Question

A heavy turntable, used for rotating large objects, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 300 N applied to its edge causes the wheel to have an angular acceleration of 0.876 rad/s2.

Answer 1

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