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If it takes 3s for a modern DVD player to stop a DVD with a rotational speed of 7490 rpm, what is the DVD’s average rotational acceleration?

User Shonique
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1 Answer

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We are asked to determine the average rotational acceleration. To do that we will use the following formula:


\alpha=(\omega_f-\omega_0)/(t)

Where "alpha " is the rotational acceleration and "omega" is the angular velocity, and "t" is time. Since the DVD stops, this means that the final angular velocity is zero. Replacing the values we get:


\alpha=(0-7490rpm)/(3s)

Solving the operation:


\alpha=2496.67(rpm)/(s)

We can convert this to radians per second squared using the following conversion factor:


1\text{rpm}=2\pi(rad)/(\min)
1\min =60s

Converting the units we get:


\alpha=2496.67\text{rpm}*\frac{2\pi(rad)/(\min)}{1\text{rpm}}*(1\min )/(60s)

Solving the operation:


\alpha=261.45\text{ rad/s}

User Harmonickey
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