Question

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 90.0 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 14.3 rev/s after 2.72 s. What is the moment of inertia of the wheel

Answer 1

Answer:

the moment of inertia of the wheel is 0.408 kg.m²

Explanation:

Given;

tangential force applied to the wheel, f = 90 N

radius of the wheel, r = 0.15 m

initial angular speed of the wheel, ω₁ = 0

final angular speed of the wheel, ω₂ = 14.3 rev/s

time of motion of the wheel, t = 2.72 s

The tangential acceleration of the wheel is calculated as;

$\alpha _t = \alpha r$

where;

$\alpha$ is the angular acceleration

$\alpha = \frac{\Delta \omega }{\Delta t} = \frac{\omega _2 - \omega_1}{t_2-t_1}\\\\where;\\\\\omega_2 \ is \ the \ final \ angular \ speed = 14.3 \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 89.86 \ rad/s$

$\alpha = \frac{89.86 - 0}{2.72} = 33.04 \ rad/s^2\\\\\alpha _t = \alpha r\\\\\alpha _t = 33.04 \ rad/s^2 \times 0.15 \ m\\\\\alpha _t = 4.96 \ m/s^2$

The mass of the wheel is calculated as;

F = ma

m = F/a

m = (90)/(4.96)

m = 18.15 kg

The moment of inertia of the wheel is calculated as;

I = mr²

I = 18.15 x (0.15)²

I = 0.408 kg.m²

Therefore, the moment of inertia of the wheel is 0.408 kg.m²