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Easy Rotational Motion Question

  • Thread starter GreenPrint
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  • #1
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Homework Statement



A grindstone has a moment of inertia of 1.6*10^-3 kg m^2. When a constant torque is applied, the flywheel reaches an angular velocity of 1200 rev/min in 15 s. Assuming it started from rest, find (a.) the angular acceleration; (b.) the torque applied; (c.) the angle turned through the 15 s; (d.) the work W done on the flywheel by the torque.
Answer: (a.) 8.38 rad/s^2 (b.) 0.0134 N m (c.) 942 rads (d.) 12.6 J


Homework Equations





The Attempt at a Solution



I'm having some issues with part (a).
I used ω = ω_0 + alpha*t
and solved for alpha
alpha = (ω-ω_0)/t
alpha = (1200 rev/min * (2 pi)/rev * 60 min/s)/15 s
I get alpha is about 30,159 1/s^2

I don't see what I'm doing wrong. Thanks for any help which you can provide me.
 

Answers and Replies

  • #2
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For part (a), you're on the right track. Because the applied torque is constant, this means that angular velocity increases linearly. This should allow you to work backwards using [itex]\omega_{0} , \omega_{f} ,[/itex] and [itex]\Delta t[/itex].

Your problem is in the calculations. You've got 60min/s, which is incorrect. It should be 1min/60s. See if that puts you closer to your expected value and then substitute it back in to check the validity of your answer by solving for [itex]\omega_{f}[/itex] from [itex]\omega_{0} , \delta t , [/itex] and [itex]\alpha[/itex].
 
  • #3
Doc Al
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alpha = (1200 rev/min * (2 pi)/rev * 60 min/s)/15 s
You messed up the conversion from rev/min to rad/sec.
 
  • #4
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ah i feel dumb thanks
 
  • #5
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for part (c) I used
θ = θ_0 + ω_0*t + (alpha*t^2)/2
θ = 1200 rev/min * (2pi)/rev * min/(60 s) * 15 s + (8.38 1/s^2 * (15 s)^2)/2 ≈ 2,828
which I guess is wrong because the answer is 942
I don't see what I'm doing wrong here, I checked that the units canceled out
 
  • #6
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Well the answer you derived is just about 3x the answer the book gives you. See if you can find where you might have accidentally multiplied something by 3 (or got a constant 3x its correct value or forgotten to divide by 3. Your approach looks right as best I can tell so just double check your inputs and whatnot.
 
  • #7
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I have checked it and can't seem to find anything wrong with it, yet the answer is wrong.
 
  • #8
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Hmm. I can't think of what might be the issue. Maybe try integrating over the specified time interval and using your known initial values and whatnot? I don't know why that would change since your alpha is constant, but it couldn't hurt. Beyond that, I can't think of anything other than just making sure you're using the right values everywhere.
 
  • #9
Doc Al
Mentor
44,867
1,114
for part (c) I used
θ = θ_0 + ω_0*t + (alpha*t^2)/2
θ = 1200 rev/min * (2pi)/rev * min/(60 s) * 15 s + (8.38 1/s^2 * (15 s)^2)/2 ≈ 2,828
which I guess is wrong because the answer is 942
I don't see what I'm doing wrong here, I checked that the units canceled out
For one thing, it starts from rest so ω_0 = 0.
 
  • #10
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Ya I don't think integrating is needed because the angular acceleration is constant. And ya I checked to make sure all my values were correct and they appear to be. Do you think the supplied answer is wrong?
 
  • #11
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For one thing, it starts from rest so ω_0 = 0.
thanks
 
  • #12
81
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Derp. I can't believe I missed that too!
 

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