# Rollercoaster problem - kinematics & forces

1. May 20, 2013

### AgentRedfield

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

A small car of mass m is released at height h on a steel track. The car rolls down the track and through a loop of radius R. At the end of the track, the car rolls off the track, which is positioned at a height H above the floor. Neglect friction and the small amount of rotational motion of the wheels of the car. Solve in terms of h, m, R, H, and g.

(a) Find the velocity of the car at point B (bottom of the loop).

(b) Find the velocity of the car at point C (top of the loop).

(c) Determine the height h at point A such that the car just barely makes contact with the loop at point C as it goes through the loop.

(d)When the car is moving at minimum speed, what provides the centripetal force on the car:
i. at point B?
ii. at point C?
iii. at point D (side of the loop)?

(e) Determine the distance from the end of the track that the car will land on the floor.

2. Relevant equations

$mgh + \frac{1}{2}mv^2$
$a=\frac{v^2}{2}$
$F_net=\frac{mv^2}{2}$
$F_N \pm F_g = F_C$
$x=x_o +v_o t +\frac{1}{2}at^2$

3. The attempt at a solution

(a) I think I understood this, I used the first equation above and got: $v=\sqrt{2gh}$

(b) I started to solve this as follows: $F_N + mg = \frac{mv^2}{2}$
The problem is that when solving for v I can't figure out what to substitute for the normal force. My intermediary answer is $v=\sqrt{\frac{(F_N + mg)R}{m}}$

(c) The height at point C = 2R. I used $mgh + \frac{1}{2}mv^2 = constant$ which I think would make the answer be h = 2R.

(d)
i. I think it would be -mg because that's the only downward force.
ii. I'm having the same trouble with this as I am with part b. The normal force combined with the force of gravity is making my substitutions became circular.
iii. The normal force equals the centripetal force so I believe it would be $\frac{2mgh}{R}$

(e)
$x=x_o +v_o t +\frac{1}{2}at^2$ and $t=\sqrt{\frac{2H}{g}}$ so the answer would be $x=\sqrt{2gh}\sqrt{\frac{2H}{g}}$

4. Conclusion

Parts b, c, & d have me the most confused so any help with understanding them would be very appreciated. Problems that require the use of only certain variables such as this one have given me the most trouble in physics so any tips or tricks you have when approaching them would be awesome as well. Thank you very much for your time.

2. May 20, 2013

### Simon Bridge

Welcome to PF;
(a)
"The first equation above" is an incomplete sum - what do they add up to?
The physics here is that the gravitational potential energy lost dropping height h is gained in kinetic energy.

(b)
It's the same principle here - except the height isn't h and some energy gets stored in the rotational motion.

(c)
You need to use circular motion for this one.

(d)
i. at point B, the centripetal force points which way: up or down?
What provides a force in that direction? Hint: not gravity.

3. May 20, 2013

### Basic_Physics

The mechanical energy is conserved in this instance since the only force that can act along the path of the motion is gravity, which is a conservative force. Friction can be ignored and the normal force do not act along the path. So

EM1 = EM2

for any two points along the path of the small car. The mechanical energy is the sum of the potential energy EU and the kinetic energy EK or

EM = EU + EK

4. May 20, 2013

### AgentRedfield

Thanks for the welcome, so I know that $mgh+\frac{1}{2}mv^2 = mgh+\frac{1}{2}mv^2$

I just forgot to write it all out because I've never used this latex syntax thing before and got distracted by it. Anyway that is what I used for part a, did I do that correctly?

For b I just realized that the height is equal to 2R. Using the above equation I get $v=\sqrt{2g(h-2R)}$ however you mentioned energy getting stored. Am I missing an equation for centripetal energy?

Part c: I went through my book and realized that since it is barely making contact with the loop then normal force can be zero. This means $mg=\frac{mv^2}{r}$ I solved for v then substituted in my answer for v from part a and after solving for I got $h=\frac{R}{2}$

Part d:
i. Ah, so the centripetal force is towards the center, in this case up. That force would be the normal force. So, given that $ƩF=\frac{mv^2}{r}$ and $ƩF=F_N - F_g$ at the bottom then the answer should be $\frac{2mgh}{R} + mg$ Is that correct?
ii. Gravity is providing the force so it would just be $mg$
iii. Force of gravity drops out at part D so the only force is the normal force so $F_N=\frac{2mgh}{R}$

Am I doing/understanding this correctly?

5. May 20, 2013

### haruspex

All ok until:
The equation's fine but the answer's wrong. h must exceed 2R.
I'm not sure what the question means here. Does it mean the minimum speed to stay in contact at each of these points (i.e. 3 different h values) or taking h to be as in part c? Anyway, it doesn't ask for the magnitude of the force, so I think your descriptive answers are fine.

6. May 20, 2013

### Simon Bridge

$E_R=\frac{1}{2}I\omega^2$
But if you work this out for the car it just turns into the equation for kinetic energy - I think you are safe.
Check if you like. $v_\perp = R\omega$.

So you seem to have made a breakthrough anyway.
That's the way to think about it - notice that for d they only want what provides the centripetal force, they don't care about the magnitude.

Good on yer - you are doing very well.

Some tips -

The summation symbol is \sum ;) you could also write \Sigma but if you use \sum you can get it to format properly next to the thing being summed and you can add the range of the sum like S_n=\sum_{i=1}^n x_i to give you $$S_n=\sum_{i=1}^n x_i$$

Practically everything special has a latex representation. It's usually easier to just type them in.
It you use "tex" tags, instead of itex, you can get your fractions to expand out full size and the equation get's it's own line.

7. May 21, 2013

### Basic_Physics

You can use your result from part b to calculate the height at which the car should be released to get this speed at the top.

8. May 22, 2013

### AgentRedfield

I'd just like to thank everyone again for your awesome help. So everything I've got nailed down except....

Part c: The relevant equation is

$mgh+\frac{1}{2}mv^2 = mgh+\frac{1}{2}mv^2$

Now the cart must reach point c which is at the height 2R. Logically to reach this point of potential energy the cart must start at a point of higher potential energy to fulfill the above equation. So conceptually the answer h≥2R makes sense. I have two questions though. Can h be equal to 2R or must it exceed it? This is an ideal system so I think it should be able to just barely reach point c. My other question is that I solved for h algebraically again and still got an answer of $\frac{R}{2}$ Why does this happen?

9. May 22, 2013

### haruspex

You showed in post #4 that v must be > 0 here, so not all the original PE has gone back into PE. So h > 2R.
Can't tell without seeing your working.

10. May 22, 2013

### Basic_Physics

You've got the speed at C :g = v2/R

use your equation from part b to get the height it should be released from to obtain this speed at C

11. May 22, 2013

### AgentRedfield

Ok, I figured out part c. I was using the velocity solution I found in part a not the velocity equation in part b. The answer is $h=\frac{5}{2}$

Thanks Everyone!