Question

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 82.0 N is applied to the rim of the wheel. The wheel has radius 0.150 m . Starting from rest, the wheel has an angular speed of 12.8 rev/s after 3.88 s. What is the moment of inertia of the wheel?

Answer 1

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

$\tau=F\times r$

$\tau=82.0\times0.150$

$\tau=12.3\ N-m$

Now, The angular acceleration

$\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}$

$\dfrac{d\omega}{dt}=20.73\ rad/s^2$

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

$\tau=I\times\dfrac{d\omega}{dt}$

$I=\dfrac{I}{\dfrac{d\omega}{dt}}$

$I=\dfrac{12.3}{20.73}$

$I= 0.593\ kg-m^2$

Hence, The moment of inertia of the wheel is 0.593 kg-m².