circular motion under gravity

luca-deltodesco · 2009-01-15 09:19:26

One end of a light inextensible string is attached to a fixed point O. The other end of the string is attache4d to a particle P of mass 0.2kg which moves in a vertical circle of radius 0.3 m. Air resistance is ignored. When the particle has speed 4m[sup]-1[/sup] the string makes an acute angle θ with the downward vertical. At this instanct the magnitude of the transvese component of acceleration of P is 6.3 ms[sup]-2[/sup]. Show that θ = 40° approximately.

At a later instant the string makes an angle of 30° with the upward vertical. Calculate the tension in the string at this instant.

v = 4, θ = 40°

The answer however says it should be 2.57

JaneFairfax · 2009-01-15 09:59:23

The speed wont be in the second case. To find the new speed, use conservation of energy.

Now

luca-deltodesco · 2009-01-15 10:02:18

I understand how you found T, but I still don't understand what is wrong with my method, i have integrated to get an equation for the centripetal force in terms of theta, which because of integrating has the constant which i find by plugging in a known pair for the centripetal force and theta.

JaneFairfax · 2009-01-15 10:21:20

I see where you went wrong.

In the second case, where , you were still taking whereas you should be taking .

goes with but for you should be taking .

luca-deltodesco · 2009-01-15 10:24:31

I still don't understand where my thinking has gone wrong... (maybe i'm just very tired)

but, i get the equation

whch i integrate with respect to theta to get

i calculate k' with a v,theta pair as you usually would after an integration, and that should allow me to calculate the centripetal force for any theta? equating that to the equation for centripetal force with tension, i find the tension such that i never have to calculate 'v' itself?

luca-deltodesco · 2009-01-15 10:28:56

although seemingly, using that equation involving k', i don't get 6.4 for v if plug in 150 for theta... what is wrong there?
seemingly there must be something wrong with the maths in my previous post.

Last edited by luca-deltodesco (2009-01-15 10:31:16)

JaneFairfax · 2009-01-15 10:45:15

AH YES!! I think Ive got it at last!

This

luca-deltodesco wrote:

is wrong. It should be

Hence your

should be

And your

should be

That should work now.

luca-deltodesco · 2009-01-15 10:48:21

aha! thank god i wasn't going crazy then! thanks jane

JaneFairfax · 2009-01-15 10:53:03

No problem. That was rather a good spot-the-problem exercise, actually. Once youve found out where the problem is, a problem like this one should be no problem at all.

luca-deltodesco · 2009-01-15 11:43:18

I never even questioned that that bit could be wrong as it gave me the correct answer for the first part of the question I just need to be more careful in doing these things to make sure i don't leave myself prone to causing them.

JaneFairfax · 2009-01-15 12:15:21

In the first part, is actually .

So the equation should strictly speaking be

.

Math Is Fun Forum

#1 2009-01-15 09:19:26

circular motion under gravity

#2 2009-01-15 09:59:23

Re: circular motion under gravity

#3 2009-01-15 10:02:18

Re: circular motion under gravity

#4 2009-01-15 10:21:20

Re: circular motion under gravity

#5 2009-01-15 10:24:31

Re: circular motion under gravity

#6 2009-01-15 10:28:56

Re: circular motion under gravity

#7 2009-01-15 10:45:15

Re: circular motion under gravity

#8 2009-01-15 10:48:21

Re: circular motion under gravity

#9 2009-01-15 10:53:03

Re: circular motion under gravity

#10 2009-01-15 11:43:18

Re: circular motion under gravity

#11 2009-01-15 12:15:21

Re: circular motion under gravity

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