physics ??

Question

the combination of an applied force and friction force produces a constant total torque of 36.0 N.m on a wheel rotating about a fixed axis .the applied force acts for 6.00s.during this time the angular speed of the wheel increases from 0 to 10.0 rad/sec.the applie force is then removed,and the wheel comes to rest in 60.0 sec.
a)find the moment of intertiA OF THE WHEEL.
b)find the magnitude of the frictinal torque.
c)find the total number of revolutions of the wheel.

Answer 1

The rotational acceleration is the change in angular velocity divided by time (a constant torque will produce a constant acceleration); this is then 10/6 rad/sec^2. The relation between torque and rotational acceleration is T = I*a, where T is the torque, I the moment of inertia, and a the acceleration. You find I from T/a. When the applied force is removed, only the frictional torque remains to slow the wheel down from 10 to 0, This happens in 60sec, so the acceleration is 10/60 rad/sec^s. Using the same formula, T = I*a, but now you have I and you can find T.

The total no of revolutions of the wheel is 2π∫w*dt. w = a*t in both intervals, so the number of revolutions in each interval is 2π*(.5at^2) = πat^2. Use the values you got for a and t in each case above to get the total no of revolution.

Answer 2

Let T be the torque, I the moment of inertia, a the angular acceleration, w the angular velocity, t the time, and d the angular displacement

A) w2 = w1 + at
10 = 0 + a(6)
a = 5/4

T = I * a
(36.0) = I (5/4)
I = 28.8 kg * m^2

B) w2 = w1 + at
0 = 10 + a(60)
a = -1/6

T = I * a
T = (28.8)(-1/6)
T = -4.8

Magnitude is 4.8 N * m

C) There are two parts to the rotation of the wheel- when there are both the applied and frictional forces, and when there is only the frictional force. We will handle each separately and add them up afterwards.

d1 = w0 t + 1/2 at^2
d1 = (0)(6) + 1/2 (5/4)(6)^2 = 22.5 rad

d2 = w0 t + 1/2 at^2
d2 = (10)(60) + 1/2 (-1/6)(60)^2 = 300 rad

Sum is 322.5 radians.

322.5 radians * (1 revolution / 2 pi radians) = 51.3 revolutions