Tension in a string swinging a rock.

In summary: I'm not sure what you're trying to say. But in the above situation, the weight at the top of the swing should be zero because the net force is equal to the centripetal force, which is zero. So the tension would be equal to the weight at that point, which is zero. Does that make sense?In summary, a rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane. The speed required for the rock to have a weight of 0 at the top of the swing is 2.71 m/s. The tension in the string at the bottom of the swing is 3.918N. The angle of the string to
  • #36
Hi rl.bhat,

Well, it's the same thing. You are using cosvn of energy, and so am I. Only I did it for the second part, because there was no need to use it for the first part. Your answers are the same as mine.

The only thing that still nags me is question C, as I have written in my earlier post. If the point where one end of the string is fixed is not moved, then you’ll get a different answer for the angle.
 
Physics news on Phys.org
  • #37
Tension

I have rewritten the part A because I can demonstrate the problem to the students by using a pendulum.
I have treated the part 3 indipendent of A and B. beacuse from vertical circular motion we cannot shift to conical pendulum. I can demonstrate this also to the students. In all the three cases the point of suspension is the same.
 
Last edited:
  • #38
Well, it's nagging me because of that reason. Actually, the question C is not properly formulated. If you treat C independently from A and B, and just keep the KE of C as of B, you’ll get one answer. The PE + KE is more in this case than from case B.

But if you can somehow apply some force to the rock perp to its velocity at the point where it’s at the bottom of the vertical swing, it won’t change the total energy but maybe change the direction so that now it swings horizontally like a conical pendulum. In this case the angle is different. I have no idea whether such a thing can be done or not.
 
Back
Top