Posted by Oisin on Tuesday, July 20, 2010 at 8:53am.
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?

physics  drwls, Tuesday, July 20, 2010 at 10:07am
(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)

physics  Oisin, Tuesday, July 20, 2010 at 10:12am
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?

physics  drwls, Tuesday, July 20, 2010 at 11:07am
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)

physics  Oisin, Tuesday, July 20, 2010 at 11:17am
no i mean to convert 45degrees into rads you multiply by 2pi
thanks for your help

physics  drwls, Tuesday, July 20, 2010 at 12:04pm
45 degrees is pi/4 radians. You multiply by pi/180 radians per degree, NOT 2 pi

physics  Oisin, Tuesday, July 20, 2010 at 4:40pm
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
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