Question

In the figure A & B are two blocks of mass 4 kg and 2 kg respectively attached to the two ends of a light string passing over a disc C of mass 40 kg and radius 0.1 m. The disc is free to rotate about a fixed horizontal axes, coinciding with its own axis. The system is released from rest and the string does not slip over the disc. Then,

• A. the linear acceleration of mass B is $10/13m/s^2$
• B. the number of revolutions made by the disc at the end of 10 sec.from the start is $5000/26\pi$
• C. the tension in the string segment supporting the block A is $480/13N$
• D. None of these.

Answer 1

Answer: (A), (B), (C)

the linear acceleration of mass B is $10/13m/s^2$
the number of revolutions made by the disc at the end of 10 sec.from the start is $5000/26\pi$
the tension in the string segment supporting the block A is $480/13N$

Block A has higher weight than Block B, so A will have a downward acceleration and B will in upward direction of same magnitude because of constrained, let $T_1$ and $T_2$ are tensions in left and right hanging part of string.

Applying Newton's Law of motion.

$M_a\times g-T_1=M_{a}a\to \left(1\right)$ ; a= acceleration of Block A and B

$T_2-M_b\times g=M_{b}a\to \left(2\right)$

$T_{1}R-T_{2}R=I\frac{a}{R}\to \left(3\right)$;

$I=\frac{1}{2}{M}_C{R}^2$; $I$ = moment of inertia of disc C about its axis;

On solving above three equations and putting given values of parameters in result :

(a)

$4g-\frac{1}{2}\times 40\times a-2g=6a$

$a=\frac{10}{13}\left(\frac{m}{s^2}\right)$

(b) Angular acceleration of disc,$\alpha =\frac{a}{R}$

$\alpha =\frac{\frac{10}{13}}{0.1}$

$\alpha =\frac{100}{13}$

$\alpha =\frac{100}{13}\left(\frac{rad}{s^2}\right))$

On applying Newton motion's equation:

$\alpha =\frac{100}{13}\left(\frac{rad}{s^2}\right))$; $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$; ${\omega }_0=0$;

$\theta =0+\frac{1}{2}\times \frac{100}{13}\times {10}^2$;

$\theta =\frac{5000}{13}$(rad); No of revolutions in 10 sec = $\frac{\frac{5000}{13}}{2\pi }$

= $\frac{5000}{26\pi }$

(c)

From equation (1)

$4g-T_1=4a$ ; $a=\frac{10}{13}\left(\frac{rad}{s^2}\right)$

$T_1=\frac{480}{13}$ N