Tricky question

  • Thread starter n_ds
  • Start date
  • #1
11
0
Hi

I am working on an assignment and I honeslty am completely baffled with this question. I have no idea how to go about trying to figure it out as its purely conceptual based. If anyone has any info that might help, I am all ears. The problem is as follows and I am attaching the diagram as well if it helps:



A block of mass m slides down an inclined plane into a loop-the-loop of radius r . (a) Neglecting friction, what is the minimum speed the block must have at the highest point of the loop in order to stay in the loop? [Hint: What force must act on the block at the top of the loop to keep the block on a circular path?] (b) At what vertical height on the inclined plane (in terms of the radius of the loop) must the block be released if it is to have the required minimum speed at the top of the loop?

I appreciate any help you can offer.

Thanks

TN
 

Attachments

Answers and Replies

  • #2
berkeman
Mentor
58,180
8,237
If you had used the homework posting template (please get in the habit of doing that), one of the sections asks what equations and/or priciples apply to your question. What can you tell us about that?
 
  • #3
11
0
I did not post any formulas because I am not sure what I can use. Like this question doesent use any numbers, you are only supplied with generalizations of certain aspects. Like friction is negligaable. Also since the cart is at a height you can use potential energy formula. But I still can't find a formula that i could use to solve for speed. I know it has to do with centrifugal force but I am just not sure. That is why i am asking.
 
  • #4
berkeman
Mentor
58,180
8,237
Well at least you're starting to use some of the right keywords. Show us some equations for the KE and PE of an object, and tell us about the total energe TE. What equations govern circular motion? What equation would you use to figure out the forces on an object that is exhibiting uniform circular motion?
 
  • #5
Remember that at the top of the loop [tex]mg=m\frac{v^2}{r}[/tex]
 

Related Threads on Tricky question

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
718
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
16
Views
7K
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
2K
Top