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Old Mar 15th 2019, 02:24 AM   #1
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angular acceleration

Hi I need some help with this:

The grinding disc of a grinding machine starts from a rest and has a constant angular acceleration for the first 10 rotations.

a) It uses 0.87 s on its second rotation. how long does it use on the first rotation?

Her i found out that it use 0.62s on the first rotation!

b) What is the angular acceleration of the disc for the 10 rotations. give the answer in radians / s ^ 2?

is this the right formula to use?

∝=(ω_1-ω_0)/(t_1-t_0 )

Do i need to find out the time it use on the 10 rotasion?
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Old Mar 15th 2019, 11:21 AM   #2
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Let the constant, a, be the constant angular acceleration in rotations per seconds. Then starting from rest, the angular velocity after t seconds is at and the actual rotation is (a/2)t^2. One rotation will be completed after (a/2)t^2= 1 so t= sqrt(2/a). Two rotations will be completed in (a/2)t^2= 2 so t= sqrt(4/a). The time used on the second rotation is sqrt(4/a)- sqrt(2/a)= (2- sqrt(2))/sqrt(a)= 0.87. So sqrt(a)= (2- sqrt(2))/0.87= 0.586/0.87= 0.67 rotations per second per second. As above the time taken for one rotation sqrt(2/a)= sqrt(2/0.67)= 1.73 seconds.
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Old Mar 15th 2019, 11:35 AM   #3
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This can perhaps be approached more as a maths problem.

The acceleration is linear:
standard equation for a linear function is y=mx+c
for this case it starts from rest so c=0
replace m with the rate of acceleration (a),
x with time (t) and y with rotation rate (w)
so w=at

but you want to know the rotation position, so you need to integrate the rate.

p = 1/2at^2
(note p is in radians; I will use "pi" to represent 3.14...)

Now it gets a little but awkward,
you are given the time between p=2pi and p=4pi
(i.e. the time between the end of the first rotation and the end of the second rotation)

so 4pi-2pi = 1/2at2^2 - 1/2at1^2

does this give you enough to carry on from?

(HallsOfIvy types faster than I do)
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Old Mar 16th 2019, 02:40 AM   #4
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aaah okey
Thank you so much for the help
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