The Student Room Group

Circular Motion

Please could someone look over my working, I made a mistake somewhere in Q13.
I can't clearly follow what you're doing, as you seem to have used v for the velocity at the top and for the velocity at the bottom. If you're going to use the same letter, then you need to use a suffix to distinguish which is which.

Note also that TBT_B is not the tension in the string when the ring is at the bottom. It is the radial component of the tensions in the string, which is hanging in a plane perpendicular to that of the ring's motion.

Also, KEtKEBKE_t\neq KE_B. The whole thing is rather confusing. Your t's and E's are difficult to distinguish between as well.
Reply 2
Is this working easier to follow..
Yep - a vast improvement.

In your diagram, you only have half the picture. The ring is hanging in the middle of the string, so there will be half the string on one side of the ring, and half on the other, and so two lots of tension supporting the ring.

2TBsinθ=5mgrr+mg\displaystyle 2T_B\sin\theta=\frac{5mgr}{r} +mg

Your next line talking about max tension occurs when radius is a max doesn't make sense to me. Radius doesn't change. Sin theta is going to be 4/5. I suspect your interpretation of the setup is causing the problem.

Here's my attempt at a diagram.
Untitled.jpg
Reply 4
Original post by ghostwalker
Yep - a vast improvement.

In your diagram, you only have half the picture. The ring is hanging in the middle of the string, so there will be half the string on one side of the ring, and half on the other, and so two lots of tension supporting the ring.

2TBsinθ=5mgrr+mg\displaystyle 2T_B\sin\theta=\frac{5mgr}{r} +mg

Your next line talking about max tension occurs when radius is a max doesn't make sense to me. Radius doesn't change. Sin theta is going to be 4/5. I suspect your interpretation of the setup is causing the problem.

Here's my attempt at a diagram.
Untitled.jpg


Aha, you're right, I misinterpreted the setup. I've got it now :biggrin:

Quick Reply

Latest