pendulum restoring force and acceleration

[dudewheresmymd]

TPR workbook says:

If F=-mgsigntheta=ma, then at equilibrium (theta=0) , pendulum's acceleration = 0? not constantly accelerating as mentioned above, correct?

 

[syoung]

It is constantly experiencing gravity (pointed down), as well as Tension from the string/rope that pulls it towards the anchored point.

 

[dudewheresmymd]

It is constantly experiencing gravity (pointed down), as well as Tension from the string/rope that pulls it towards the anchored point.

Right but at equilibrium (theta = 0), acceleration would be 0 right?

 

[syoung]

Why do you think that?

 

[Chrisz]

Right but at equilibrium (theta = 0), acceleration would be 0 right?

At equilibrium position, the pendulum has zero acceleration, it is where acceleration start to change sign, by which it means it begin to decelerate. The tpr explanation is indeed not so correct, but one thing to be sure is that it always experiences tension and gravity.

 

[dudewheresmymd]

Why do you think that?

Because it would have to be 0 for it to switch from positive to negative acceleration at least at one point in the cycle?

 

[dudewheresmymd]

At equilibrium position, the pendulum has zero acceleration, it is where acceleration start to change sign, by which it means it begin to decelerate. The tpr explanation is indeed not so correct, but one thing to be sure is that it always experiences tension and gravity.

The analagous of F=-kx is F=-mgsintheta for pendulum right?

 

[Chrisz]

The analagous of F=-kx is F=-mgsintheta for pendulum right?

Yes, they are both conservative system, potential energy is transformed into kinetic, and then kinetic transformed into potential. At equilibrium position (spring), F starts to change direction, which means acceleration changes direction, which in turn means deceleration. Velocity changes direction only at the extreme end points where, KE=0

 

[syoung]

Because it would have to be 0 for it to switch from positive to negative acceleration at least at one point in the cycle?

I think you are confusing that with velocity.

 

[SuperSneaky24]

I think it depends on how you look at the direction. From a purely left and right point of view, yes acceleration would be zero at the middle bottom of the swing as it changes direction. I think everyone else is referring to the centripetal acceleration view of things?

 

[SuperSneaky24]

Ah I should probably explain my reply better.
I think you're viewing the pendulum in only acceleration in the x direction. Yes, the direction changes in the x-axis, so x-acceleration would be zero at the bottom. However, remember that there is always tension and gravity acting on the pendulum, so y-acceleration is not zero and will never be in ideal situations. Therefore, total acceleration will never be zero. This y-acceleration gives the centripetal force needed to move the pendulum.

(and I'm referring to the x-axis and y-axis at the instantaneous moment when the pendulum is at the bottom, at its equilibrium position)

 

[Nv2late]

Ah I should probably explain my reply better.
I think you're viewing the pendulum in only acceleration in the x direction. Yes, the direction changes in the x-axis, so x-acceleration would be zero at the bottom. However, remember that there is always tension and gravity acting on the pendulum, so y-acceleration is not zero and will never be in ideal situations. Therefore, total acceleration will never be zero. This y-acceleration gives the centripetal force needed to move the pendulum.

To add to what SuperSneaky wrote, not to be confused the y-acceleration to be just gravity. The centripetal force is the net forces of all the forces, which points to the center, hence the acceleration at the bottom of the path, is sum of -F(grav) and +Tension. If the question asks what is the direction of the net acceleration (centripetal acceleration) that provides the motion, then the answer is up. But what is the restoring force at theta=zero, then the answer should be zero; F(restoring)=mg(theta) (small angle sin(theta)=theta). *Remember the object feels a constant force of (mg) wherever it is; it is the component of that force that changes as theta changes.

This will get you think a little bit more. If you have a spring and mass question and the question asks, what is the acceleration that provides the motion (or what is the restoring acceleration) at the equilibrium point, then what is your answer?...It should be zero. (Then another question can ask, what is the total acceleration of the system at the equilibrium point? The answer is not 10m/s^2 (gravity) because the mass would sink into the table)

One more thing, (-)acceleration or (+) acceleration should be thought of in the frame of velocity, therefore, direction of motion. (-)mg(theta) and (+)velocity would slow the object down and blah blah. Understand these concepts would ensure easy time with kinematics as well.

Hope this helps

 

[NextStepTutor_1]

Lastly, to add to everyone else as well, for circular motion problems like this pendulum remember that the net acceleration the object will feel is centripetal acceleration and the net force is Force centripetal. Otherwise, the circular motion would not be maintained. Remember that in all of the problems you do for circular motion--there is a net acceleration and net force to maintain circular motion.