Question

A disk of radius 25.0cm turns about an axis through the center. The pull on the string produces a linear acceleration a(t)=At on the ball The disk starts from rest and after 3 seconds, linear a(3)=1.80m/s2. Find A and then write an expression for the angular acceleration α(t).

Answer 1

The value for A is A= 0.6

The angular acceleration is
`\alpha (t) = 2.4 \ t \ m/s^2`

Step-by-step explanation:

From the question we are told that

The radius of the disk is
`r = 25.0 \ cm = 0.25 \ m`

The linear acceleration is
`a(t) = At`

At time
`t = 3 \ s`

`a(3) = 1.80 \ m/s^2`

Generally angular acceleration is mathematically represented as

`\alpha(t) = (a(t))/(r)`

Now at t = 3 seconds

a(3) = A * 3

=> 1.80 = A * 3

=.> A = 0.6

So therefore

a(t) = 0.6 t

Now substituting this into formula for angular acceleration

`\alpha (t) = (0.6 t )/(R)`

substituting for r

`\alpha (t) = (0.6 t )/(0.25)`

`\alpha (t) = 2.4 \ t \ m/s^2`