$A$ flywheel is used with an engine because it

$A$ thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement$(s)$ is/are correct,when the rod makes an angle $60^{\circ}$ with vertical? [$g$ is the acceleration due to gravity]
$(1)$ The radial acceleration of the rod's center of mass will be $\frac{3g}{4}$
$(2)$ The angular acceleration of the rod will be $\frac{3\sqrt{3}g}{4L}$
$(3)$ The angular speed of the rod will be $\sqrt{\frac{3g}{2L}}$
$(4)$ The normal reaction force from the floor on the rod will be $\frac{Mg}{16}$

$A$ thin uniform rod of mass $M$ and length $L$ has its moment of inertia $I_1$ about its perpendicular bisector. The rod is bent in the form of a semicircular arc. Now its moment of inertia through the centre of the semicircular arc and perpendicular to its plane is $I_2$. The ratio of $I_1 : I_2$ will be

$A$ rigid uniform bar $AB$ of length $L$ is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time,the angle made by the bar with the vertical is $\theta$. Which of the following statements about its motion is/are correct?
$[A]$ The midpoint of the bar will fall vertically downward
$[B]$ The trajectory of the point $A$ is a parabola
$[C]$ Instantaneous torque about the point in contact with the floor is proportional to $\sin \theta$
$[D]$ When the bar makes an angle $\theta$ with the vertical,the displacement of its midpoint from the initial position is proportional to $(1-\cos \theta)$

Two discs have moments of inertia $I_{1}$ and $I_{2}$ about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds $\omega_{1}$ and $\omega_{2}$ respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

$A$ block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially,the right edge of the block is at $x=0$,in a coordinate system fixed to the table. $A$ point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block,its position is $x$ and the velocity is $v$. At that instant,which of the following options is/are correct?
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{m R}{M+m}$.
$[B]$ The position of the point mass is: $x=-\sqrt{2} \frac{mR}{M+m}$.
$[C]$ The velocity of the point mass $m$ is: $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
$[D]$ The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.