Question

3. A rigid uniform disk (moment of inertia I = 1/2 m R²) has a mass m = 10 Kg and radius R = 2 m, it is free to rotate around a fixed axis passing through its center and is at rest at t = 0 s. A constant force F = 20 N is applied halfway between the center and the edge of the disk, in a direction perpendicular to the disk radius.

(i) Calculate the torque, the angular acceleration, the angular velocity and the rotation angle of the disk after t = 10 s. τ=20 Nm ; α=1 s^-2 ; ω=10 s^-1 ; Δθ=50 (rad)

(ii) Calculate the kinetic energy and the angular momentum of the disk at t = 10 s. K=1000 J ; L = 200 Js

(iii) Show that the relationships between torque, work, kinetic energy and angular momentum are numerically satisfied. W = τ∗Δθ = 20 Nm*50 = 1000 J ; Kf-Ki = 1000 J

τ∗Δt = 20 Nm*10s = 200 Js ; Lf-Li = 200 Js

THE ANSWERS ARE IN BOLD, COULD YOU PLEASE SHOW THE WORKING NEEDED FOR THIS QUESTION?

Answer 1

Given, m= 10 kg; R= 2 m; F = 20 N; distance from the axis to the point of application of force,r = R/2 = 1 m

initial angular velocity, w_o= 0 rad/s

Moment of inertia of the disk, I = 0.5mR² = 0.5*10*2² = 20 kg-m²

i) given, t= 10 s

Torque, T = F*r = 20*1

T= 20 N-m

Also, torque,T = I* { is the angular acceleration}

20 = 20*

= 1 rad/s²

now for the motion for 10 seconds, using,

w = w_o + t

w = 0 + 1*10

w = 10 rad/s

now, using, = w_o*t + 0.5*t²

= 0*10 + 0.5*1*10²

= 50 radians

ii) kinetic energy of a rotating disk is, K = 0.5*I*w² = 0.5*20*10²

K = 1000 J

angular momentum, L = I*w = 20*10

L = 200 J-s

iii) Work done, W = T* { is the angular momentum}

W = 20*50

W= 1000 J

According to Work energy theorem, work done(W) is equal to the change in kinetic energy,

W = K_f-K_i = 1000J

Angular momentum, L = T*t = 20*10 = 200 J-s

also from second part, L= 200 J-s

Hence verified