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## Homework Statement

3. A solid wooden disk is suspended as shown: it has mass 7.70 kg and radius

0.219 m . It is rotated from rest about its vertical axis, reaching an angular velocity of 3.69 rad/s in 15.1 s.

http://i41.tinypic.com/2u9mgzs.jpg

Calculate

(a) the average angular acceleration

(b) the angular momentum.

(c) the rotational kinetic energy

A 0.17 kg metal block is gently placed on the disk 0.140 m from the rotational axis.

(d) Show that the new angular velocity is 3.62 rad/s

(e) Calculate the new rotational kinetic energy.

The wooden cylinder is very slowly accelerated, and when the new angular velocity is 4.02 rad/s the block is just about to slide.

Calculate

(f) the frictional force on the block.

(g) the coefficient of static friction μs between the block and the disk.

## Homework Equations

a = [itex]\Delta[/itex][itex]\omega[/itex]/[itex]\Delta[/itex]t

I(Disk) = MR^2/2

L=I[itex]\omega[/itex]

Krot = 1/2I[itex]\omega[/itex]^2

W2 = I/I2 x [itex]\omega[/itex]

## The Attempt at a Solution

(a) a = [itex]\Delta[/itex][itex]\omega[/itex]/[itex]\Delta[/itex]t = 3.69/15.1 = 0.244 rad/s^2

(b)

I(Disk) = MR^2/2 = (7.70x0.219^2)/2 = 0.185 kg m^2

L=I[itex]\omega[/itex] = 0.185x3.69 = 0.681 kg m^2 s^-1

(c) Krot = 1/2I[itex]\omega[/itex]^2 = 1/2(0.185)(3.69)^2 = 1.257J

(d)

I know L1 = L2, can we prove that this is the case if we didn't have given the value of angular velocity?

[itex]\tau[/itex]NET = 0 (Not sure how I can prove this)

I2 = I1 + MR^2 (Inertia of block)

I2 = 0.185 + 0.17(0.14)^2 = 0.188 kg m^2

[itex]\omega[/itex]2 = (I1/I2)x[itex]\omega[/itex] = (0.185/0.188)x3.69 = 3.62 rad/s

(e)

Krot = 1/2I[itex]\omega[/itex]^2 = 1/2(0.188)(3.62)^2 = 1.23J

(f) & (g) no idea how to work out, can anyone please give me a hand?