A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.50 rev/s in 3.00 s .

Question

a)What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00?

b)What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 3.00?

Answer 1

Hello Lexi;

I would say that the tangential acceleration is constant in both cases as it speeds up uniformly.

Angular Velocity ω = 4.5 rev/s = 4.5 x 2π

time = 3s

Therefore the angular acceleration α is 9π ÷ 3 = 3π rad/s²

The tangential acceleration is a = αr ==> 3π x 0.06m = 0.18π m/s² has to be the tangential acceleration. Hence, we would expect both choices have the same answer as given 0.18π m/s²

Let me know what you think....

Best

Bilal