A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is
\[ \frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}= mgh =mg\frac{3v^{2}}{4g}= \frac{3}{4}mv^{2} \]
\[ \frac{1}{2}I\omega^{2}= \frac{1}{4}mv^{2} \]
Solving I = MR²/2 so, Object is a disc or hollow cylinder.
A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is
Given, rotational kinetic energy is 40% of total energy. so,
\[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \]
Solving ,
\[ I = \frac{2}{5}mR^{2} \]
Object is Solid Sphere.
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)
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 $x = 0.6\text{ m}$.
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.