Question

As shown below, a uniform disc is mounted to a fixed axle and is free to rotate without friction. A thin uniform rod is rigidly attached to the disc so that it will rotate with the disc. A block is attached to the rod.

Disc : mass = 3m, radius = R, moment of inertia about center $I_D=\frac{3}{2}m{R}^2$

Rod : mass = m, length = 2R, moment of inertia about one end $I_R=\frac{4}{3}m{R}^2$

Block : mass 2m

The system is held in equilibrium with the rod at an angle ${\theta }_0$ to the vertical, as shown above, by a horizontal string of negligible mass one end attached to the disc and the other to a wall.

The string is now cut, and the disc-rod-block system is free to rotate. As disc rotates, the rod passes the horizontal position shown below. The linear speed of the end of the rod for the instant the rod is in the horizontal position is : (The arrow represent the direction of vector quantities)

• A. $\sqrt{\frac{48}{13}gRcos{\theta }_0}\downarrow$
• B. $\sqrt{\frac{12}{13}gRcos{\theta }_0}\downarrow$
• C. $\sqrt{\frac{12}{13}gRcos{\theta }_0}\leftarrow$
• D. $\sqrt{\frac{48}{13}gRcos{\theta }_0}\to$

Answer 1

The correct option is A

$\sqrt{\frac{48}{13}gRcos{\theta }_0}\downarrow$

From work-energy theorem between initial position and horizontal,

$2mg\left(2R\right)cos{\theta }_0+mgRcos{\theta }_0=\frac{1}{2}[\frac{3}{2}m{R}^2+\frac{4}{3}m{R}^2+2m(2R{)}^2]{\omega }^2$

$\Rightarrow 5mgRcos{\theta }_0=\frac{1}{2}[\frac{3}{2}+\frac{4}{3}+8]m{R}^2{\omega }^2\Rightarrow 5mgcos{\theta }_0=\frac{1}{2}\times \left(\frac{9+8+48}{6}\right)R{\omega }^{2}andv=r\omega =2R\omega$