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AH

Carnegie Mellon University

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Problem 96

A child places a picnic basket on the outer rim of a merry-go-round that has a radius of 4.6 $\mathrm{m}$ and revolves once every 30 $\mathrm{s}$ .

(a) What is the speed of a point on that rim? (b) What is the lowest

value of the coefficient of static friction between basket and

merry-go-round that allows the basket to stay on the ride?

Answer

(a) $v=0.96 \mathrm{m} / \mathrm{s}$

(b) $\mu_{s}=0.02$

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## Discussion

## Video Transcript

so here the for party the distance traveled in one revolution would simply be equal to the circumference or we consider it. Their conference would be too pi r, which would be equal to two pi times 4.6 meters, which is approximately 29 meters. So that constant velocity would simply be 29 meters divided by 30 seconds. This is giving us 300.96 meters per second. Approximately we can apply for part B. We can apply Newton's second law and we can say that the force of friction, the static off the static force of friction would be equal to the and the squared over r. And this would be equal to M times 0.20 and we're going to in this case, we know that in this situation, the force normal is equaling mg. And this is simply due to the application of Newton's second law in the ex direction. So if we were to, I quit this Ah, we can say that the force of friction static with the equal to Mu S o m g. This would be equal to M v squared over r and essentially, uh, this would be equal to point 20 times. Um, now EMS cancel out and we're getting that. The coefficient of static friction would simply be equal to a 200.20 divided by 9.80 meters. Rather weak unjustly. 9.0. Because hear cowfish. Of course, the coefficient of static friction would be unit list, and we find the coefficient of static friction to be equal to 0.0 to 1. So this would be our answer for part B. That is the end of the solution. Thank you for watching.

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