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Help with basic circular motion question please?

Really should be able to do this :frown:

Question is
Pr2) You hold a massless string of length L= 1m vertically, which has a mass equal to 0.50
kg attached on its end.
How much is the minimum initial velocity in the horizontal direction that should be given to
the mass so that it can go around in a vertical circle?


I tried it by saying the the kinetic energy given to it would have to = the potential energy at the top of the circle so it could reach the top of the circle.

So mv2/2 = mgh, cancelled out to get v = sqrt(2gh) using h as 2 times the lenght of the string (so 2m). This worked out as 6.3ms-1

Also tried equation the centripetal force mv2/R to the Force when the mass is at the top of the circle, just mg? but got an even worse answer.

According to my lecturer the answer is 7, but I really don't know how she got this. Any help would be much appreciated! :smile:
Reply 1
The problem with your first approach is that you are assuming that all of the initial KE has become GPE at the top of the circle. This means that the mass is not moving at the top of the circle. If this were the case, the string would have gone slack before this point, and the mass would no longer be moving in a circle.

You have to do two things here. One is indeed to put together a conservation of energy equation, including the kinetic energy at the top of the circle, and noting that the mass is not moving around the circle at a constant speed. The other is to think about the centripetal force on the mass at the top of the circle; what it consists of, what you can equate this to, and what condition is necessary for the mass to still be moving in a circle. This will give you two equations, which together, should give you the answer.

Sorry this is a bit vague, but you'll gain much more from having another go that you will from seeing a spelt out solution. One clue, though; it doesn't matter what the value of the mass is, the answer will be the same for any mass.
Reply 2
Original post by Pangol
The problem with your first approach is that you are assuming that all of the initial KE has become GPE at the top of the circle. This means that the mass is not moving at the top of the circle. If this were the case, the string would have gone slack before this point, and the mass would no longer be moving in a circle.

You have to do two things here. One is indeed to put together a conservation of energy equation, including the kinetic energy at the top of the circle, and noting that the mass is not moving around the circle at a constant speed. The other is to think about the centripetal force on the mass at the top of the circle; what it consists of, what you can equate this to, and what condition is necessary for the mass to still be moving in a circle. This will give you two equations, which together, should give you the answer.

Sorry this is a bit vague, but you'll gain much more from having another go that you will from seeing a spelt out solution. One clue, though; it doesn't matter what the value of the mass is, the answer will be the same for any mass.


Ok, think I finally got it! the top of the circle is it KE+GPE = mv2/2 +mgh and at the bottom just KE? I'd already worked out that by equating centripetal force with gravitational force at the top of the circle I get mv2/R = mg so v2 = gR. Sub that into the energy equation for the top of the circle to get mgR/2 + mg(2R) = 2.5mgR. It starts from 0J of KE and gets to that amount so that's the change in KE, so equate it to mv2/2 to get the change in KE. End up with V=sqrt(5gR) =sqrt(5*9.8*1)=7 :biggrin: Just wanted to check that's right and I haven't just fluked it? Thanks for your help :smile:
Reply 3
That looks OK to me. The interesting thing is ending up with a minimum initial speed of 5gr\sqrt{5gr}, which as you've shown, is independent of the mass.
Reply 4
Original post by Pangol
That looks OK to me. The interesting thing is ending up with a minimum initial speed of 5gr\sqrt{5gr}, which as you've shown, is independent of the mass.


Ah thanks a lot :smile: is the minimum v always 5gR\sqrt{5gR} in circular motion then?

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