A small object of mass m starts at rest at the position shown and slides along the frictionless loop-the-loop track of radius R.What is the smallest value of y such that the object will slide without losing contact with the track?
A)R/4
B)R/2
C)R
D)2R

Question

A small object of mass m starts at rest at the position shown and slides along the frictionless loop-the-loop track of radius R.What is the smallest value of y such that the object will slide without losing contact with the track? 
A)R/4 
B)R/2 
C)R 
D)2R

Quizplus · Accepted Answer

For the object to remain in contact with the track, the net force on it must be directed towards the center of the loop. At the top of the loop, the net force is $mg-n$, where $n$ is the normal force, and at the bottom of the loop, the net force is $mg+n$. For the object to not lose contact with the track, the normal force must be greater than or equal to zero. Therefore, we need to find the minimum height at which the normal force becomes zero.

At the top of the loop, for the object to remain in contact with the track, we need:
$mg-n \geq 0$
$n \leq mg$

At the bottom of the loop, for the object to remain in contact with the track, we need:
$mg+n \geq 0$
$n \geq -mg$

Combining the two conditions, we have:
$-mg \leq n \leq mg$

The normal force is maximum at the bottom of the loop and minimum at the top of the loop. When the normal force at the top of the loop becomes zero, the object will start losing contact with the track. Setting $n = 0$ in the first condition, we have:
$mg-n = mg = \frac{mv^2}{R}$
Solving for $v$, we get:
$v = \sqrt{gR}$

At the top of the loop, the object's potential energy is maximum and therefore its total energy is also maximum. At the bottom of the loop, the object's potential energy is minimum and therefore its total energy is also minimum. Using conservation of energy, we can relate the object's minimum height at the top of the loop with its maximum height at the bottom of the loop:
$mgy_{min} + \frac{1}{2}mv_{min}^2 = mgy_{max} + \frac{1}{2}mv_{max}^2$
$mg(y_{min}-y_{max}) = \frac{1}{2}m(gR + gR)$
$y_{min}-y_{max} = R$
$y_{min} = y_{max} + R$

Therefore, the minimum height at which the object will slide without losing contact with the track is:
$y_{min} = R$
Thus, the answer is $\boxed{	extbf{(C)}}.$

A Small Object of Mass M Starts at Rest at the Position