In an amusement park ride called The Roundup, passengers stand inside a 16.0 m -

Question

In an amusement park ride called The Roundup, passengers stand inside a 16.0 m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.70 s . If a rider's mass is 51.0 kg , with how much force does the ring push on her at the top of the ride?

Suppose the ring rotates once every 4.70 s . If a rider's mass is 51.0 kg , with how much force does the ring push on her at the bottom of the ride?

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

Explanation / Answer

diameter of ring d=16m

radius r=d/2 =8m

mass of the rider m=51 kg

time period T=4.7 sec

w=2pi//T =2pi/4.7 =1.34 rad/sec

a)

at the top of the ride,

force N=m*w^2*r- m*g

=51*(1.34^2*8-9.8)

=232.8 N

b)

at the bottom of the ride,

force N=m*w^2*r+ m*g

=51*(1.34^2*8+9.8)

=1232.4 N

c)

if the rider is falling

N=0

m*w^2*r=mg

w=sqrt(g/r)

2pi/T=sqer(g/r)

===>

longest rotation period is,

T=2pi*sqrt(r/g)

=2pi*sqrt(8/9.8)

=5.68 sec

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.