Is this Right? Circular Motion Problem

**Pn0yS0ld13r** · January 4th 2009, 04:38 AM

Is my answer right?

A coin with a diameter of 2.4cm is dropped on edge onto a horizontal surface. The coin starts out with an initial angular speed of 18 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular accleration of magnitude 1.9 rad/s^2, how far does the coin roll before coming to a rest?

Using the equation:

$\omega^{2}=\omega^{2}_{0}+2\alpha \Delta \theta$ where $\omega$ and $\omega_{0}$ are the final and initial angular speeds respectively; $\alpha$ is the angular acceleration; and $\Delta \theta$ is the angular displacement.

$0=(18)^{2}+2(-1.9)\Delta \theta$

So $\Delta \theta = 85.26 \text{rad}$ .

Now $\Delta \theta = \dfrac{\Delta s}{r}$ where $\Delta s$ is the change in arc length and r is the radius.

$85.26=\dfrac{\Delta s}{0.012}$

So $\Delta s=1.023 \text{m}$ . Is this right?

**skeeter** · January 4th 2009, 05:20 AM

looks fine to me

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