Potential energy and turning points

[nightshade123]

[SOLVED] potential energy and turning points

Homework Statement

A block slides on the frictionless loop the lopp track shown in this img, what is the min height at which it can start from rest and still make it around the loop

The Attempt at a Solution

I haev solved this problem TWO ways, and i can't decide which way is correct

the eqn...

U0+K0 = K+U
where U = potential energy and K = kinetic energy
first

m * g * h + (1/2) * m * v^2 >= m * g * 2 * R

final answer = h = 2R
the other way is

m * g * h >= (1/2) * m * v^2 + m * g * 2 * R

final answer is h = (5*R) / 2??

 

[nightshade123]

if i am way off let me kno, thx

 

[awvvu]

What are you assuming v to be? 0?

I don't think the answer is just 2R. You have to do some analysis on the forces, since at the top, the normal force should be zero, and you find the velocity for which that is required. If velocity were zero at the top, it wouldn't make it around the loop, just fall off.

The 5/2 R answer looks correct.

 

[nightshade123]

you can solve the problem for velocity and plug the velocity function in and solve for h, and that is how you derive the formula, and for the 2R velocity cancels out so that wouldn't make sense, thanks

 

[nightshade123]

can someone else sovle the two equations i listed at the top of the page and tell me if they got the same thing, i keep solving the 2nd one and getting h=h-2R+2R which is saying the min height is the min height of the radius...

 

[uber_dzl]

well you are starting from rest with U_i so it makes sense to me that what ever happens after would only be distributions of the energy between kintetic and potential so I would set the equation to what I am starting with

 

[uber_dzl]

i just did this exact problem in my homework and got the correct answer but i don't kno how to help without breaking a forum rule and giving u the answer

 

[nightshade123]

dont worry about it i got it already, had to open a diffrent thread because the titel was wrong