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Need confirmation with my way of solution.

  1. Dec 21, 2004 #1
    I am facing problem on question regarding rotational motion.I have my solution and my question in the attachment that followed.

    Attached Files:

  2. jcsd
  3. Dec 21, 2004 #2
    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:

    [tex]\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}[/tex]

    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,

    [tex]\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}[/tex]

    Now, if it takes [itex]1.\overline{3}[/itex]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?
  4. Dec 21, 2004 #3
    Confirm with the number of revolutions made

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

    Attached Files:

  5. Dec 21, 2004 #4


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    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 [itex] \theta [/itex] 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.

  6. Dec 22, 2004 #5
    Thanks for your confirmation and mark with my mistake.Now I have understand what the question want.Thanks again.
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