Connecting energy conservation to centripetal acceleration in a loop

[UrbanXrisis]

This is really giving me trouble. The problem is on the following site:

http://home.earthlink.net/~suburban-xrisis/index.mht

OKay, what I thought of was mgh=.5mv^2
so therefore g(5r/2)=.5v^2

I'm not sure how to connect it all to a=v^2/r

any ideas?

 

[Chen]

(Probably by the time I'm done writing this someone would have linked you to a post that already explains this problem, but what the hell!)

At the top of the loop, the centripetal force is the combination of two forces - the normal force and the weight of the object. For the circular motion to continue, this condition must be met:

$$F_c = N + mg = ma_c = m\frac{v^2}{r}$$

$$a_c = \frac{N}{m} + g = \frac{v^2}{r}$$

The lower we drop the object from, the lower it velocity will be when it reaches the top of the loop. However, it has a minimal value below which the object will simply fall of the track. As you can see in the formula above, the minimum value of the centripetal acceleration at the top is g, in which case there is normal force. Therefore the minimal velocity at the top is:

$$ma_c = mg = m\frac{v_{min}^2}{r}$$

$$v_{min}^2 = gr$$

That's the speed of the object at 2r above the ground. Thanks to conservation of energy we see that:

$$\Delta E_p + \Delta E_k = (2mgr - mgh) + (\frac{1}{2}mv_{min}^2 - 0) = 0$$

$$4gr - 2gh + v_{min}^2 = 0$$

Now just substitute v_min² and solve for h.

 

[UrbanXrisis]

Thanks, that explains a lot! I have a new question:

There is a frictionless track with 3 hills. Hill “A” is “h” in height, following hill “A” is the second hill, hill “B” that is also “h” in height. The third hill following hill “B” is hill “C” which is h/2 in height. Hill C slants to the flat ground labeled D and E. There is a car of mass “m” that is currently on the top of hill “A” traveling with speed “v1.”

1.The speed at the top of hill B would be v1 as well correct? But how would I determine it with equations? Mgh=mgh=.5mv^2?

2.The speed at the top of hill C would be two times v1 right? If this is true, how would I determine it with equations? mgh=.2mv^2 …then mgh/2=.2mv^2 so you have to multiply 2 to the velocity?

3.When the car reaches point D and E, the momentum would be 2mv1 correct? If I got 2v1 from question #2, then I sub it in with p=mv=2mv1?

4.To determine the average breaking force “F” needed to stop the car “d” distance would be W=Fd so F=W/d=.5m(2v1)^2/d?

Questions can be better observed on:

http://home.earthlink.net/~suburban-xrisis/pics02.jpg

thanks

 

[e(ho0n3]

This is really giving me trouble. The problem is on the following site:

http://home.earthlink.net/~suburban-xrisis/index.mht

OKay, what I thought of was mgh=.5mv^2
so therefore g(5r/2)=.5v^2

I'm not sure how to connect it all to a=v^2/r

any ideas?

Funny, I just did this problem. Here is the link https://www.physicsforums.com/showthread.php?t=27932.
You can also browse down in the forums section and look for a thread with the title: Roller Coaster Problem.

e(ho0n3

 

[e(ho0n3]

1.The speed at the top of hill B would be v1 as well correct? But how would I determine it with equations? Mgh=mgh=.5mv^2?

2.The speed at the top of hill C would be two times v1 right? If this is true, how would I determine it with equations? mgh=.2mv^2 …then mgh/2=.2mv^2 so you have to multiply 2 to the velocity?

3.When the car reaches point D and E, the momentum would be 2mv1 correct? If I got 2v1 from question #2, then I sub it in with p=mv=2mv1?

4.To determine the average breaking force “F” needed to stop the car “d” distance would be W=Fd so F=W/d=.5m(2v1)^2/d?

Seems you are applying the laws incorrectly. Use conservation of mechanical energy properly and you'll get the answers. As for problem 4, you don't need to use work to find the force. You can just use the velocity calculated in problem 3 to find the average acceleration which will give you your average force.