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Mechanics M2 - Circular Motion

marcus2001

Struggling on this problem:

A particle of mass m, is attached to one end of a light, inextensible string of length l. The other end of the string is fixed at P. The particle moves in a horizontal circle of radius r at a constant speed v. (In the diagram, the particle is basically following the path of the circular base of a cone, and the point P is the tip of the cone.)

Show that the tension in the string, T, is given by

T= \dfrac{mgl}{\sqrt{l^2 - r^2}}

.

What I did do was:

Let the angle between the vertical and string be a.
Equating vertical components:

Tcos(a)=mg

Equating horizontal componets with centripetal force

Tsin(a)=\dfrac{mv^2}{r}

r=lsin(a)

then

Tsin(a)=\dfrac{mv^2}{lsin(a)}

Then in order to eliminate the sine's and cosine's:

sin^2(a)=\dfrac{mv^2}{Tl}

cos^2(a)=1 - \dfrac{mv^2}{Tl}

from resolving vertical components

cos^2(a)=(\dfrac{mg}{T})^2

.

Therefore

(\dfrac{mg}{T})^2 = 1 - \dfrac{mv^2}{Tl}

But rearranging this to find T doesn't seem to work out, am I heading along the right tracks? Cheers in advance

Reply 1

bcrazy

Original post by marcus2001

What I did do was:

Let the angle between the vertical and string be a.
Equating vertical components:

Tcos(a)=mg

Can't see anything wrong with your working but you have over-complicated your approach. Surprisingly, the bit of your working quoted above is pretty much all you need as you can get an expression for cos(a) with a bit of Pythagoras and trigonometry and it works out quite quickly.

You went to a lot of trouble to eliminate the angle but this was in fact something that you knew plenty about in terms of r and l.