As shown in the figure,a mass $m = 500 \ g$ hangs from the rim of a wheel of radius $r = 20 \ cm$. When released from rest,the mass falls $2.0 \ m$ in $8 \ s$. Then the moment of inertia of the wheel is .......... $kg \cdot m^2$. $(g = 10 \ m/s^2)$

The linear velocity of a rotating body is given by $\overrightarrow v = \overrightarrow \omega \times \overrightarrow r,$ where $\overrightarrow \omega$ is the angular velocity and $\overrightarrow r$ is the radius vector. If the angular velocity of a body is $\overrightarrow \omega = \hat i - 2\hat j + 2\hat k$ and the radius vector is $\overrightarrow r = 4\hat j - 3\hat k,$ then find the magnitude of linear velocity $|\overrightarrow v |$.

$A$ binary star system consists of two stars,one of which has double the mass of the other. The stars rotate about their common centre of mass:

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:

$A$ uniform rod of mass $m$ and length $l$ is suspended by means of two identical inextensible light strings as shown in the figure. The tension in one string immediately after the other string is cut is . . . . . . . ($g$ is the acceleration due to gravity)

Two particles of equal mass $m$ at $A$ and $B$ are connected by a rigid light rod $AB$ of length $L$ lying on a smooth horizontal table. An impulse $J$ is applied at $A$ in the plane of the table and perpendicular to $AB$. Then the velocity of the particle at $A$ is: