# Rotational Mechanics Problem (Rotating solid wooden disk)

1. Nov 14, 2011

### savva

I have the working for part d) but I don't understand certain parts, so my problems lie with part d), f) and g)

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
a = $\Delta$$\omega$/$\Delta$t
I(Disk) = MR^2/2
L=I$\omega$
Krot = 1/2I$\omega$^2
W2 = I/I2 x $\omega$

3. The attempt at a solution

(a) a = $\Delta$$\omega$/$\Delta$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$\omega$ = 0.185x3.69 = 0.681 kg m^2 s^-1

(c) Krot = 1/2I$\omega$^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?
$\tau$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

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

(e)
Krot = 1/2I$\omega$^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?

2. Nov 14, 2011

### hmzone

Although the wooden disk is rotating, the block's motion is circular, that means there must be a force (centripetal force) that keeps it going in circles. In this particular example, it's the friction.

Fμ = v2 / R

Last edited: Nov 14, 2011
3. Nov 14, 2011

### savva

I tried using the formula given a few different ways, but can't seem to get the correct answer:

so, Fμ = v^2 / R

using v=ωr = 4.02x0.14 = 0.5628 ms^-1

We have R = 0.14m from previous part

Subbing that in we get

Fμ = (0.5628)^2/0.14 = 2.26N

The answer for this part is 0.385N

Why am I not getting this answer?

4. Nov 14, 2011

### hmzone

Sorry my fault..

It's fμ = m v2/r

I forgot to mention the mass. :tongue: Newton's law is F = ma after all, and the acceleration in this one is the centripetal acceleration that keeps the block moving in circles a = v^2/r

Now that you know the friction, you should be able to calculate the coefficient of fiction

5. Nov 14, 2011

### savva

Yep, I got both answers correct now. I'll post up my working tomorrow. Thanks alot for your help!