The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle $theta$ without slipping and slipping down the incline without rolling is:
(2014)
Acceleration without slipping is $a_1 = \frac{g sin\theta}{1 + I/(mR^2)} = \frac{5}{7}g sin\theta$. Acceleration with pure slipping is $a_2 = g sin\theta$. The ratio $a_1/a_2$ is $5/7$.
Small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is:
(2013)
Using energy conservation, initial kinetic energy equals potential energy at max height: $\frac{1}{2}mv^2 \left(1 + \frac{I}{mR^2}\right) = mgH$. Substituting $H = \frac{3v^2}{4g}$, we get $1 + \frac{I}{mR^2} = 2$, which gives $\frac{I}{mR^2} = 1$. This corresponds to a ring.
A disc is rolling the velocity of its centre of mass is $v_{\text{cm}}$ then which one will be correct:
(2001)
For pure rolling, the velocity of the topmost point is $$v*{\text{cm}} + \omega R = 2v_{\text{cm}}$$ and the point of contact is $$v_{\text{cm}} - \omega R = 0$$.
For a hollow cylinder & a solid cylinder rolling without slipping on an inclined plane, then which of these reaches earlier on the ground:
(2000)
Acceleration of a rolling body is given by $$a = \frac{g \sin\theta}{1 + I/MR^2}$$. Since the solid cylinder has a smaller moment of inertia ratio than the hollow cylinder, its acceleration is greater, so it reaches the bottom first.
A solid sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on the inclined plane, then:
(1993)
The acceleration on an inclined plane is inversely proportional to $1 + I/MR^2$. Solid sphere has the lowest moment of inertia coefficient ($2/5$), giving it maximum acceleration and shortest time to reach the bottom.
The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height $h$ from rest without sliding is:
(1992)
Using conservation of mechanical energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For a solid sphere ($I = \frac{2}{5}MR^2$), solving yields $v = \sqrt{\frac{10}{7}gh}$.
If a sphere is rolling, the ratio of the translational energy to total kinetic energy is given by:
(1991)
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 $E_t / E = 5:7$.
A planet is moving in an elliptical orbit around the sun. If $T$, $V$, $E$ and $L$ stand respectively for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following is correct?
(1990)
For a bound elliptical orbit, total energy $E$ is always negative. $T$ and $V$ vary with distance, and $L$ is conserved in both magnitude and direction as Torque is Zero. Gravitational force is passing through Center of Rotation so Toque is zero.