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Mass m is connected to the end of a cord at length R above its rotational axis (the axis is parallel to the horizon, the position of the mass is perpendicular to the horizon). It is given an initial velocity, V0, at a direction parallel to the horizon. The initial state is depicted at position A in the image.

The forces working on the mass are MG from the earth and T the tension of the cord.

How can I calculate the tension of the cord when the mass is at some angle $\theta$ from its initial position (position B in the image)?

image.

Here's what I thought:

Since the mass is moving in a circle then the total force in the radial direction is T - MG*$\cos\theta$ = M*(V^2)/R

and so T = MG*$\cos\theta$+M*(V^2)/R

but since MG applies acceleration in the tangential direction then V should also be a function of $\theta$ and that is where I kind of got lost. I tried to express V as the integration of MG*$\sin\theta$, but I wasn't sure if that's the right approach.

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2 Answers 2

Some hints to get you started.

The mass is always on the circle. The radius is constant. Therefore the tension supplied must always be equal to the centrifugal forces acting on the mass. This is the composition of the rotational centrifugal force and gravity.

$$F_{tension} = F_{rot}+F_{grav}$$

From the point of view of the equations this is a pendulum - you will obtain a set of equations of the form

$$\ddot\theta(t) = k \sin \theta(t)$$

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Imagine the initial velocity being $v_0 = 0\,{\rm m/s}$ (i.e. bungee jumping). How does that fall into your answer? –  Marek Jan 6 '11 at 13:13
    
The assumption is that the cord is always stretched to R, so v>v_min(theta). But you are right, this is also a case that should be considered (I doubt that this is what the OP is looking for though) –  Ebenezer Sklivvze Jan 6 '11 at 13:48

If you know the motion path, then you know what the acceleration of the mass should be at any angle $\theta$

Position $$\vec{r}=[\mbox{-}R\,\sin\theta,R\,\cos\theta]$$

Velocity $$\vec{v}=[\mbox{-}R\dot{\theta}\,\cos\theta,\mbox{-}R\dot{\theta}\,\sin\theta]$$

Acceleration $$\vec{a}=[R\dot{\theta}^{2}\,\sin\theta-R\ddot{\theta}\,\cos\theta,\mbox{-}R\dot{\theta}^{2}\cos\,\theta-R\ddot{\theta}\,\sin\theta]$$

The Sum of the forces acting on the mass are $$\sum\vec{F}=[T\,\sin\theta,\mbox{-}T\,\cos\theta]-[0,m\, g]$$

Newtons law of motion is used to solve for the angular acceleration $\ddot{\theta}$ and the tension $T$ (must be positive).

$$\ddot{\theta}=\frac{g}{R}\,\sin\theta$$

and

$$T=m\,R\,\dot{\theta}^2-m\,g\,\cos\theta$$

So to get the tension, you not only need the angle $\theta$, but also the angular speed $\dot{\theta}$.

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