Question 128: Centre of Mass and Rotational Motion

Question

A solid sphere of mass ' $m$ ' and radius ' $r$ ' is allowed to roll without slipping from the highest point of an inclined plane of length ' $L$ ' and makes an angle $30^{\circ}$ with the horizontal. The speed of the particle at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant. Then the new speed of the sphere at the bottom of the plane is $v_2$. The ratio $v_1^2: v_2^2$ is

Accepted Answer

Let's analyze the problem step-by-step.
Energy conservation:
When the sphere rolls without slipping, its gravitational potential energy converts into both translational and rotational kinetic energy. The energy conservation equation is given by:
$$mgL\sin	heta = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$
where:
$L\sin	heta$ is the vertical height,
$I$ is the moment of inertia, and
$\omega$ is the angular speed.
Moment of Inertia and Rolling Condition:
For a solid sphere, the moment of inertia about its center is:
$$I = \frac{2}{5}mr^2$$
Since the sphere rolls without slipping, the linear speed $v$ and the angular speed $\omega$ are related by:
$$v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r}$$
Total Kinetic Energy:
Substituting the rolling condition into the rotational kinetic energy gives:
$    \begin{aligned}    K &= \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2ight) \left(\frac{v}{r}ight)^2 \      &= \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2ight) \frac{v^2}{r^2} \      &= \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \      &= \frac{7}{10}mv^2    \end{aligned}    $
Finding $v^2$ in Terms of $L$ and $\sin	heta$:
Equate the initial potential energy to the final kinetic energy:
$$mgL\sin	heta = \frac{7}{10}mv^2$$
Solving for $v^2$:
$$v^2 = \frac{10}{7}gL\sin	heta$$
Apply to the Two Cases:
For $	heta = 30^\circ$, since $\sin(30^\circ) = \frac{1}{2}$:
$$v_1^2 = \frac{10}{7}gL\left(\frac{1}{2}ight) = \frac{5}{7}gL$$
For $	heta = 45^\circ$, since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$:
$$v_2^2 = \frac{10}{7}gL\left(\frac{1}{\sqrt{2}}ight)$$
Calculate the Ratio $\frac{v_1^2}{v_2^2}$:
$    \begin{aligned}    \frac{v_1^2}{v_2^2} &= \frac{\left(\frac{5}{7}gLight)}{\left(\frac{10}{7}gL\frac{1}{\sqrt{2}}ight)} \    &= \frac{5}{10}\sqrt{2} \    &= \frac{\sqrt{2}}{2}    \end{aligned}    $
Expressing this ratio as $v_1^2 : v_2^2$, we have:
$$v_1^2 : v_2^2 = 1 : \sqrt{2}$$
Conclusion:
According to the options given, the correct answer is:
Option C: $1:\sqrt{2}$.

Answer

$1: 3$

Answer

$1: 2$

Answer

$1: \sqrt{2}$

Answer

$1: \sqrt{3}$

Centre of Mass and Rotational Motion

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