Question

In an amusement park ride called The Roundup, passengers stand inside a 19.0 m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane

a)Suppose the ring rotates once every 4.50 s . If a rider's mass is 55.0 kg , with how much force does the ring push on her at the top of the ride?

b)Suppose the ring rotates once every 4.50 s . If a rider's mass is 55.0 kg , with how much force does the ring push on her at the bottom of the ride?

c)What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Answer 1

Diameter of the rotating ring D = 19 m

Radius of the rotating ring r = D / 2 = 19 / 2 = 9.5 m

Mass of the rider M = 55 kg

Time taken for one revolution is T = 4.5 s

Only forces that act on the rider are the force of gravity W and a normal force N

This normal force is equal to the force applied by the ring on the rider.

Let a be the net acceleration of the rider.

   be the angular speed of the rider.

A) At the top , both the normal force and weight act downwards.

- N - W = m a ........... ( 1 )

But a = - ² r

So - N - W = - M ² r

N = M ² r - W

= M ² r - Mg

= { M × ( 2 π / T )² × r } - Mg

= { 55 × ( 6.28 / 4.5 )² × 9.5 } - ( 55 × 9.8 )

= 1017.60 - 539

= 478.60 N

B) At the bottom , normal force acts upwards and force of gravity downwards.

N - W = Ma ........... ( 2 )

Here a = + ² r

N = Ma + Mg

= { 55 × ( 2 π / 4.5 )² × 9.5 } - ( 55 × 9.8 )

= 1017.60 + 539

= 1556.60 N

C ) To calculate the longest time period , we must set the normal force to zero in the equation ( 1 ) so that a = g.

W = Ma

Mg = Ma

g = a

g =  ² r

g = ( 2 π / T )² × r

T² = 4 π² r / g

= ( 6.28 × 6.28 × 9.5 ) / 9.8

T² = 38.23

T = 38.23

T = 6.18 seconds

So the required time period is T = 6.18 s