A popular circus act features daredevil motorcycle riders encased in the \"Globe

Question

A popular circus act features daredevil motorcycle riders encased in the "Globe of Death" (see the figure below), a spherical metal cage of diameter 17 ft. Assume a speed of 24mi/h for both tricks.

(a) A rider of mass 72 kg on a 125-cc (95-kg) motorcycle keeps his bike horizontal as he rides around the "equator" of the globe. What coefficient of friction is needed between his tire and the cage to keep him in place?


(b) How many loops will the rider make per second?
loops/s

(c) The same rider performs vertical loops in the globe. What force does the cage need to withstand at the top and the bottom of the rider's loop?

Explanation / Answer

v= 24mi/h =10.73m/s

d= 17ft

r=d/2= 8.5ft = 2.59m

a) Applying Newton’s second law horizontally (into the page)

Fn = Fc = mv^2/r = (167*10.73^2)/2.59 = 7423.63 N

Applying Newton’s second law vertically,

Ff – mg= 0

Ff=mg

us*Fn= mg

us*7423.63 = 167*9.8   => us=0.22

b) w= v/r = (10.73m)/( 2.59) = 4.14 rad/s = 39.53 rev/s = 39.53loops/s

c)

At the top,

Newton’s second law along vertical,

-Fn-mg= - Fc

Fn+mg=Fc

Fn= Fc – mg = mv^2/r – mg = (167*10.73^2)/2.59 - 167*9.8 = 5787.03 N

Thus Ftop= Fn= 5787.03 N

At the bottom,

Newton’s second law along vertical,

Fn-mg= Fc

Fn= Fc+mg = (167*10.73^2)/2.59 + 167*9.8 = 9060.23 N= Fbottom

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