3
$\begingroup$

A simple pendulum has a bob with a mass of $0.50$ kg. The cord has a length of $1.5$ m, and the bob is displaced $20^\circ$.

I am trying to use this expression to find the maximum velocity of the bob.

$$\omega^2_f = 2 \alpha\Delta\theta$$

I get the following expression:

$$\frac{v^2}{L^2} = 2L\frac{mg\sin\theta}{I}\Delta\theta$$

Is this expression correct to solve the question?

$\endgroup$
10
$\begingroup$

First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $\frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-\cos\theta)$, so $$mgL(1-\cos\theta)=\frac 12 mv^2$$ Notice that the velocity is independent of mass

$\endgroup$
8
  • $\begingroup$ But, I would like to solve this particular manner the professor solved it that way already. $\endgroup$ Mar 3 '19 at 2:10
  • $\begingroup$ Secondly, everything was given to you.in the detail. $\endgroup$ Mar 3 '19 at 2:11
  • 4
    $\begingroup$ You did not explain what the $\alpha$ or $I$ or $\omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $\alpha$, which is not the case here. The torque depends on the angle between gravity and the cord. $\endgroup$
    – Andrei
    Mar 3 '19 at 2:15
  • $\begingroup$ Isn't the force is $mg\sin\theta$ $\endgroup$ Mar 3 '19 at 2:19
  • $\begingroup$ No, the force is $mg$ always pointing downwards. The torque is $mgL\sin\theta$ $\endgroup$
    – Andrei
    Mar 3 '19 at 2:22
3
$\begingroup$

The equation only holds when angular acceleration $\alpha$ is a constant. However in this case $\alpha=\frac{g\sin\theta}{l}$, which depends on $\theta$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.