Question

If R = 12 cm, M = 310 g, and m = 40 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

Answer 1

Answer:

$v = 1.42 m/s$

Explanation:

While mass is falling downwards there is no frictional loss so here we can use mechanical energy conservation

So change in gravitational potential energy = gain in kinetic energy of the system

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

for uniform cylinder we will have

$I = \frac{1}{2}MR^2$

now we have

$mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$

$mgh = \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2$

$mgh = \frac{1}{4}Mv^2 + \frac{1}{2}mv^2$

$mgh = (\frac{M}{4} + \frac{m}{2})v^2$

now we have

$v^2 = \frac{mgh}{(\frac{M}{4} + \frac{m}{2})}$

$v^2 = \frac{40(9.81)(0.50)}{(\frac{310}{4} + \frac{40}{2})}$

$v = 1.42 m/s$