How can one verify that an object is undergoing rotation about a fixed axis?

$A$ uniform disc of radius $R$ and mass $M$ is free to rotate only about its axis. $A$ string is wrapped over its rim and a body of mass $m$ is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is

$A$ ball of mass $m$ is released from inside a smooth wedge of mass $m$ as shown in the figure. What is the speed of the wedge when the ball reaches point $B$?

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $V(r) = kr^2 / 2$,where $k$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $R$ about the point $O$. If $v$ is the speed of the particle and $L$ is the magnitude of its angular momentum about $O$,which of the following statements is (are) true?
$(A)$ $v = \sqrt{\frac{k}{2m}} R$
$(B)$ $v = \sqrt{\frac{k}{m}} R$
$(C)$ $L = \sqrt{mk} R^2$
$(D)$ $L = \sqrt{\frac{mk}{2}} R^2$

The mass of the pulley is $m$ and its radius is $R$. Assume the pulley to be a uniform disc. The pulley is free to rotate about an axis passing through its centre and perpendicular to its plane. The string is massless and inextensible,and it is wrapped on the pulley. There is no slipping between the string and the pulley. The length of the string which is not wrapped on the pulley is $2R$. $A$ block of mass $m$ is released from the position as shown in the figure. The impulse exerted by the string on the pulley at the moment the string becomes taut is $J$. The value of $\frac{2m \sqrt{gR}}{J}$ is equal to:

$A$ ladder of length $L$ is slipping with its ends against a vertical wall and a horizontal floor. At a certain moment,the speed of the end in contact with the horizontal floor is $v$ and the ladder makes an angle $\alpha = 30^o$ with the horizontal. If $dv/dt = 0$,then the angular acceleration of the ladder when $\alpha = 45^o$ is: