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Rotating Computer Disk Drive

by PirateFan308
Tags: angular acceleration, physics, rotating
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PirateFan308
#1
Nov18-11, 09:06 PM
P: 94
1. The problem statement, all variables and given/known data
A computer disk drive is turned on starting from rest and has constant angular acceleration.
If it took 0.690s for the drive to make its second complete revolution, how long did it take to make the first complete revolution?
What is the angular acceleration?


2. Relevant equations
w2 = w02+2[itex]\alpha[/itex]∅

∅=w0t+(0.5)[itex]\alpha[/itex]t2


3. The attempt at a solution
I'm trying to first find the angular acceleration, and from there I can find out how long it took to make the first complete revolution.

First revolution:
w0=0 t=? [itex]\alpha[/itex]=[itex]\alpha[/itex] w=w ∅=2∏

Second revolution:
w0=w t=0.690s [itex]\alpha[/itex]=[itex]\alpha[/itex] ∅=2∏

Second revolution:

∅=w0t+(0.5)[itex]\alpha[/itex]t2
2∏ = w0(0.690)+(0.5)[itex]\alpha[/itex](0.690)2
[itex]\alpha[/itex]=[itex]\frac{2∏-w0(0.69)}{(0.5)(0.69)^2}[/itex]

First equation:
w2=w02+2[itex]\alpha[/itex](2∏)
([itex]\frac{2∏-w0(0.69)}{(0.5)(0.69)^2}[/itex])([itex]\frac{2∏-w0(0.69)}{(0.5)(0.69)^2}[/itex])=2[itex]\alpha[/itex](2∏)
I rearranged to get:
4∏2=[itex]\alpha[/itex]((4∏(0.69)2)+(π(0.69)2)-((0.5)(0.69)2))
[itex]\alpha[/itex]=5.45
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dynamicsolo
#2
Nov18-11, 10:55 PM
HW Helper
P: 1,663
Quote Quote by PirateFan308 View Post


First revolution:
w0=0 t=? [itex]\alpha[/itex]=[itex]\alpha[/itex] w=w ∅=2∏

Second revolution:
w0=w t=0.690s [itex]\alpha[/itex]=[itex]\alpha[/itex] ∅=2∏

Second revolution:

∅=w0t+(0.5)[itex]\alpha[/itex]t2
2∏ = w0(0.690)+(0.5)[itex]\alpha[/itex](0.690)2
[itex]\alpha[/itex]=[itex]\frac{2∏-w0(0.69)}{(0.5)(0.69)^2}[/itex]
I agree up to this point.


First equation:
w2=w02+2[itex]\alpha[/itex](2∏)
([itex]\frac{2∏-w0(0.69)}{(0.5)(0.69)^2}[/itex])([itex]\frac{2∏-w0(0.69)}{(0.5)(0.69)^2}[/itex])=2[itex]\alpha[/itex](2∏)
I rearranged to get:
4∏2=[itex]\alpha[/itex]((4∏(0.69)2)+(π(0.69)2)-((0.5)(0.69)2))
[itex]\alpha[/itex]=5.45
Your "velocity-squared" equation is all right, and [itex]\omega_{0} = 0 [/itex] . But doesn't that then lead to

[tex]\omega^{2} = 0^{2} + 4 \alpha \pi \Rightarrow \omega^{2} = 4 \pi \cdot \frac{2 \pi - \omega (0.69)}{(0.5)(0.69)^2} , [/tex]

with [itex]\omega[/itex] being the angular speed at the start of the second revolution? (In fact, your expression in parentheses has units of rad/(sec2), so you can't mean to insert it into the first equation for [itex]\omega [/itex] .)

(And I don't see a much tidier way of doing this, in view of the information given and the fact that there are two unknowns.)

BTW, don't forget to answer the first question about how long the first revolution took.
cepheid
#3
Nov19-11, 12:33 AM
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P: 5,197
I figured out a couple of different ways to solve this. They both lead me to the same quadratic equation for t2/t1, where t1 is the time required for the first revolution, and t2 is the time required for the second revolution (which we know to be 0.69 s).

Method 1:

Angular displacement between t = 0 and t = t1:

[tex] \Delta \theta = \frac{1}{2}\alpha (\Delta t)^2 \Rightarrow 2 \pi = \frac{1}{2}\alpha t_1^2 [/tex]

[tex]\Rightarrow \alpha = \frac{4\pi}{t_1^2} [/tex]

Angular displacement between t = 0 and t = t2:

[tex] 4 \pi = \frac{1}{2}\alpha (t_1+t_2)^2 [/tex]

[tex] 8 \pi = \frac{4\pi}{t_1^2} (t_1+t_2)^2 [/tex]

[tex] 2 = \frac{t_1^2 + 2t_1t_2 + t_2^2}{t_1^2} [/tex]

[tex] 2 = 1 + 2\frac{t_2}{t_1} + \left(\frac{t_2}{t_1}\right)^2 [/tex]

Letting x = t2/t1, this becomes the quadratic equation:

[tex] x^2 + 2x - 1 = 0 [/tex]

which you can solve.

Method 2:

Angular displacement between t = 0 and t = t1:

[tex] \Delta \theta = \frac{1}{2}\alpha (\Delta t)^2 \Rightarrow 2 \pi = \frac{1}{2}\alpha t_1^2 [/tex]

[tex]\Rightarrow \alpha = \frac{4\pi}{t_1^2} [/tex]

Angular displacement between t = t1 and t = t2:

[tex] 2\pi = \omega_1 t_2 + \frac{1}{2}\alpha t_2^2 [/tex]

But [itex] \omega_1 [/itex] (the speed at the end of the first revolution) is just [itex] \alpha t_1 [/itex]. Also, if you substitute in the expression for alpha, you get:

[tex] 2\pi = 4\pi \frac{t_2}{t_1} + 2\pi \frac{t_2^2}{t_1^2} [/tex]

[tex] 1 = 2x + x^2 [/tex]

Same quadratic as before. Since you know t2 (it's given), once you solve for x, you have solved for t1. Discard the solution to the quadratic in which the ratio is negative.

PirateFan308
#4
Nov19-11, 09:18 AM
P: 94
Rotating Computer Disk Drive

Wow, thanks guys! That helped a lot!
PirateFan308
#5
Nov19-11, 09:19 AM
P: 94
One question, how did you format your equations so that they looked so much neater and easier to read?
cepheid
#6
Nov19-11, 12:02 PM
Emeritus
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PF Gold
cepheid's Avatar
P: 5,197
Quote Quote by PirateFan308 View Post
One question, how did you format your equations so that they looked so much neater and easier to read?
http://www.physicsforums.com/showthread.php?t=546968


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