Finding the initial velocity of a vertical circular motion [closed]

A particle P is suspended from a fixed point O by a light inextensible string of length a. When hanging at rest under gravity at A it is given a horizontal velocity u. The particle moves freely in a vertical circle and the string slackens when OP makes an angle of $\frac{π}{3}$ with the upward vertical.

Find u in terms of a and g

Here is an aided picture I drew to help me: I applied the conservation of energy (Took A as zero P.E level)

$$\frac{mu^2}{2}= mg(a+a \cos \frac{\pi}{3}) + \frac {mv^2}{2}$$

$$u^2=3ga+v^2$$

How do I find $v^2$ in order to find $u$. Please help. Thanks.

closed as off-topic by ACuriousMind♦, Kyle Kanos, Brandon Enright, Qmechanic♦Oct 23 '14 at 23:50

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• You have to use the fact that the string slackens at the point in question. – garyp Oct 23 '14 at 19:28
• @garyp I can't think of a way to use that :/ – The Artist Oct 23 '14 at 19:38

The key is the statement that the string slackens. This means that the entirety of the centripetal force needed to make the bend of radius $a$ comes from the component of the weight which is along $a$. From kinematics, you know that the centripetal acceleration at the top point is $$a_c=\frac{v^2}{a},$$ and you know that the component of gravity which is along the radial direction is $$mg\cos(\pi/3)$$ Applying Newton's Second Law to the radial direction should result in $$\boxed{v^2=ga\cos(\pi/3)}$$ if I got the maths right. The key insight is that the rod is no longer providing any of the forces required to achieve the observed motion$-$it might as well not be there.

There are two forces acting on the mass (omitting any frictional forces). The first one is the centrifugal force. The second one is the gravitational force.

At the moment the string slackens, the sum of the forces in the direction of the string will be zero. In a formula:

m*v^2/a - m * g / cos(pi/3) = 0

From here you can work out the relation between v^2 and g and a. This will lead to your answer. Just make a drawing of the forces acting on the mass and you will see it.

• Thank you alotttttt :) I wish I could upvote, but I have a reputation <15 – The Artist Oct 23 '14 at 19:56
• No worries. Note that Bryson S. is correct of course that the component of grav force is mgcos(pi/3). I made a slight mistake there. – Ruudc Oct 23 '14 at 20:02
• Centrifugal force is a fictitious force. If you were to draw a free body diagram for the particle at any point, you would see two forces: gravity and the force the during exerts on the particle. The sum of this two forces is the centripetal force – Sean Oct 23 '14 at 22:46
• I agree. So in a formula that would be. $F_{s} = F_{cp} - F_{g} * cos(\pi/3) = 0$ ? – Ruudc Oct 24 '14 at 8:48