A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom:
(2003)
Using conservation of energy, potential energy equals total kinetic energy: $Mgh = \frac{3}{4}Mv^2$. Solving for velocity gives $v = \sqrt{\frac{4}{3}gh}$.
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)
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.
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 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 $v = \sqrt{\frac{10}{7}gh}$.
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.