The Pulley in the fugure has a mass M=10kg and radius R=0.5m. It is uniform soli

Question

The Pulley in the fugure has a mass M=10kg and radius R=0.5m. It is uniform solid disk with moment of inertia I=MR2/2. A massles corl is wrapped around it and a tention force FT is applied. the pulley is initially at rest. After FT is applied for 10s, the angular speed of the pulley i w=50 rad/s.

a)What is the pulley's angular acceleration? What is the linear velocity of a point on the rim of the pulley after FT is applied for 10 seconds?

b) What is the only Force producing a torque, calculate FT.

C)After FT is applied for 10 s, what is the pulley's kinetic energy?

Explanation / Answer

a) Use the standard rotational kinematics equation

= (0) + t, since (0) = 0 (starts from rest), we get

= /t = (50 rad/s)/(10s) = 5 rad/s^2

Since the radius is 0.5m, and = 50rad/s, v = r = (50 rad/s)*(0.5m) = 25 m/s

b) First, calculate the moment of inertia

I = (1/2)MR^2 = (1/2) (10kg)(0.5m)^2 = 1.25kgm^2

Now, = Fr = I*, so

F = I*/r = (1.25kgm^2)*(5 rad/s^2)/(0.5m) = 12.5 N

c) We know that = 50rad/s after 10s, so

K = (1/2)I*^2 = (1/2)(1.25kgm^2)(50 rad/s^2) = 1.56 kJ of energy

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