Minimum Height for Roller Coaster Loop Success

[rsala]

Homework Statement

problem should be solved somewhat with energy conservation.

A car in an amusement park ride rolls without friction around the track shown in the figure . It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.

What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)?

Homework Equations

conservation of energy
centripetal force $$\frac{v^{2}}{R}$$

The Attempt at a Solution

Energy at point A
U = $$mgh_{max}$$
K = 0
Energy at point B
U = $$mgh_{b}$$
K = $$\frac{1}{2} * mv^{2}$$

set them equal

$$mgh_{max} = mgh_{b} + \frac{1}{2} * mv^{2}$$

all masses cancel out
$$gh_{max} = gh_{b} + \frac{1}{2} * v^{2}$$

move all terms with gravity to the right side, and factor g
$$gh_{max} - gh_{b} = \frac{1}{2} * v^{2}$$
$$g(h_{max} - h_{b}) = \frac{1}{2} * v^{2}$$

with $$\frac{v^{2}}{R} = g$$ remove all g from equation. because i need V^2/r to be equal to g or the coaster won't make it past b,,, is my thinking wrong?

$$\frac{v^{2}}{R}(h_{max} - h_{b}) = \frac{1}{2} * v^{2}$$

solve for H-max

$$H_{max} = H_{b} + \frac{R}{4}$$

height at b is 2R of course.

$$H_{max} = 2R + \frac{R}{4}$$

simplify

$$H_{max} = \frac{9R}{4}$$

Wrong answer, mastering physics says, off by a multiplicative factor, of course that's mastering physics for , your wrong start all over.

any ideas?

 

[rsala]

also i tried the answer 2R, since energy should be conserved,, and it should make it back up the distance it went down,, i think.

didnt work

 

[Doc Al]

The Attempt at a Solution

Energy at point A
U = $$mgh_{max}$$
K = 0
Energy at point B
U = $$mgh_{b}$$
K = $$\frac{1}{2} * mv^{2}$$

set them equal

$$mgh_{max} = mgh_{b} + \frac{1}{2} * mv^{2}$$

all masses cancel out
$$gh_{max} = gh_{b} + \frac{1}{2} * v^{2}$$

move all terms with gravity to the right side, and factor g
$$gh_{max} - gh_{b} = \frac{1}{2} * v^{2}$$
$$g(h_{max} - h_{b}) = \frac{1}{2} * v^{2}$$

with $$\frac{v^{2}}{R} = g$$ remove all g from equation. because i need V^2/r to be equal to g or the coaster won't make it past b,,, is my thinking wrong?

$$\frac{v^{2}}{R}(h_{max} - h_{b}) = \frac{1}{2} * v^{2}$$

All good.

solve for H-max

$$H_{max} = H_{b} + \frac{R}{4}$$

Redo this step.

 

[rsala]

i don't understand. can you explain?

 

[Doc Al]

Why don't you explain how you got R/4 in that last equation.

 

[rsala]

i understand now, answer is 5r/2
thanks doc al

 

[Danesh123]

I was faced with this problem. But I was not given v at the top of the loop. From the way you layed your work down it seems that you were. Can someone please help me. It is the same problem in essence, but I need the v at the top of the loop in order to continue. What is known is that v not = 0. As otherwise. The cart falls off of the loop.

Thanks.

 

[Feldoh]

No v cancels out, look over his work a bit more.

 

[Doc Al]

I was faced with this problem. But I was not given v at the top of the loop. From the way you layed your work down it seems that you were. Can someone please help me. It is the same problem in essence, but I need the v at the top of the loop in order to continue. What is known is that v not = 0. As otherwise. The cart falls off of the loop.

The key to this problem is to figure out the minimum speed at the top of the loop. Of course you're not given it. Hint: Use Newton's 2nd law.

 

[Danesh123]

right, my bad. I see now. Pretty stupid to have missed it actually. when you introduce v2 .