Radius Rad

a wheel of mass 0.52 kg and radius 46 cm is spinning with an angular velocity of 21 rad/s.?

You then push your hand against the edge of the wheel, exerting a force F on the wheel as shown in the figure below. If the wheel comes to a stop after traveling 1/4 of a turn, what is F?

Angular momentum is defined as:

L = I*w

I = moment of intertia = m*R^2 (for a hoop or wheel)
w = angular velocity = 21 rad/s
R = radius of the wheel = 0.46 m
m = mass of the wheel = 0.52 kg

L = m*R^2*w

The force applies a torque to the wheel and this causes a change in the angular momentum of the wheel.

T = torque = dL/dt ... I am using "d" to represent just a change)
And T = F*R so F*R = dL/dt
F = (1/R)(dL/dt)

dL is the change in momentum and since the wheel is brought to a stop we can write:
dL = m*R^2*w

dt is the time the wheel takes to stop and this is done after 1/4 a turn. We can get the time this takes by:
Period = 2*pi/w
We want 1/4 of this so it takes pi/(2*w) seconds to stop

Now put dL and dt into the above equation for the force:
F = (1/R)(m*R^2*w)/[pi/(2*w)]
F = 2*m*R*w^2/pi

Using our numbers:
F = (2*0.52*0.46*21^2)/pi N
F = 67.1552 N

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