Pendulum

[pm1]

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I read a couple of threads about this but I just want to make sure that I got it.
Could someone please verify if my thought process is correct? Thanks!

Where is Tension the highest in a swinging pendulum?

Equilibrium

Because at equilibrium
Fnet = T - mg ------- since there is centripetal motion, Fnet = mv^/2
mv^2/2 = T - mg
T = mv^2 + mg

At a point other than equilibrium
T = mg cos(theta)

Since at equilibrium there is the centripetal force to be counted in it is where T is the highest. Also, at points other than equilibrium only a component of mg is taken into account (cos).

However, I am still not quite certain why can we not equal Fnet to centripetal force at points other than equilibrium?? Is it because it is not at constant speed?

Thank you!

 

[pm1]

bump!

anyone? N1st or rjosh33 ? 😀

 

[ouroboros39]

I think that most of what you have is correct, as well as the reasoning - although the tension should include the radius, so it should be T=mv^2/r + mg (this comes from circular motion equation a=v^2/r)

And I'm not sure what your second question is, but you can find the force from the centripetal acceleration at other points, its just that it would be smaller due to the smaller velocity - since after the equilibrium position, you begin to change Kinetic Energy to Gravitational Potential Energy.

 

[N1st]

I read a couple of threads about this but I just want to make sure that I got it.
Could someone please verify if my thought process is correct? Thanks!

Where is Tension the highest in a swinging pendulum?

Equilibrium

Because at equilibrium
Fnet = T - mg ------- since there is centripetal motion, Fnet = mv^/2
mv^2/2 = T - mg
T = mv^2 + mg

At a point other than equilibrium
T = mg cos(theta)

Since at equilibrium there is the centripetal force to be counted in it is where T is the highest. Also, at points other than equilibrium only a component of mg is taken into account (cos).

However, I am still not quite certain why can we not equal Fnet to centripetal force at points other than equilibrium?? Is it because it is not at constant speed?

Thank you!

I think it would be reasonable to say the tension T would be mv^2/r + mgcos(theta), because mgcos(theta) is the component of the gravitational force parallel to the pendulum string, and mv^2/r is the centripetal force needed. Using this, the tension would be greatest when the pendulum is at the bottom both because its v is highest at that point, and cos(0) = 1, so gravity is fully contributing. I think you can use the centripetal force equation even if the velocity isn't constant?

 

[pm1]

thank you guys!

yes, I guess I can use it but as ouroboros39 pointed out the v is going to be smaller at points other than equilibrium, thus it will be smaller. And may be at the apex because all the energy will have been converted into potential, no kinetic, then v=0 and then we just use mgcos(theta).

thank you!