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Rotational motion: Conservation of energy doesn't work...

  • #1
New user's thread moved to the Homework Help forums, so no Template is shown. They have been reminded to show their work and readking so far...
http://www.animations.physics.unsw.edu.au/jw/rotation.htm#rolling

I have set up an apparatus similar to what the above link says (the first bit about brass object with shaft). So basically, the shaft is in contact when the brass is first rolling, then it suddenly accelerates when the edge of the brass contacts the surface.

The thing I was curious was why this particular set-up translates the object much slower in comparison to when you just roll an object down a normal ramp, and I don't understand the explanation the link provides me with.

My guess was that the object receives less torque from the friction force when only the shaft is in contact with the surface. This makes it harder for the object to rotate faster and actually move, so it's slow. Whereas when the whole edge of the object is in contact, there's more friction force providing torque to the object, and hence it goes down faster. Is this correct?

Also, I believed that the conservation of energy (mgh =Translational KE + Rotational KE) would stay true for this apparatus. So I tried to test this myself, and created a similar apparatus without the horizontal surface part (only when the shaft is in contact with the surface).

The values I got were:
Mass = 216.7g for the solid object with a shaft
Radius of Object = 6 cm
Radius of Shaft = 1 cm
Distance Travelled = 80cm
Time = 8.35s

However, when I actually tried out the set-up and measured the time and velocity, that doesn't seem to be the case. My theoretical moment of inertia for the object (1/2MR^2) was much smaller than the moment of inertia I calculated from the conservation of energy. Can someone please tell me what I am missing?

If possible, I would really appreciate a mathematical representation of the situation.

Thanks all!

PS: I have attached an image that describes the set-up the link is talking about just in case.
 

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  • #2
CWatters
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http://www.animations.physics.unsw.edu.au/jw/rotation.htm#rolling

I have set up an apparatus similar to what the above link says (the first bit about brass object with shaft). So basically, the shaft is in contact when the brass is first rolling, then it suddenly accelerates when the edge of the brass contacts the surface.

The thing I was curious was why this particular set-up translates the object much slower in comparison to when you just roll an object down a normal ramp, and I don't understand the explanation the link provides me with.
The relationship between linear velocity (v) and angular velocity (w) is set by the radius r of the rolling surface...

w = v/r
so
v/w = r

So if r is smaller the ratio of v to w must be smaller.

Also, I believed that the conservation of energy (mgh =Translational KE + Rotational KE) would stay true for this apparatus. So I tried to test this myself, and created a similar apparatus without the horizontal surface part (only when the shaft is in contact with the surface).

The values I got were:
Mass = 216.7g for the solid object with a shaft
Radius of Object = 6 cm
Radius of Shaft = 1 cm
Distance Travelled = 80cm
Time = 8.35s

However, when I actually tried out the set-up and measured the time and velocity, that doesn't seem to be the case. My theoretical moment of inertia for the object (1/2MR^2) was much smaller than the moment of inertia I calculated from the conservation of energy. Can someone please tell me what I am missing?

If possible, I would really appreciate a mathematical representation of the situation.

Thanks all!

PS: I have attached an image that describes the set-up the link is talking about just in case.
I would expect conservation of energy to hold unless there was some slipping.

Perhaps show us your working.
 
  • #3
haruspex
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the conservation of energy (mgh =Translational KE + Rotational KE) would stay true for this apparatus.
It won't, and it is not necessarily through skidding.
When the surfaces come into contact you have a sudden impact - a partial coalescence even - in the tangential direction. You can compute the new velocities from angular momentum alone, and hence calculate the loss in mechanical energy.
 
  • #4
CWatters
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Contact? Sudden impact? I thought he he was just talking about COE down the ramp? No horizontal track at the bottom.
 
  • #5
haruspex
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Contact? Sudden impact? I thought he he was just talking about COE down the ramp? No horizontal track at the bottom.
You are right, I paid too much attention to the experiment described in the link, and the fact that the text there erroneously attributes the loss of KE to slippage.
 
  • #6
The relationship between linear velocity (v) and angular velocity (w) is set by the radius r of the rolling surface...
I know I'm late, but thanks a lot!! I now understand the reason behind the acceleration in the apparatus.

Could you check if my calculations are correct? I tried fixing it accordingly.
***The calculations are for the section where the shaft, attached by disks in the center, is the surface that is rolling, not the disc***

mgh = 1/2mv^2 + 1/2Iw^2 (forgot to mention h = 18cm)

From,
Mass = 216.7g for the solid object with a shaft
Radius of Object = 6 cm
Radius of Shaft = 1 cm
Distance Travelled = 80cm
Time = 8.35s

mgh = 0.2167kg * 9.8ms^-2 * 0.18m = 0.38 J
1/2mv^2 = 1/2 * 0.2167kg * (0.80m / 8.35s * 2)^2 = 0.0040 J

Thus,
1/2Iw^2 = mgh - 1/2mv^2 = 0.38 J - 0.0040 J = 0.3760J

--> Iw^2 = 0.3760J * 2 = 0.7520J

Since w = v/r, and the shaft is the surface of contact, r = 0.01m where instantaneous v = 0.80m / 8.35s * 2 = 0.192ms^-1
--> w = 0.192 / 0.01 = 19.2rads^-1
---> w^2 = (19.2)^2 = 368.64

I = 0.7520J / w^2 = 0.7520J / 368.64 = 0.002 kg m^2 (rounded)

So the moment of inertia I calculated from the data is around 0.002 kgm^2.

The theoretical moment of inertia I calculated is from I = 1/2MR^2, I = 0.5* 0.2167kg*(0.06m)^2 = 0.00039 kgm^2 (I just ignored the shaft's moment of inertia because it had really small mass.)

The reason why the calculated moment of inertia is bigger would be because there must have been energy loss due to heat, meaning that the rotational kinetic energy would be smaller than the calculated 0.752J, right?
 
  • #7
CWatters
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I got quite a different value for the MOI but was in a rush so I probably made a mistake. I will try and have another look later on.

I started with these two equations...

mgh = 1/2mv^2 + 1/2Iw^2
w=v/r
and then substituted to eliminate w.

mgh = 1/2mv^2 + 1/2Iv^2/r^2
mgh = 1/2v^2 (m + I/r^2)

Then I rearranged to give an equation for I and only then substituted the numbers.
 
  • #8
haruspex
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Are you sure the ramp has constant slope?
 
  • #9
By constant slope you mean that the top end of the ramp was at a same height? Then yes. When the disc rolled, the ramp didn't shift at all and stayed the same way it was.
 
  • #10
haruspex
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By constant slope you mean that the top end of the ramp was at a same height? Then yes. When the disc rolled, the ramp didn't shift at all and stayed the same way it was.
No, I mean was it a straight line, same gradient all the way? Your result could be explained if the slope was gentle at first, getting steeper further down.
 
  • #11
CWatters
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So the moment of inertia I calculated from the data is around 0.002 kgm^2.
I've rechecked my calculations and agree with this figure.
 
  • #12
No, I mean was it a straight line, same gradient all the way? Your result could be explained if the slope was gentle at first, getting steeper further down.
Yes the ramp was a straight line. To be exact it was made up with two metersticks. So i placed wooden blocks in the top and bottom part between the two metersticks, taped it. So there would be a 15cm ish gap where the bigger radius part of the disk was rolling (without touching any surface except air) and the shaft would roll on the surface of the meter stick. Im pretty sure this is considered as constant gradient as you mentioned
 

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