# Solving rotational motion without torque

1. Aug 24, 2015

### Zynoakib

1. The problem statement, all variables and given/known data
A potter’s wheel—a thick stone disk of radius 0.500 m
and mass 100 kg—is freely rotating at 50.0 rev/min.
The potter can stop the wheel in 6.00 s by pressing a
wet rag against the rim and exerting a radially inward
force of 70.0 N. Find the effective coefficient of kinetic
friction between wheel and rag

2. Relevant equations

3. The attempt at a solution
Actually, I could get the answer using angular acceleration and torque, but I could do the same thing with Newtonian Laws and I want to know why.

Tangential velocity of the wheel:
50 rev/ min = 2.618 m/s

u = 2.618 m/s, v = 0, a = ?, t = 6
v = u + at
0 = 2.618 + 6a
a = 0.436 m/s^2

Deceleration force
F = ma = (100)(0.436) = 43.6N

The friction coefficient
43.6 = (coefficient)(70)
coefficient = 0.623, which is exactly the double of the correct answer.

Thanks!

2. Aug 24, 2015

### andrevdh

not all of the mass is moving at the calculated deceleration (or speed)

3. Aug 24, 2015

### Zynoakib

Why and how is that? The whole wheel is spinning.

4. Aug 25, 2015

### haruspex

It's all moving at the same angular speed, but you have taken it all to be moving at the same linear speed. What equation relates the two?

5. Aug 25, 2015

### Zynoakib

v = rw?

Do you mean I can only use angular velocity to calculate angular motion and newtonian laws to calculate linear motion?

6. Aug 25, 2015

### haruspex

Yes, that's the right equation. All parts of the disc have the same $\omega$, but they do not all have the same r.

7. Aug 25, 2015

### Zynoakib

So I eventually need to convert the tangetial velocity back to angular velocity. Ok I get it now

Thanks!

8. Aug 25, 2015

### haruspex

I'm not quite sure what you mean by that.
Anyway, it would be more usual to apply standard theory regarding moments of inertia. You are aware of that?

9. Aug 25, 2015

### Zynoakib

Yes, I know. But when you calculate circular motion, it is ok not to use angular velocity or angular acceleration, so I was wondering if I can do the same thing in rotational motion. You know what I saying?

10. Aug 25, 2015

### haruspex

Ok, but it will turn out to be equivalent to reinventing the concept of moment of inertia. How will you now calculate the KE of the disc?

11. Aug 25, 2015

### Zynoakib

Moment of inertia = 1/2MR^2 = 0.5(100)(05)^2 = 12.5 kgm^2
KE = 1/2(12.5)(2.618 / 0.5)^2 = 171 J

12. Aug 25, 2015

### andrevdh

I get that 50 rpm is 5.24 rad/s?

13. Aug 25, 2015