PErfIS A 40- 30 sec kg child rides on a Ferris Wheel of radius 20 m. The time fo
ID: 1884782 • Letter: P
Question
PErfIS A 40- 30 sec kg child rides on a Ferris Wheel of radius 20 m. The time for the Ferris Wheel to make a full rotation is at ehe forces acting on the child at the top of the ride? Show these forces in the figure to the right. (b) Which force is the greatest? What is the direction of the net force? (c) What is the magnitude and direction of the acceleration of the child at the top? (d) Apply Newton's 2 law to determine the normal force exerted by the seat on the child at the top. The normal force is the 'effective' weight of the child. How does this compare with the child's actual weight?Explanation / Answer
Velocity of child
V= 2*3.14* 20/ 30= 4.187 m/s
a)
There are two forces that are acting on the child, gravitational force and centripital force.
Fg= mg= 40*9.8= 392 N
Fc = mv^2/r= 40*(4.187^2)/20= 35.06 N
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b)
Gravitational force is greater than centripital force
Net force directed to the down direction.
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c)
a= F/m= (392-35.06)/40= 8.92 m/s^2
Directed downwards
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d)
Normal force,
N= 8.92*40= 356.94 N
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e) F= 356.94/392= 0.91 * weight
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Plz post other problems seprsepera
Comment in case any doubt.. good luck
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