What is the minimum speed required at the bottom to perform a vertical loop, if the radius of the depth well in a circus is 25m.
Please solve the numerical
" the radius of the depth well in a circus is 25m"
We take this to mean the radius of the vertical loop and solve.
Let the minimum speed be vmin, then the kinetic energy is, mvmin2/2.
If we consider the bottom point as reference for potential energy zero, then potential energy at highest point is,
mg(2R) = 2mgR
At the highest point the centripetal force is provided by weight of the performer and bike,
mv2/R = mg
v2 = Rg
Hence KE at the highest point is mRg/2 and potential energy 2mgR
So total energy = mRg/2 + 2mgR = 5mgR/2,
This must be equal to the total energy at bottom, KE + PE = mvmin2/2 + 0
mvmin2/2 = 5mgR/2
vmin = (5gR)