A turntable is set rotating with an angular velocity of 30rad/s. The turntable has a mass of 2kg and a radius of 0.2m (Assume the turntable rotates on its axis with no friction and can be modeled as a disk.
a.) What is the rotational inertia of the turntable
ans: 0.04kgm^2
b.) What is the angular momentum:
ans: 1.2kgm^2/s
c.) A small lump of clay with a mass of 0.5kg is dropped on the turntable and sticks to it at a distance of 0.1m from the axis of rotation (Assume the clay can be modeled as a point mass) What is the angular velocity afterwards?
ans: 26.67rad/s
d.) A rubber pad is the pressed against the outer rim of the disk as a brake and the disk/clay combination comes to a stop in 8s. What is the angular acceleration?
ans: -3.33rad/s^2
e.) What is the average force exerted by the brake?
ans: 0.1499Nm
f.) THIS IS THE ONLY ONE I CAN'T FIGURE OUT
How much energy was dissipated by the brake?
Do I use 1/2mv^2? That's the only way I can think of doing it...and answers will be GREATLY appreciated, thank you!
a.) What is the rotational inertia of the turntable
ans: 0.04kgm^2
b.) What is the angular momentum:
ans: 1.2kgm^2/s
c.) A small lump of clay with a mass of 0.5kg is dropped on the turntable and sticks to it at a distance of 0.1m from the axis of rotation (Assume the clay can be modeled as a point mass) What is the angular velocity afterwards?
ans: 26.67rad/s
d.) A rubber pad is the pressed against the outer rim of the disk as a brake and the disk/clay combination comes to a stop in 8s. What is the angular acceleration?
ans: -3.33rad/s^2
e.) What is the average force exerted by the brake?
ans: 0.1499Nm
f.) THIS IS THE ONLY ONE I CAN'T FIGURE OUT
How much energy was dissipated by the brake?
Do I use 1/2mv^2? That's the only way I can think of doing it...and answers will be GREATLY appreciated, thank you!
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f.) The same as the total energy in the disk: E = ½I*w² = ½*.04*30² = 18 J
e.) You have calculated the torque applied by the brake; the FORCE is the torque divided by the radius: F = .15/.2 = 0.75 N
e.) You have calculated the torque applied by the brake; the FORCE is the torque divided by the radius: F = .15/.2 = 0.75 N