Q > v_P$. Then translational kinetic energy: $$K_{\text{trans}} = \frac{1}{2}mv^2$$ Since masses are same and $v_Q > v_P$, $$K_{\text{trans},Q} > K_{\text{trans},P}$$ Therefore **C is false**. --- 4. **Angular speed at the bottom** For rolling without slipping, $$\omega = \frac{v}{R}$$ Since both have the same radius and $v_Q > v_P$, $$\omega_Q > \omega_P$$ Therefore **D is true**. --- 5. **Final evaluation of options** - **A:** False - **B:** False - **C:** False - **D:** True So the correct option is: $$\boxed{D}$$ --- 6. **Comparison with stored correct answer** Stored correct answer: **D** My derived answer: **D** They agree.

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View written solutionFree

Correct answer: D

Rolling down an incline: acceleration formula

For a rigid body rolling without slipping down an incline of angle $\theta$ ,

$a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$

where:

$m$ = mass,
$R$ = radius,
$I$ = moment of inertia about its axis.

Compare moments of inertia of P and Q

Both cylinders have the same mass and same radius.

Cylinder P has most mass near the surface $\Rightarrow I_P$ is larger.
Cylinder Q has most mass near the axis $\Rightarrow I_Q$ is smaller.

Hence,

$I_P > I_Q$

Therefore,

$a_P = \frac{g\sin\theta}{1 + \frac{I_P}{mR^2}} < \frac{g\sin\theta}{1 + \frac{I_Q}{mR^2}} = a_Q$

So cylinder Q has larger linear acceleration and reaches first.

Thus:

A is false
B is false

Translational kinetic energy at the bottom

Since both start from the same height, total loss of potential energy is the same:

$mgh = K_{\text{trans}} + K_{\text{rot}}$

For rolling without slipping,

$K_{\text{trans}} = \frac{1}{2}mv^2, \qquad K_{\text{rot}} = \frac{1}{2}I\omega^2$

and

$v = \omega R$

So,

= \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right)$$ Thus, $$v^2 = \frac{2gh}{1 + \frac{I}{mR^2}}$$ Hence smaller $I$ gives larger $v$. So $v_Q > v_P$. Then translational kinetic energy: $$K_{\text{trans}} = \frac{1}{2}mv^2$$ Since masses are same and $v_Q > v_P$, $$K_{\text{trans},Q} > K_{\text{trans},P}$$ Therefore **C is false**. --- 4. **Angular speed at the bottom** For rolling without slipping, $$\omega = \frac{v}{R}$$ Since both have the same radius and $v_Q > v_P$, $$\omega_Q > \omega_P$$ Therefore **D is true**. --- 5. **Final evaluation of options** - **A:** False - **B:** False - **C:** False - **D:** True So the correct option is: $$\boxed{D}$$ --- 6. **Comparison with stored correct answer** Stored correct answer: **D** My derived answer: **D** They agree.

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