Question

A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kgm2 and a radius of 12.5 cm. A mass of 423 grams is attached to the free end of the string. With the string vertical and taut, the mass is gently released so it can descend under the influence of gravity. As the mass descends, the string unwinds and causes the pulley to rotate, but does not slip on the pulley. What is the speed (in m/s) of the mass after it has fallen through 1.25 m

Answer 1

Answer:

the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s

Explanation:

Given that :

Mass attached to the free end of the string, m = 423 g = 0.423 kg

Moment of inertia of pulley, I = 0.0352 kg m²

Radius of the pulley, r = 12.5 cm = 0.125 meters

Depth of fallen mass, h = 1.25 m

Acceleration due to gravity, g = 9.8 m/s²

Change in potential energy = mgh

= 0.423 × 9.8 × 1.25

=5.18175 J

From the question, we understand that the change in potential energy is used to raise and increase the kinetic energy of hanging mass and  the rotational kinetic energy of pulley.

As such;

$5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2$

where;

$\omega$  is the angular velocity of the pulley

v is the velocity of the mass after falling 1.25 m

where:

$v = r \omega$

$\omega = \frac{v}{r}$

replacing $\omega = \frac{v}{r}$  into above equation; we have:

$5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I( \frac{v}{r})^2$

$5.18175= (\frac{1}{2} m + \frac{1}{2}* (\frac{I}{r^2})v^2 \\ \\ v^2 = \frac{5.18175}{0.5*0.423+0.5*\frac{0.0352}{0.1252}} \\ \\ v^2 = 3.873047 \\ \\ v = 1.968 \ m/s$

Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s