Question

A solid brass ball of mass 22 g and radius 8.51 mm rolls smoothly along a loop-the-loop track when released from rest from a tall ramp leading to the loop-the-loop. The circular loop has a 17 cm radius.

(a) What is the minimum height from which you can release the ball so that it will go around the loop-the-loop without falling off?

Answer 1

To determine the minimum height from which the solid brass ball should be released in order to go around the loop-the-loop without falling off, we can use the conservation of mechanical energy. The minimum height is 3/5 times the radius of the loop-the-loop.

Step-by-step explanation:

To determine the minimum height from which the solid brass ball should be released in order to go around the loop-the-loop without falling off, we need to consider the conservation of mechanical energy. At the bottom of the loop, all the potential energy is converted to kinetic energy. Using the equation for the conservation of mechanical energy, we can solve for the height:

mgh = (1/2)mv^2 + (1/2)Iω^2

Where m is the mass of the ball, g is the acceleration due to gravity, h is the height, v is the velocity, I is the moment of inertia, and ω is the angular velocity. Since the ball is rolling smoothly, we can use the relationship between linear and angular velocity: v = ωr, where r is the radius of the ball.

Plugging in the given values and solving for h:

mgh = (1/2)(mv^2) + (1/2)(2/5)(mr^2)v^2

h = (3/5)r

Therefore, the minimum height from which the ball should be released is 3/5 times the radius of the loop-the-loop.