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Topic: Please Rewind (Read 2151 times) |
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denis
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Please Rewind
« on: Apr 27th, 2008, 5:45pm » |
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I got this question on a first year university Physics assignment 25 years ago. I thought it was an interesting question because the answer does not depend on the mass of the tape roll.
Easy-Medium difficulty for a Physics major.... Hard for a non physics students.
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A length L of flexible tape is tightly wound. It is then allowed to unwind as it rolls down a steep incline that makes an angle  with the horizontal, the upper end of the tape being tacked down.
Find the time T it takes for the tape to completely unwind.
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| « Last Edit: Apr 28th, 2008, 1:07pm by denis » |
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william wu
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Re: Please Rewind
« Reply #1 on: May 20th, 2008, 3:52pm » |
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I'm not a physics student, but I'll try. My approach is naive and most likely wrong  However, I'd be very happy to have someone explain why it's wrong, if it is.
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Let's consider any particular instant in time, t. At that moment, the tape roll has mass M(t) and radius R(t); henceforth, for brevity, I will just refer to these as M and R.
Let f denote the friction force the tape roll experiences at that moment.
From the equations F=ma and  = I  , we have
(1) Mg sin - f = Ma, where a denotes linear acceleration down the slope.
(2) f R = I a/R, where I is the moment of inertia of the tape roll
We model the tape roll as a solid disk. Thus, I = MR^2, and (2) becomes
(edit: Actually, the moment of inertia should be 1/2 MR^2. See my post later for corrections that account for this error.)
(2) f = M a.
Combining (1) and (2) yields
(3) a = (g/2) sin
an answer which is apparently independent of both M and R.
As time progresses, we will have M(t) > M(t +  ), and R(t) > R(t + ). But by equation (3) above, that does not affect the acceleration, which is constant. Hence we can use kinematics to finish the problem:
(4) L = (1/2) a t^2
or
t =  {2L/a} = 2 { (L/g) csc }.
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| « Last Edit: May 21st, 2008, 11:49am by william wu » |
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temporary
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Re: Please Rewind
« Reply #2 on: May 20th, 2008, 9:00pm » |
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I think you would need the density to know exactly how flexible the tape is.
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denis
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Re: Please Rewind
« Reply #3 on: May 20th, 2008, 9:02pm » |
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William,
For someone who is not physics major, you did well indeed. Your approach is correct and well thought out.
You just need a small adjustment: recheck the moment of inertia formula you used for the disk. The formula you used is for a cylindrical shell with open ends which assumes the shell thickness is negligible (which is not the case for the solid disk).
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| « Last Edit: May 20th, 2008, 9:41pm by denis » |
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denis
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Re: Please Rewind
« Reply #4 on: May 20th, 2008, 9:04pm » |
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on May 20th, 2008, 9:00pm, temporary wrote:| I think you would need the density to know exactly how flexible the tape is. |
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You can assume no friction or resistance arises from the tape unrolling other than inertial forces.
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| « Last Edit: May 20th, 2008, 9:50pm by denis » |
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william wu
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Re: Please Rewind
« Reply #5 on: May 20th, 2008, 10:15pm » |
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Thanks denis!
Scaling factor corrections:
- I = 1/2 MR^2, and thus a = 2/3 g sin
- t = { 2L/a } = { (3L/g) csc }
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| « Last Edit: May 20th, 2008, 10:15pm by william wu » |
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temporary
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Re: Please Rewind
« Reply #6 on: May 20th, 2008, 10:26pm » |
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on May 20th, 2008, 9:04pm, denis wrote:
You can assume no friction or resistance arises from the tape unrolling other than inertial forces. |
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But that doesn't give me the density.
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ThudnBlunder
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Re: Please Rewind
« Reply #7 on: May 21st, 2008, 3:48am » |
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on May 20th, 2008, 10:26pm, temporary wrote:
But that doesn't give me the density. |
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You must be a member of DENSA.
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denis
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Re: Please Rewind
« Reply #8 on: May 21st, 2008, 8:40am » |
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on May 20th, 2008, 10:15pm, william wu wrote:Thanks denis!
Scaling factor corrections:
- I = 1/2 MR^2, and thus a = 2/3 g sin
- t = { 2L/a } = { (3L/g) csc }
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Yep!
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