0
$\begingroup$

At what point of a pendulum's swing is its acceleration the greatest? What does this tell you about where the forces act in a pendulum?

Is my answer correct?

The force that causes the acceleration of the pendulum is due to the weight of the pendulum. The weight is constant but if the pendulum is still, hanging in a direct downwards position, then the weight does not make the pendulum swing. This is because the force is vertical and the pendulum's possible motion is horizontal when it's at the bottom. If we pull the pendulum some distance to the side and release it, then it will swing. That's because when the pendulum is taken to the side, a component of the weight acts in the direction that the pendulum can move - along the arc.

It's often convenient to take a vector (like the pendulum's weight) and resolve it into 2 components that are perpendicular. For a pendulum, the components of the weight that help us are along the string (or rod) and perpendicular to the string. A perpendicular to the string would be tangential to the arc of possible motion. Give the symbol Wt to the component that is tangential to the arc. When we take it to the side (so far that the string (or rod) was horizontal), then the weight would of course point downwards and that direction would be tangential to the arc of the pendulum's possible movement. In that case, Wt is equal to the weight. So, the entire weight provides force for the acceleration of the pendulum.

However big the swing is, the pendulum's acceleration is greatest when the component of the pendulum's weight lies tangential to the arc of possible motion.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

Do you need to give a literal explanation? If not, Isn't it easier to use the equation $\overrightarrow a=-\omega^2 \overrightarrow x$?

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ I agree. I think this is what the answer is supposed to be. (To justify that $\omega^2$ is maximum at the bottom, conservation of energy would suffice.) $\endgroup$
    – Tunococ
    Commented Aug 3, 2013 at 11:36
2
$\begingroup$

Here we can think in another way: if we deal with mathematical pendulum, it satisfies the equation

$$\frac{d^2}{dx^2}y+k^2 y=0 $$

It's well known that the general solution is:

$$y(x)=A\sin kx+B\cos kx =C\sin \left(\omega +kx\right),$$ the values $A,B$ depend on amplitude $C$ of pendulum (it's no care with what pendulum we deal--mechanical, electrical or something else, but for convenience we will talk about mechanical). Hence, its' acceleration is the second derivative:

$$a(x)=-Ck^2\sin\left(\omega +kx\right) $$ Maximum absolute value holds when $\omega +kx=\frac{\pi}{2}+\pi k$--this means that it is the same moment when amplitude is maximal--it is the point when potential energy is the biggest, which holds in the point of maximal angle of diversion of pendulum from the zero-amplitude point

Improve this answer
$\endgroup$

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