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
E) 0

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 
E) 0

Quizplus · Accepted Answer

The smallest value of y such that the object will slide without losing contact with the track is when the normal force is zero at the top of the track. At this point, the only force acting on the object is gravity. Therefore, the centripetal force required to keep the object on the track is equal to the force of gravity. 
$m \frac{v^2}{R} = mg$
$v^2 = gR$
$v = \sqrt{gR}$
At the top of the track, the velocity of the object is zero (since it momentarily stops moving). Therefore, the total energy of the object is equal to its potential energy at the top of the track. 
$mg(2R-y) = \frac{1}{2}mv^2$
$mg(2R-y) = \frac{1}{2}m(gR)$
$2R-y = \frac{1}{2}R$
$y = \frac{3}{2}R$
However, this is the minimum height required for the object to make it around the loop. The question asks for the minimum height required for the object to make it around the loop without losing contact with the track. Therefore, we must subtract off the radius of the object to get the answer. 
$y_{min} = \frac{3}{2}R - R = \frac{1}{2}R$
Therefore, the answer is B.

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