Question

1.      A motorcycle is traveling at a speed of 10 m/s over a speed bump that has a radius of curvature of 0.2 m. Does the motorcycle jump off of the bump or stay on the ground?

Answer 1

The extremeum case is that motorcycle will jump that is when the centrifugal force will exceed the gravity

mg=mv^2/r

g<v^2/r,the motor cycle will jump

here

V^2/r=100/0.2=500>9.8

so motor cycle will jump.

But for accurate approximation the friction should be taken care

Answer 2

YES THE MOTOTRCYCLE WILL JUMP OFF THE BUMP BECAUSE IT IS MOVING AT 10m/s AND BUMP RADIUS OF CURVATURE IS ONLY 0.2m . THE CENTRIFUGAL FORCE WILL HELP THE BIKE TO REMAIN ON THE BUMP AND HENCE IT WILL JUMP OFF THE BUMP

for this the force of gravity should be less then the force exerted by bike

so force exert by gravity   f=mg

force of kinetic energy    \(KE=1/2M V^{2}\)   

ON COMPARING KE(KINETIC ENERGY)=FORCE BY GRAVITY

  mg=    \(1/2 M V^{2}\)   

     

CLEARLY g IS SMALLER THAN FORCE EXRERTED BY BIKE

HENCE IT WILL JUMP OFF THE BUMP