Question

A student of weight 627 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force on the student from the seat is 563 N. (a) What is the magnitude of at the lowest point? If the wheel's speed is doubled, what is the magnitude FN at the (b) highest and (c) lowest point?

Answer 1

a)Net force = 691 N

b)Net force = 371 N

c)Net force = 883 N

Step-by-step explanation:

Weight= 627 N

At the highest point

Net force = Centripetal force + weight

At the highest ,Net force = 563 N.

So

Centripetal force(Fc) = 563 - 627 N

Fc= - 64 N

Negative sign means force is upward.

But at the lowest point

Fc= 64 N

Net force = Centripetal force + weight

Net force = 64 + 627

Net force = 691 N

When speed will become double

We know that

`F_c=m\omega ^2r`

Then the force Fc will become 4 times.

Fc'= 4 Fc

F'c= 4 x 64 N

F'c= 256 N

So new Centripetal force'= 256 N

At the highest point

Net force = Centripetal force' + weight

At the highest point

Net force =-256 + 627

Net force = 371 N

At the lowest point

Net force = Centripetal force' + weight

Net force = 256 + 627

Net force = 883 N