# angular acceleration

#### star89

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?

#### HallsofIvy

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.

• 2 people

#### Woody

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)

• 2 people

#### star89

aaah okey Thank you so much for the help #### CaptainMarvel199

Well what you have found is wrong.The object accelerates at a steady rate so during the second rotation, it would have more speed so it would have bigger frequency(rotations/second).

#### topsquark

Forum Staff
Well what you have found is wrong.The object accelerates at a steady rate so during the second rotation, it would have more speed so it would have bigger frequency(rotations/second).
Ummm... What the OP calculated was incorrect, but the later posts fixed that.

Please use the "quote" feature or somehow otherwise let us know which post you are commenting on. That will clear up the confusion.

-Dan