Question

In the figure above, a small (treat as a point mass) block of mass m is on top of a frictionless cylinder so that its center of mass is a distance R from the axis of the cylinder. It is given a nudge so that it slides with negligible initial speed down the side of the cylinder. Express your answers in m, R, g and theta. A) When its angular position is theta as shown, what is its speed (assuming that it is still on the cylinder)? B) What is the magnitude of the normal force exerted on the block by the cylinder at this point? C) For what value of theta will the block leave the cylinder?

Answer 1

A) We can use the energy conservation to find the speed at the angular position $\theta$
Initially it has the potential energy m g R
Where R is the radius of the cylinder .We can assume that the initial kinetic energy is zero
At the angular position $\theta$ ,the potential energy is m g R cos $\theta$
The kinetic energy of the mass at this position is (1/2) m v²
Thus equating the energies
m g R = m g R cos $\theta$  + (1/2) m v²
g R (1 - cos $\theta$) = (1/2) v²
v = sqrt (2 g R (1 - cos $\theta$))
B) The magnitude of the normal force at the angular position $\theta$ is
N = m g cos $\theta$
C) The block will leave the cylinder when the centripetal force becomes greater than the component of gravitational force acting on it
m v² / R = m g cos $\theta$_f
where  $\theta$_f is the angle at which the block leaves the cylinder
The velocity v at the position can be found from the energy conservation as done before
v =  sqrt (2 g R (1 - cos $\theta$_f))
Substituting this in the above equation
m x 2 g (1 - cos $\theta$_f) = m g cos $\theta$_f  
2 - 2 cos $\theta$  = cos $\theta$
cos $\theta$   = 2 /3
$\theta$  = 48.2⁰