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# ssm A pitcher throws a curveball that reaches the catcher in 0.60 s. The ball curves because it is spinning at an average angular velocity of 330 rev/min (assumed constant) on its way to the catcher’s mitt. What is the angular displacement of the baseball (in radians) as it travels from the pitcher to the catcher?

## 21 $\mathrm{rad}$

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##### Andy C.

University of Michigan - Ann Arbor

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##### Jared E.

University of Winnipeg

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### Video Transcript

What important thing to keep in mind with this problem is that when we have rotational motion, that's coupled with translational motion. The pitcher throws the ball and it moves sideways. But it's also rotating Is that that translational motion and the rotation are completely separate from each other. A lot like when we do to the emotion problems where motion along each dimension is independent of each other. Like if we have ballistic trajectories along X and Y, the rotation is completely separate from the translation, so we can ignore the fact that it's a baseball pitcher throwing the ball across some distance and only pay attention to the rotation of the ball. So we're given a time that the ball is rotating of 0.6 seconds, which is good. We want everything to be in S I u nits as much as possible and then the ball has a rotational velocity of 330 revolutions per second. We're sorry revolutions per minute, and that's something that we're gonna need to convert in order to be able to multiply. Ultimately, we want the total angle displaced by the ball, which is gonna be from this equation analysis too, if we had a translational velocity that I had some ex displacement. So it's good to make connections between formulas that we already know in love when we're moving into rotational motion. So we need to convert Omega into as I units in order to move on. So when you convert a value to different units, you need to use unit conversion. So omega we can write as 330 revolutions per minute times. Now we want to multiply by something that is equivalent to one. So it has the same thing on the top and the bottom of a fraction that has the same number on the top of the bottom. So we want to cancel revolutions. So in one revolution we have two pi radiance. So that's just the same, is multiplying by one. Now we also need to convert the minutes into seconds. So we have another fraction and to to cancel out the minutes, we have to have something in the numerator. So one minute is equal to 60 seconds, and now the minutes cancel the Revolution's cancel and we're gonna be left with a number that's ingredients per second. And when we multiply this out 330 times two pi divided by 60. What we end up getting is 34.6 radiance per second. Now, if we multiply this by the time 0.6 seconds, we end up getting a number in radiance. So for our final answer, we get 20 point. It's don't 0.8 radiance. But since we were only given two significant figures in the statement of the problem, we should only write our solution to the same number of significant figures. So this is 21 radiance.

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##### Andy C.

University of Michigan - Ann Arbor

LB
##### Jared E.

University of Winnipeg

Lectures

Join Bootcamp