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A 60 kg physics student is working to pay his tuition fees by performing in a travelling carnival. He rides a motorcycle inside a hollow steel sphere of radius 15m. The surface of the sphere is full of holes so the audience can see in. After gaining sufficient speed, he travels in a vertical circle. What minimum velocity must he have at the top of the circle if the tires of the motorcycle are not to lose contact with the steel sphere?
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The mass cancels in the equations so you can use the acceleration rates only.
The centripetal acceleration at the top must equal or exceed that of gravity ( say 9.8 m/s^2 )
So:
9.8 = v^2 / 15
So:
v = square root ( 15 * 9.8 )
v = 12.124 m/s
Note :
Strictly speaking the radius at the centre of gravity of the bike + rider should be used.
the net force acting on him at the top is his weight + the reaction force of the sphere pushing down. since he is going at the minimum velocity to just maintain contact with the sphere, the reaction force is zero. therefore the only force acting on him is due to gravity.
so acceleration is 9.8m/s^2
and radius is 15m
a = v^2 / r
v^2 = ar
v^2 = 9.8 * 15
v^2 = 147
v = 12.12 m/s