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?

Question

Quizplus · Accepted Answer

The minimum height at which the object will slide without losing contact with the track is when the normal force is zero at the top of the loop. At this point, the only force acting on the object is gravity pulling it towards the center of the loop. Using conservation of energy, we can set the potential energy at the starting point equal to the kinetic energy at the top of the loop:

$mgy = \frac{1}{2}mv^2$

where y is the height of the starting point above the top of the loop (measured along the y-axis), v is the speed of the object at the top of the loop, and g is the acceleration due to gravity.

The speed of the object at the top of the loop can be found using conservation of energy again. At the starting point, the only energy is potential energy, so

$mgy = \frac{1}{2}mv_0^2$

where $v_0$ is the speed of the object at the starting point. Since the track is frictionless, the total mechanical energy of the object is conserved, so we can write

$mgy = \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

Combining these equations, we get

$y = \frac{v^2}{2g} = \frac{R}{2}$

So the smallest value of y that will allow the object to slide without losing contact with the track is y = R/2. The correct answer is therefore B.

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