Question

How much work is required to accelerate a 1440kg cylinder from rest to 1.00 revolution per second? Assume a solid cylinder of radius 7.50m. The moment of inertia of a cylinder is (1/2)mr2. Answer in J.

Answer 1

Given,

m = 1440 kg ; w0 = 0 rad/s ; w = 1 rev/s = 6.28 rad/s ; r = 7.5 m

We know that, the work required will be the diffrence of intial and final rottational KE

W = 1/2 I (w^2 - w0^2)

I = 1/2 m r^2 = 0.5 x 1440 x 7.5^2 = 40500 kg-m^2

W = 0.5 x 40500 x (6.28^2 - 0^2) = 798627.6

Hence, W = 798627.6 J = 7.98 x 10^5 J

[Incase of rolling without slipping]

W = 1/2 I w^2 + 1/2 m v^2

v = r w = 7.5 x 6.28 = 47.1 m/s

W = 0.5 x 40500 x 6.28^2 + 0.5 x 1440 x 47.1^2 = 2395882.8 J

Hence, W = 2.4 x 10^6 J