What Is the Minimum Height for a Block to Complete a Frictionless Loop?

[nightshade123]

[SOLVED] Loop the Loop


1. 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 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 can't 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?

 

[Doc Al]

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.

 

[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?

 

[Doc Al]

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

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

 

[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...

 

[nightshade123]

mg = (mv^2) /R

you can find min speed so it doesn't fall off the track this way

R*g = v^2

 

[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

 

[Doc Al]

Good!

 

[nightshade123]

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