Kinematics Problem - interpretation problem and other way of solving it

Question

I have the following problem :

A point $P$ of mass $m$ is constrained to move on a frictionless circle of radius $l$, centred at $O$, in a vertical plane in a gravitational field with acceleration $g$. If the particle is released from rest when $OP$ makes an angle $\theta_0$ with the downward vertical, calculate the speed at which the particle passes the lowest point.

I think I solved it with energy. Setting the zero of potential energy at its lowest point I simply get that at the point of release I have $E = mg(l-l\cos\theta_0 )$ (because the cosine is negative in the upper half and positive in the lower half). Also at the lowest point I have $E=\frac{1}{2}mv^2$ hence I have $v^2 = 2gl(1-\cos(\theta_0))$

I wanted to solve this using Newton's laws, so I tried polar coordinates.

Setting first the usual Cartesian axes, I wrote $\underline{W} = -mg\underline{j}$ to be the weight and I used the formula $\underline{j} = \cos(\theta)\underline{e_\theta}+\sin(\theta)\underline{e_r}$ I get $\underline{W} = -mg\cos(\theta)\underline{e_\theta}-mg\sin(\theta)\underline{e_r}$. The only component that counts towards the velocity in the motion is the tangential one, hence $\underline{e_\theta}$.

Hence I get $m\ddot{\theta} = -mg\cos(\theta)$ from Newton's Second Law. However I think this is the usual pendulum equation which can only be solved for small oscillations using Taylors. So I must have done something wrong? Certainly I can find a solution for any oscillation (By oscillation I mean by releasing the particle at any point of the circle, maybe excluding the highest point). The last step is the one I am most doubtful about but I really can't see how to get the same result as above.

Answer 1

Consider the following angular position of the pendulum at which it has an angular velocity $\omega$

The angular acceleration of the pendulum is given by

$$\alpha=\frac{g}{l}\sin\theta$$

The moment of inertia of the pendulum is given by $I=ml^2$, and the torque acting in the pendulum bob is $$\tau=mgl\sin\theta$$

And we know that torque is given by $\tau=I\alpha$. Hence, we get

$$\tau=mgl\sin\theta\implies ml^2\cdot\alpha=mgl\sin\theta\implies \alpha=\dfrac{g}{l}\sin\theta$$

And if you are not familiar with the notion of moment of inertia, then just consider the relation b/w the length of arc and the angle subtended by that arc, which is

$$\theta=\dfrac{L}{l}, \text{ where $L$ is the length of the circular arc}$$

Differentiate the equation twice and you get $$\alpha=\dfrac{a_t}{l}\text{, where $a_t$ is the tangential acceleration}$$

Now we do know that angular acceleration in terms of angular velocity and angular position can be written as $\alpha=\omega\dfrac{d\omega}{d\theta}$.Hence, we get

$$\omega\dfrac{d\omega}{d\theta}=\dfrac{g}{l}\sin\theta\implies \int{\omega\cdot d\omega}=\int{\left(\dfrac{g}{l}\sin\theta \right)(-d\theta)}$$

Setting proper limits for the above integration gives you a relation b/w $\theta$ and $\omega$ from which you can easily find the relation b/w time $t$ and angular position $\theta$

The proper limits for finding the velocity of the object at the lowest point is as follows

$$\int_{0}^{\omega_l}{\omega\cdot d\omega}=-\int_{\theta_0}^{0}{\dfrac{g}{l}\sin\theta\cdot d\theta}\\ \implies \dfrac{{\omega_{l}}^2}{2}=\dfrac{g}{l}(1-\cos\theta_0)\\ \implies v^2=\sqrt{2gl\left(1-\cos\theta_0\right)}$$

Answer 2

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$Hint$: Try to find a relation between $\dot{\theta}$ and $u$. Then multiply the equation you get from Newton's second law by $\dot{\theta}$, use appropriate differentiation rules and integrate.