What Is the Minimum Height for a Cart to Safely Negotiate a Loop-the-Loop?

[gknowels]

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 = 7 m, as shown in the figures. Assume that friction can be ignored. Also assume that, 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. (Note: This is different from the conditions needed to "just negotiate" the loop.) You may treat the cart as a point particle.

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

F=ma
K=1/2mv^2
U=mgh

The Attempt at a Solution

First I found the centripetal acceleration as Acp(m)=.8m+mg
then found velocity from Acp=(v^2)/r
from here i found K=1/2mv^2 and added it to U=mgh. This value must then equal mg(start height) and subtract 2r to get h

the answer I arrived at was 3.79 m, which the system said was incorrect. So I am stumped at this point. Can anyone please help me figure out where I have made a mistake? Thanks so much, and I love the site.

 

[rl.bhat]

First I found the centripetal acceleration as Acp(m)=.8m+mg
Try this one: Acp(m)=.8mg+mg

 

[gknowels]

Thank you for your help, that indeed gave me the correct solution. A simple oversight by me. Thanks again