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.

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

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.