04%) Problem 7: An elementary student of mass m 38 kg is swinging on a swing. Th

Question

04%) Problem 7: An elementary student of mass m 38 kg is swinging on a swing. The length from the top of the swing set to the seat is L = 4.4 m. The child is attempting to swing all the way around in a full circle ©theexpertta.com 33% Part (a) What is the minimum speed, in meters per second, the child must be moving with at the top of the path in order to make a full circle? Grade Summarv Deductions Potential 0% 100% Submissions Attempts remaining: 3 (10% per attempt) detailed view sin coS tan HOME cotan) asin() acos) atan)acotanO sinho O tanhO cot Degrees Radia 123 0 cosh cotanh() END ns BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: Feedback: 33% Part (b) Assuming the child is traveling at the speed found in part (a), what is their apparent weight, W, in newtons, at the top of their path? (At the top, the child is upside-down.) 33% Part (c) What is the child's apparent weight, in newtons, at the bottom of their path if they have the velocity from part (a) at the top?

Explanation / Answer

at the top of the path , in order to make a full circle, the tension in the rope should
be higher than 0, atleast.

let tension be T.

then writing force balance equation:

T+m*g=m*v^2/r

where v=speed

r=radius of the path=length of the rope=4.4 m

so putting T=0 so that we can find the minimum value for v,

we get v=sqrt(r*g)=6.567 m/s

part b:

centripetal acceleration=v^2/r=g

so net acceleration=0

hence apparent weight will be 0.

part c:

let at the bottom of the path, speed is v.

take bottom point in the loop to be potential referrrence point

i.e. at the bottom point of the loop, potential energy=0 J

conserving energy:

initial kinetic energy=final kinetic energy+final potential energy

==>0.5*m*v^2=0.5*m*6.567^2+m*g*2*4.4

==>v=14.683 m/s

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