Need confirmation with my way of solution.

1. Dec 21, 2004

Sanosuke Sagara

I am facing problem on question regarding rotational motion.I have my solution and my question in the attachment that followed.

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• rotational.doc
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2. Dec 21, 2004

mjfairch

Firstly, let me comment that you don't need to do all that extra work converting everything into rad/s. You can simply leave things in rev/min, as long as you're consistent. The 1.3 answer is the time in minutes for it to come to a rest, not the number of revolutions. Assuming constant angular acceleration:

$$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{120\textrm{rev/min} - 150\textrm{rev/min}}{1/3\textrm{min}} = -90 \frac{\textrm{rev}}{\textrm{min}^2}$$

where the negative rate makes sense as it's slowing down. Now, to compute the time it takes to come to a stop, assume it's angular position and time is 0 after then end of that first 20 seconds. Then,

$$\omega = \omega_0 + \alpha t \implies 0 = 120\frac{\textrm{rev}}{\textrm{min}} + -90\frac{\textrm{rev}}{\textrm{min}^2}t \implies t=\frac{120\frac{\textrm{rev}}{\textrm{min}}}{-90\frac{\textrm{rev}}{\textrm{min}^2}} \implies t=1.\overline{3} \textrm{min}$$

Now, if it takes $1.\overline{3}$minutes to stop from an initial angular velocity of 120 rev/min at a constant acceleration of -90 rev/min/min, then how many revoutions will it take to stop?

3. Dec 21, 2004

Sanosuke Sagara

Confirm with the number of revolutions made

I have my calculation in the attachment that followed.Do comment if I have any calculation error.

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• Doc3.doc
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4. Dec 21, 2004

dextercioby

Nope,it's wrong.The last 2n3 lines of your document are wrong.Using these special units (revolutions/minute) will defintely enable u to get to the answer right away.That's because the result for $\theta$ is the final answer,as it already put in terms of revolutions.Atually the angle is 80revolutions*2\pi radians/revolution approx 500 radians.
So the final answer to yor question is 80 revolutions.This is the number the problem is waiting from u.

Daniel.

5. Dec 22, 2004

Sanosuke Sagara

Thanks for your confirmation and mark with my mistake.Now I have understand what the question want.Thanks again.