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Angular speed of merry-go-round

  • Thread starter cdotter
  • Start date
  • #1
305
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Homework Statement


A playground merry-go-round has a radius 2.40 m and a moment of inertia 2100 kgm^2 about a vertical axle through its center, and it turns with negligible friction.

A child applies an 18.0 N force tangential to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s interval?


Homework Equations


[tex]\vec \tau = \vec r \times \vec F[/tex]

The Attempt at a Solution



I have absolutely no idea what to do. I know the answer is .309 rad/s. From playing around with the numbers I know that (2.40 m * 18.0 N * 15.0 s)/2100 kg*m^2 = 0.309 rad/s, but I don't know why. I can't find any sort of relationship between what I have and what I need.
 

Answers and Replies

  • #2
305
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Immediately after clicking post, it clicked in my mind. Wow. For anyone else who is having trouble with this problem:

[tex]
\begin{array}{l}
\vec \tau = \vec r \times \vec F\\
\sum \vec \tau = I \vec \alpha\\
\omega_f = \omega_i + \alpha t\\\\
\vec \tau = (2.40 m)(18.0 N)(sin 90) = 43.2 N \cdot m\\
\sum \vec \tau = 43.2 N \cdot m = (2100 kg \cdot m^2)(\alpha)\\
\Rightarrow \alpha = 0.021 rad/s^2\\
\omega_f = 0 + (0.021 rad/s^2)(15.0 s)\\
\Rightarrow \omega_f = 0.309 rad/s
\end{array}
[/tex]
 
Last edited:

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