Circular motion question and problem...

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dedicate

took mcat, now applying
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First of all, would a question like this even be asked on the MCAT? The passage does not go into the details about this question. It's obviously one of those questions attached to a passage that can stand alone. So I will only post the question.

It is from EK, but really EK covers almost nothing about circular motion. In fact, the equation they say you need for this answer was not even in the lesson. They only say you need to know a = v^2/r
and
F = mv^2 / r

Do you need to know more than just that for the MCAT?

Also, how do you solve the question attached?....

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So velocity is m/s, and they tell you the velocity of a point on the surface of the ball is equal to the speed of the air at the surface. You need to find out how fast a single point on the golf ball spins one rotation. This one rotation has a length given by the golf ball's circumference, pi*d. So velocity is simply given as distance/time:circumference/time:circumference*frequency. (The inverse of time is frequency.)

Pi*d*frequency = m/s
 
yessira is right. This problem has nothing to do with circular motion or its given formulas. Other than understanding what frequency is, this problem is more basic mathematics than physics, and yes, knowledge of this kind is fair game for the real MCAT.

The ball spins at 60Hz, which means it makes 60 rotations per second.

1 rotation = circumference of ball = pi*d = (3.14)*(4.3 cm) = approx. 12 cm

60 rotation = (12 cm)*(60) = 720 cm

The closest answer is 8 m/s. Remember we rounded down on both pi and diameter, so the answer will be greater than what we get. And remember to convert centimeters to meters.

note: This happens all the time on the MCAT. Just because a passage discusses a certain topic, it does not mean that all the questions will be answered by formulas from that one topic. A single passage can require knowledge from multiple areas and formulas. Some questions require no prior knowledge at all and can be answered strictly from passage information.
 
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