A uniform cube of mass $2.0 \mathrm{kg}$ rests or a rough horizontal surface with coefficient of friction $\mu =0.2$ A force $F$ is applied on the cube as shown. The range of value of $F$ for which the cube will slide without toppling is given by $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

The figure shown is an overhead view of a thin uniform rod of length $L$ and mass $M$ rotating horizontally at angular speed $\omega$ about an axis passing through its centre. A particle of mass $m$ initially attached to the end $A$ is ejected from the rod with velocity $V_p$ which is perpendicular to the rod at the instant of ejection. If $\omega =10.0 \mathrm{rad} {\mathrm{s}}^{-1}, L=1.0m, m=M$ and $V_p$ is $4 \mathrm{m} {\mathrm{s}}^{-1}$ more than the speed of rod end $A$ just after the ejection, then the value of $V_p$ is 

Uniform rod of mass $M$, length $2L$ is suspended with the help of two ideal strings as shown. Complete arrangement is in equilibrium in vertical $x-y$ plane. Find value of $F_0$ so that string $1$ remains vertical (assuming rod is horizontal).

Ends $A$ and $B$ of a rigid uniform rod of mass $M$ parallel to $Y$-axis have velocities ${\overrightarrow{v}}_A=v_0\hat{i},{\overrightarrow{v}}_B=2v_0\hat{i}$ as shown. A point mass $M$ is also attached at the end $B$ of rod, then the angular momentum of system about point $O$ will be

Consider a rigid body which consist of uniform hemispherical shell of radius $R$ and uniform cylindrical shell of height $L$ and radius $R$ (material of hemispherical shell and cylindrical shell are same). Find maximum value of $L$ so that complete rigid body remains in stable equilibrium in the position shown (neglect friction).

Horizontal force $F_0$ is applied on a uniform circular disc of mass $M$ and radius $R$ which can rotate in vertical $x-y$ plane about an axes passing through $O$ as shown. What is the acceleration of bottom most point $P$ on the disc at this instant?

Consider the shown diagram. A fixed vertical smooth rod $AB$ is kept in front of a uniform disc of mass $m$ and radius $R$ which is rolling without slipping on a rough horizontal surface. Velocity of centre of disc is $\sqrt{3} \mathrm{m} {\mathrm{s}}^{-1}$. A rod $PQ$ of length $2 \mathrm{m}$ is connected with disc at $Q$, (point $Q$ is at a vertical distance of $\frac{R}{2}$ from centre of disc) by pin joint and other end of rod $PQ$ can freely slide over smooth vertical rod $AB$. At this instant rod $PQ$ makes an angle $\theta =150°$ with rod $AB$ as shown. Find the angular velocity of rod at this instant.

A uniform circular disc of mass $M\left(=1\mathrm{kg}\right)$ and radius $R\left(=1m\right)$ is fixed to vertical axis $AB$ with the help of a rod of length $L$ as shown. The disc rolls on a horizontal surface and is free to rotate about its centre. If there is no slipping at the point of contact of disc and horizontal surface then kinetic energy of the disc will be (given ${\omega }_0=2\mathrm{rad} {\mathrm{s}}^{-1}, L=1 \mathrm{m}$)

A uniform circular disc of mass $M$ and radius $R$ is having a point mass $M$ at its circumference is performing rolling without slipping over a rough horizontal surface as shown. Acceleration of point $O$ is $a_0$ towards right. If $N$ is the normal reaction on disc by the horizontal surface at this instant then