Sliding down a Track, loop, Conservation problem

[iamtrojan3]

Homework Statement

A small cube (m=0.450 kg) is at a height of 393 cm up a frictionless track which has a loop of radius, R = 58.95 cm at the bottom. The cube starts from rest and slides freely down the ramp and around the loop. Find the speed of the block when it is at the top of the loop.

Homework Equations

see below

The Attempt at a Solution

s
I tried using the formula
v^2=4/3 g (0.66H-2R)
I don't know I'm really confused at these kind of question, this was obvoiusy wrong, but i think I'm close. or maybe not.

Thanks for all the help.

 

[ralilu]

First try and calculate the velocity of the block at the bottom of the ramp:
mgh = 0.5 x mass x change in velocity squared
9.8m/s^2 x change in height = 0.5 x change in velocity squared
change in height = 393 cm. initial velocity = zero
so now u can find the final velocity at the bottom of the ramp
at the bottom of the ramp velocity = speed.
the speed of the block remains the same as it goes around the loop, because its uniform circular motion.

 

[iamtrojan3]

I don't think that works.
Remember gravity is still pulling on the block as it travels up the loop, so the speed has to be different from when it entered the loop.
I believe it has something to do w/ conservation of energy, i just don't know where to start

 

[alphysicist]

First try and calculate the velocity of the block at the bottom of the ramp:
mgh = 0.5 x mass x change in velocity squared
9.8m/s^2 x change in height = 0.5 x change in velocity squared
change in height = 393 cm. initial velocity = zero
so now u can find the final velocity at the bottom of the ramp
at the bottom of the ramp velocity = speed.
the speed of the block remains the same as it goes around the loop, because its uniform circular motion.

I don't think that's right, because the problem does not say the the block is going in uniform circular motion around the loop.

I don't think that works.
Remember gravity is still pulling on the block as it travels up the loop, so the speed has to be different from when it entered the loop.
I believe it has something to do w/ conservation of energy, i just don't know where to start

That's right; energy of the block is conserved as it goes around the track. Since energy is conserved, that means you can pick any two points along the path, and the energy at those two points has to be the same.

So pick the the two points you care about. What is the total energy (PE+KE) at the starting point? What is the total energy at the top of the loop? (not a number, but an expression using the formulas for PE and KE)

Using conservation of energy means setting those two expressions equal to each other; then you can solve that equation for v.

 

[rabbitweed]

Umm, isn't the top of the loop where the block starts? I mean if it's sliding around purely undergravity, then it must start at the top.

If it starts at rest, then initial speed is 0.

 

[alphysicist]

Hi rabbitweed,

Umm, isn't the top of the loop where the block starts? I mean if it's sliding around purely undergravity, then it must start at the top.

If it starts at rest, then initial speed is 0.

I think you're visualizing the problem incorrectly. There is first a ramp that the block slides down (it starts 393 cm high); then after it gets to the bottom of the ramp, it then enters a loop of radius 58.95cm.