Asked by Oisin

A constant force of 20N is applied tangentially to a string wound on the rim of a 40 cm diameter wheel, initially at rest.
(i) How much work does this force do as it turns the wheel through 45„a?
(3 %)
(ii) If the wheel has mass 8.0 kg and radius of gyration 15 cm, how long does it take the wheel to rotate through this angle?
Answered by drwls
(i) 20N * 0.2 m * (angle in radians).
I cannot interpret your angle symbol, which shows up here as ,,a.

(ii) Angular acceleration rate = (Torque)/(Moment of inertia)
= (Torque)/[(Mass)*Rg^2]
The torque is 4 N*m
Angular acceleration =
4 N*m/(8*.15^2 kg m^2) = 22.2 rad/s^2

Time required to rotate angle theta =
2*(theta)/(angular acceleration rate)

Answered by Oisin
so for the first part 20N * 0.2m * 2pi is that correct?

also how did you get your torque?

(iii) What is the angular momentum of the wheel after this time?
Answered by drwls
You never said the angle turned was 2 pi. You typed 45,,a

Torque is force times lever arm (in this case, wheel radius)

(iii) Angular momentum after time T is
(Torque)*(Time) =
(Moment of inertia)*(angular velocity)

Answered by Oisin
no i mean to convert 45degrees into rads you multiply by 2pi
thanks for your help
Answered by drwls
45 degrees is pi/4 radians. You multiply by pi/180 radians per degree, NOT 2 pi
Answered by Oisin
how did u find the angular velocity?
im tryin to use v = wr but i don't know where to find the normal velocity of the rim