- #1

Trentonx

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## Homework Statement

The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 10 m, as shown in the figures. All surfaces are frictionless. In order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.8 times the weight of the cart. You may neglect the size of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.)

a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?

## Homework Equations

a_c = (V^2)/R

F = ma

V = sqrt(2gh)

## The Attempt at a Solution

Top of circle

a_c equals the sum of the other forces, so

4900+400=500((v^2)/10)

V = 10.295 m/s, a_c = 10.6 m/s/s

height to have given velocity - 5.41m

This is where I don't know. How does having centripital acceleration help in finding the height at all? Does it?