Angular acceleration of a flywheel

May 2019
50
1
Mumbai (Bombay),Maharashtra State,India
A flywheel is brought from rest up to a speed of 1500 rpm in 1 minute. What is the average angular acceleration, α?

There are four choices available to select.

1) 333 radians per second.

2)157 radians per second.

3)2.617 radians per second per second.

4)5.23 radians per second per second.

5) doubtful

I know the formulas for computing final angular position and final angular velocity. Suppose $\theta_0$= initial angular position, $\theta$= final angular postion,$\omega_0$=initial angular velocity,$\omega$=final angular velocity,$\alpha$= angular acceleration, t=time. then $1)\theta=\theta_0+\omega_0*t+\frac12\alpha*t^2$,

$2)\omega=\omega_0+\alpha*t$,

$3)\omega=\theta_0+\frac12(\omega_0+\omega)*t$,

$4)\theta=\theta_0+\omega*t-\frac12\alpha*t^2$

How to use one of these formula here? I know by changing the position of $\alpha$ we can compute it. I tried for it, but i didn't get any of the answer provided by the author.

If any member knows the answer, he may reply with correct answer.
 
Last edited:

romsek

MHF Helper
Nov 2013
6,725
3,030
California
They want average angular acceleration.

$a = \dfrac{\dot{\theta}_f - \dot{\theta}_i}{\Delta t} = \dfrac{\frac{1500}{60}2\pi - 0}{60} = \dfrac{50\pi}{60} = \dfrac 5 6 \pi ~rad/s^2 \approx 2.617 ~rad/s^2$
 
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May 2019
50
1
Mumbai (Bombay),Maharashtra State,India
They want average angular acceleration.

$a = \dfrac{\dot{\theta}_f - \dot{\theta}_i}{\Delta t} = \dfrac{\frac{1500}{60}2\pi - 0}{60} = \dfrac{50\pi}{60} = \dfrac 5 6 \pi ~rad/s^2 \approx 2.617 ~rad/s^2$
Hello,

What is rpm? Is it rounds per minute or radians per minute?
 

romsek

MHF Helper
Nov 2013
6,725
3,030
California
rpm is revolutions per minute

1 revolution is 2 pi radians
 
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