Solving rotational motion without torque

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  • #1
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Homework Statement


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

Homework Equations




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!
 

Answers and Replies

  • #2
andrevdh
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not all of the mass is moving at the calculated deceleration (or speed)
 
  • #3
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not all of the mass is moving at the calculated deceleration (or speed)

Why and how is that? The whole wheel is spinning.
 
  • #4
haruspex
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Why and how is that? The whole wheel is spinning.
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
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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?
v = rw?

Do you mean I can only use angular velocity to calculate angular motion and newtonian laws to calculate linear motion?
 
  • #6
haruspex
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v = rw?
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
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Yes, that's the right equation. All parts of the disc have the same ##\omega##, but they do not all have the same r.

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

Thanks!
 
  • #8
haruspex
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So I eventually need to convert the tangetial velocity back to angular velocity. Ok I get it now

Thanks!
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
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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?

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
haruspex
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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?
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
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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?

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
andrevdh
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I get that 50 rpm is 5.24 rad/s?
 
  • #13
haruspex
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I get that 50 rpm is 5.24 rad/s?
Which is indeed about 2.618/0.5.
 

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