Question

A mass m of 400g hangs from the rim of a wheel of radius=15cm when released from rest the mass falls 2m in 6.5 sec. Find the moment of inertia of the wheel.

Answer 1

Answer on Question #47771, Physics, Computational Physics

A mass $m$ of 400g hangs from the rim of a wheel of radius = 15cm when released from rest the mass falls 2m in 6.5 sec. Find the moment of inertia of the wheel.

Solution:

In 6.5 seconds, potential energy loss is

$$PE = mgh = 0.4 \, \mathrm{kg} \cdot 2.0 \, \mathrm{m} \cdot 9.81 \, \mathrm{m/s^2} = 7.848 \, \mathrm{J}$$

Final velocity of mass (twice the average velocity):

$$v_f = 2 \cdot \frac{2 \, \mathrm{m}}{6.5 \, \mathrm{s}} = 0.6154 \, \mathrm{m/s}$$

Final angular velocity of wheel:

$$\omega_f = \frac{v_f}{R} = \frac{0.6154}{0.15} = 4.10 \, \mathrm{rad/s}$$ $$K.E. \, \text{gain} = P.E. \, \text{loss}$$ $$\frac{m v_f^2}{2} + \frac{I \omega_f^2}{2} = PE$$

Solve for moment of inertia, $I$,

$$I = \frac{2PE - m v_f^2}{\omega_f^2}$$ $$I = \frac{2 \cdot 7.848 - 0.4 \cdot 0.6154^2}{4.10^2} = 0.925 \, \mathrm{kg \cdot m^2}$$

Answer: $I = 0.925 \, \mathrm{kg \cdot m^2}$.

http://www.AssignmentExpert.com/