Loop the Loop

1. Feb 24, 2008

nightshade123

[SOLVED] Loop the Loop

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution
I haev solved this problem TWO ways, and i cant 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 way my friend set it up and solved it

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

my friend solved it this way and got h = (5*R) / 2

but i cant seem to get what he got and i get h=h-2r+2r which says the min height has to be the min height of the radius aka h = 2R

any hints, tips, or advice?

2. Feb 24, 2008

Staff: Mentor

For one thing, the initial energy is completely potential since it starts from rest (KE = 0). So the first method doesn't make sense.

The trick is to figure out what KE the block must have at the top of the loop in order to maintain contact with the track.

3. Feb 24, 2008

nightshade123

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

you can solve for v^2 and get

v^2 = 2g(h-2R)

that would be what is required to stay on the track, so would u plug that back into

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

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

cancel mass, g cancels, .5 *2 =1

g*h >= (1/2)*(2g(h-2R)) + g*2*R

g*h >= g(h-2R)) + g*2*R

h >= h-2R + 2*R

would this be the correct eqn to solve for min height?

Last edited: Feb 24, 2008
4. Feb 24, 2008

Staff: Mentor

This is conservation of energy. It's necessary but not sufficient to solve this problem. Hint: Apply Newton's 2nd law.

5. Feb 24, 2008

Littlepig

i guess you're still not understanding what's the condition to the block doesn't feel down from the loop....

what the block has in top of the look to don't fell down? first you need to understand that...

6. Feb 24, 2008

nightshade123

mg = (mv^2) /R

you can find min speed so it dosnt fall off the track this way

R*g = v^2

Last edited: Feb 24, 2008
7. Feb 24, 2008

nightshade123

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

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

h >=1/2 R + 2 R

h >= 2.5 R

8. Feb 24, 2008

Staff: Mentor

Good!

9. Feb 24, 2008

nightshade123

thanks! i totally looked at that v^2 the wrong way

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