Physics 20 Problem

So I’ve missed the entire week of school and need some help with physics for my exam tomorrow. Anyways enough excuses, the question is this:

A vertical loop on a roller coaster has a radius of 10.0m. At what minimum speed must a car travel at the top of the loop for it to remain in contact with the track?

So if anyone is still awake and willing to help, I would greatly appreciate it. Thanks a bundle :mountie:

PS: For anyone who does attempt this, the correct answer to the problem is 9.90m/s.

Edit: Scratch that, just realized that some other equations applied to this problem, specifically:

T = 2[pi]√(l/g)
and
v=2[pi]r/T

I’ll keep this thread open in case I have some more questions :slight_smile:

Well thatis a pretty easy question to answer. Well actually it is about 9.8ms^2 rather than 9.9.

Gravity is measured to be at a constant of 9.8 ms^2. Anything above this in an opposing direction will overcome the force of gravity. There could be many solutions past this, as you could have an incoming ramp of 1km (if u are talking about aproach length)

But it needs to be at at least 9.8 (or 9.9. if you round it off to that)
to balance out :stuck_out_tongue:

Ahh looking at ur edit I can see ur using equations for centripetal motion eh? :stuck_out_tongue: hahahahah have fun… :stuck_out_tongue:

Thanks for the reply, hopefully you’ll be able to answer my new question:

A girl swings 0.5kg teddy bear on a 0.3m rope in a vertical loop with a constant period of 1s. Estimate the tension in the cord when the teddy is in its:
a) lowest position (answer: 10.8N)
b) highest position (answer: 1.01N)

I guess the real question is, what the deuce is tension? From my thinking I’m guessing it has something to do with centripetal force acting on the bear but, again, I would really appreciate some help with this :slight_smile:

Ah u just take away gravity from it :stuck_out_tongue:
remember that there is more force at the bottom of a loop :smiley:

OMG OMG

My post count says 1337 !!!

Alright, apparently I’m dumber than I look - unless I look bad in which case I would be exactly as dumb as I look and have two problems. Anyways, how would you calculate it?

Who ever said that g was necessarily equal to 9.8 ms^2?

In Pakistan it’s almost 9.9. It varies every latitude or so dude!

rofl well that is the standard that it is usually measured.:stuck_out_tongue: Bleh good point though heheh

Anyways, u just need to remember that the force at the bottom of the loop (assuming u move at a constant) is 9.8(or 9) more than at the top of the loop.

Plugin ur numbers into the formulae and u will get a nice answer :stuck_out_tongue:
( I haven’t touched centripetal motion for a while)

Me neither :stuck_out_tongue: