Answer:
I = 113.014 kg.m^2
m = 2075.56 kg
wf = 3.942 rad/s
Explanation:
Given:
- The constant Force applied F = 300 N
- The radius of the wheel r = 0.33 m
- The angular acceleration α = 0.876 rad / s^2
Find:
(a) What is the moment of inertia of the wheel (in kg · m2)?
(b) What is the mass (in kg) of the wheel?
(c) The wheel starts from rest and the tangential force remains constant over a time period of t= 4.50 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?
Solution:
- We will apply Newton's second law for the rotational motion of the disc given by:
F*r = I*α
Where, I: The moment of inertia of the cylindrical wheel.
I = F*r / α
I = 300*0.33 / 0.876
I = 113.014 kg.m^2
- Assuming the cylindrical wheel as cylindrical disc with moment inertia given as:
I = 0.5*m*r^2
m = 2*I / r^2
Where, m is the mass of the wheel in kg.
m = 2*113.014 / 0.33^2
m = 2075.56 kg
- The initial angular velocity wi = 0, after time t sec the final angular speed wf can be determined by rotational kinematics equation 1:
wf = wi + α*t
wf = 0 + 0.876*(4.5)
wf = 3.942 rad/s
