Solution Found!
A compact disk, which has a diameter of 12.0 cm, speeds up
Chapter 10, Problem 36P(choose chapter or problem)
A compact disk, which has a diameter of \(12.0 \mathrm{~cm}\), speeds up uniformly from \(0.00\) to \(4.00 \ \mathrm{rev} / \mathrm{s}\) in \(3.00 \mathrm{~s}\). What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is (a) \(2.00 \ \mathrm{rev} / \mathrm{s}\) and (b) \(3.00 \ \mathrm{rev} / \mathrm{s}\)?
Questions & Answers
QUESTION:
A compact disk, which has a diameter of \(12.0 \mathrm{~cm}\), speeds up uniformly from \(0.00\) to \(4.00 \ \mathrm{rev} / \mathrm{s}\) in \(3.00 \mathrm{~s}\). What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is (a) \(2.00 \ \mathrm{rev} / \mathrm{s}\) and (b) \(3.00 \ \mathrm{rev} / \mathrm{s}\)?
ANSWER:Step 1 of 3
a.)
We have to find the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 rev/s
The tangential acceleration is given by the expression