How Long Did It Take for the First Revolution of a Computer Disk Drive?

AI Thread Summary
The discussion revolves around calculating the time taken for a computer disk drive to complete its first revolution, given that it takes 0.680 seconds for the second revolution. Participants clarify that the drive starts from rest with constant angular acceleration, leading to the use of rotational kinematics equations. The key point is recognizing that the time for each revolution is not the same due to the constant acceleration. The challenge lies in correctly applying the equations of motion for rotational systems to find the time for the first revolution. Ultimately, the discussion emphasizes understanding the relationship between angular displacement, time, and acceleration in this context.
sophixm
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Homework Statement


A computer disk drive is turned on starting from rest and has constant angular acceleration.
If it took 0.680s for the drive to make its second complete revolution, how long did it take to make the first complete revolution? (t=?)

Homework Equations


Θ=(ct^2)/2 (i think)

The Attempt at a Solution


So, since the angular acceleration is constant, I'm assuming α=c (c being a constant).
So then the intergral with respect to time, would be ω=ct, and then Θ=(ct^2)/2
I tried solving for c using 4π(2 revolutions)=(c(.68^2))/2, and got 54.4 for c, and used that to solve for time of one revolution, but this is wrong. I guess what the problem is saying is that it took 0.680 seconds to go from 2π to 4π, so I am stuck
 
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sophixm said:

Homework Statement


A computer disk drive is turned on starting from rest and has constant angular acceleration.
If it took 0.680s for the drive to make its second complete revolution, how long did it take to make the first complete revolution? (t=?)

Homework Equations


Θ=(ct^2)/2 (i think)

The Attempt at a Solution


So, since the angular acceleration is constant, I'm assuming α=c (c being a constant).
So then the intergral with respect to time, would be ω=ct, and then Θ=(ct^2)/2
I tried solving for c using 4π(2 revolutions)=(c(.68^2))/2, and got 54.4 for c, and used that to solve for time of one revolution, but this is wrong. I guess what the problem is saying is that it took 0.680 seconds to go from 2π to 4π, so I am stuck

Take a step back. If something accelerates from rest and takes ##t_1## seconds to travel an angle ##\theta##, then how long does it take to travel a further ##\theta##?
 
sophixm said:
the problem is saying is that it took 0.680 seconds to go from 2π to 4π
Quite so. Are you familiar with the SUVAT equations for constant linear acceleration? The equations for constant rotational acceleration are strongly analogous.
 
haruspex said:
Quite so. Are you familiar with the SUVAT equations for constant linear acceleration? The equations for constant rotational acceleration are strongly analogous.
I believe so, ones like Θ=Θinitial+ωinitial(t)+(1/2)αt^2 right? I've taken some time to look at those but I'm still stuck. I'm familiar with the method of subbing in when you have two unknown variables, but i feel like I'm missing more that that. All I know is that at a time it has gone 2π radians, and 0.680 seconds from that time it has gone another 2π radians. I feel like I'm just missing something
 
sophixm said:
I believe so, ones like Θ=Θinitial+ωinitial(t)+(1/2)αt^2 right? I've taken some time to look at those but I'm still stuck. I'm familiar with the method of subbing in when you have two unknown variables, but i feel like I'm missing more that that. All I know is that at a time it has gone 2π radians, and 0.680 seconds from that time it has gone another 2π radians. I feel like I'm just missing something
Let ti, i = 0, 1, 2, be the time at which the ith rotation is completed (t0=0, of course). Likewise three 'distances' ##\theta_0, \theta_1, \theta_2##, one acceleration. What equations can you write down involving these?
 
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