(10%) Problem 5: A 2675 kg child is riding a playground merry-go-round that is r

Question

(10%) Problem 5: A 2675 kg child is riding a playground merry-go-round that is rotating at 41.1 rev/min. A 25% Part (a) what net force is actin this situation is sometimes called "centripetal force". g on her if she is standing on the merry-go-round 1.05 m from its center in newtons? The net force in Grade Summary potential 100% sind) | cos() tanO | | (! cotan0 asin acos0 atanO acotan sinho cosh0 tanh0 cotanh0 Submissions Attempts remaining: 5 (2 per attempt) view 0 Hint I give up? Feedback: 0% deduction per feedback. what net force (in N) s acting on her if she is standing on an amusement park acting on her if she is standing on an amusement park merry go-round that rotates at 2.4 rpm and 25% Part (b) she is 9.4 m from its center? 25% Part (c) How many times her weight is the force in part (a)? 25% Part (d) - How many times her weight is the force in part (b)?

Explanation / Answer

According to the concept of the circular motion

the centripetal force F=mrw^2

Given that

mass m=26.75 kg

angular velocity w=41.1 rev/min=4.304 rad/s

radius r=1.05 m

now we find the centripetal force

the centripetal force F=26.75*1.05*4.304^2=520.3 N

now we find the net force in the case 2

angular velocity w=2 rpm=0.033 rad/s

the net force F=26.75*9.4*0.033^2=0.274 N

now we find the how many times her weight in the force in part a

N=weight /centripetal force =26.75*9.8/520.3=0.5

now we find the how many times her weight in the force in part b

N=26.75*9.8/0.274=956.8

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