A constant force of 20N is applied tangentially to a string wound on the rim of a 40 cm diameter wheel, initially at rest.

Question

(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?

Answer 1

(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)

Answer 2

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?

Answer 3

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)

Answer 4

no i mean to convert 45degrees into rads you multiply by 2pi
thanks for your help

Answer 5

45 degrees is pi/4 radians. You multiply by pi/180 radians per degree, NOT 2 pi

Answer 6

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