Auniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. the disk is initially at rest, and then a constant force f = 30.0 n is applied tangent to the rim of the disk.
(a) what is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution?
(b) what is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolution?

Given that

r= 0.2 m

m= 40 kg

F= 30 N

We know that mass moment of inertia of disk given as

I = m r²/2

I= 40 x 0.2²/2

I= 0.8 kg.m²

The torque produce by force

T = F.r

T= 30 x 0.2 = 6 N.m

We also know that

T= I α

α = angular acceleration of disk

by putting the values

6 = 0.8 x α

α = 7.5 rad/s²

Given that initial speed is zero.

Lets speed is ω after 0.2 rev

ω ² = ωo+ 2 α θ

We know that

1 rev  = 2π

0.2 rev= 0.4π

θ = 0.4π

ω ² = 0+ 2 x 7.5 x 0.4 x π

ω = 4.34 rad/s

So the tangential velocity v

v= ω  r

v= 4.34 x 0.2 = 0.86 m/s

v= 0.86 m/s

Tangential acceleration (at)= α .r= 7.5 x 0.2 =1.5 m/s²

Radial acceleration(ar) = ω² r = 4.34² x 0.2 = 3.76 m/s²

resultant acceleration ,a

a = 4.05 m/s²

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over

where is the