How Does Pendulum Angle Affect Velocity?

[CollegeStudent]

Homework Statement

A mass of 200 g is attached to the end of a cord 1.20 m long. The cord is hung from a horizontal beam and the mass is free to swing making a pendulum. The mass is released from rest at an angle of 60° from vertical. What will be the velocity of the mass at an angle of 30°, and at an angle of 0°?

Homework Equations

F_c = m(v²/r)

The Attempt at a Solution

I figured that since it seems to be traveling in a half circle type of movement...I could use the fact that the length of the string would be the radius...so 1.20m

after that I did

ƩF_y = ma_y
mgcosθ = m(v²/r)
the m's cancel so
gcosθ = v²/r
v = √(rgcosθ)

And I figured that I could use that same equation to solve for each of the angles...i.e

= √((1.20(9.8cos60))
= √((1.20(9.8cos30))
= √((1.20(9.8cos0))

Would this be correct? Or am I totally off here?

 

[technician]

you are on the right track but the angles marked on your diagram are wrong !
You need to sort that out to make sense of your equations

 

[CollegeStudent]

you are on the right track but the angles marked on your diagram are wrong !
You need to sort that out to make sense of your equations

Right, sorry about that I'm just so used to doing it "from horizontal"

 

[technician]

You are on the right track.
A good check is to look at the extremes...when angle = 0, Cos0 = 1 and the velocity should be max...which it is !

 

[SammyS]

Homework Statement

A mass of 200 g is attached to the end of a cord 1.20 m long. The cord is hung from a horizontal beam and the mass is free to swing making a pendulum. The mass is released from rest at an angle of 60° from vertical. What will be the velocity of the mass at an angle of 30°, and at an angle of 0°?

Homework Equations

F_c = m(v²/r)

The Attempt at a Solution

View attachment 57778

I figured that since it seems to be traveling in a half circle type of movement...I could use the fact that the length of the string would be the radius...so 1.20m

after that I did

ƩF_y = ma_y
mgcosθ = m(v²/r)
the m's cancel so
gcosθ = v²/r
v = √(rgcosθ)

And I figured that I could use that same equation to solve for each of the angles...i.e

= √((1.20(9.8cos60))
= √((1.20(9.8cos30))
= √((1.20(9.8cos0))

Would this be correct? Or am I totally off here?

You're quite far off.

How do you get that mgcos(θ) is equal to the centripetal force? It's not.


mgcos(θ) = m v²/r can't possibly work, because for instance, at the instant that the mass is released, the velocity is zero, but θ = 60° .

Use conservation of energy.

 

[technician]

In these calculations it is better if you can use angular frequency (ω)
You will find the equations in SHM written as
displacement x = ASin(ωt)
velocity v = dx/dt = ωACos(ωt)
acceleration =dv/dt = -ω^2.ASin(ωt)

dont worry about this if you have not met it, you have done the right thing

 

[CollegeStudent]

You are on the right track.
A good check is to look at the extremes...when angle = 0, Cos0 = 1 and the velocity should be max...which it is !

Right, makes sense...the lowest point here would be the max velocity. Just like if the angle was 90° the velocity would = 0.

So okay then my equations would be correct?

= √((1.20(9.81cos60)) = 5.886m/s
= √((1.20(9.81cos30)) = 10.19m/s
= √((1.20(9.81cos0)) = 11.77m/s ( which would be the max velocity )

Look about right?

 

[CollegeStudent]

You're quite far off.

How do you get that mgcos(θ) is equal to the centripetal force? It's not.mgcos(θ) = m v²/r can't possibly work, because for instance, at the instant that the mass is released, the velocity is zero, but θ = 60° .

Use conservation of energy.

Oh..then they are not correct.

Well okay...assuming that this system is frictionless then W = 0

It goes from Gravitational P.E to Kinetic Energy so

mgh = 1/2mv²

m's cancel so

gh = 1/2v²

rearrange for v

V = √(2gh)

and h would be 1.20cosθ here?

 

[technician]

How do you get that mgcos(θ) is equal to the centripetal force? It's not.mgcos(θ) = m v2/r can't possibly work, because for instance, at the instant that the mass is released, the velocity is zero, but θ = 60° .

The angle θ = 60 for the pendulum diagram corresponds to (ωt) = 90 (the release point) in the equivalent circular motion and the amplitude A = LSinθ, ie A=LSin(ωt)

mgCosθ = mv^2/r = centripetal force is correct.
mgCosθ = ma where a = centripetal acceleration (v^2/r)...r = length of pendulum

 

[CollegeStudent]

How do you get that mgcos(θ) is equal to the centripetal force? It's not.


mgcos(θ) = m v2/r can't possibly work, because for instance, at the instant that the mass is released, the velocity is zero, but θ = 60° .

The angle θ = 60 for the pendulum diagram corresponds to (ωt) = 90 (the release point) in the equivalent circular motion and the amplitude A = LSinθ, ie A=LSin(ωt)

mgCosθ = mv^2/r = centripetal force is correct.
mgCosθ = ma where a = centripetal acceleration (v^2/r)...r = length of pendulum

While I'm not 100% on what's happening here (mainly because we haven't met the use of those equations...we are in fact going over conservation of energy in class...and what SammyS said about the release point being at 60° where the velocity = 0...then

mgcos(60) = m(v²/r)
some number = 0 ...so I actually don't see how this would work.

but then again I'm not sure how to add centripetal acceleration into that equation.

I'm 100% not sure here lol so I'll wait until you guys can come to a consensus about this one while I read up on it.

Thanks guys

 

[technician]

the source of the problem is that the angle you use to describe the pendulum is not the angle that occurs in SHM equations.
Your 60 is 90 (or 0) in the SHM equation

 

[CollegeStudent]

the source of the problem is that the angle you use to describe the pendulum is not the angle that occurs in SHM equations.
Your 60 is 90 (or 0) in the SHM equation

I think that your use of SHM stands for harmonic motion? (simple harmonic motion I think it was?)

While we haven't reached that yet in lecture...I remember reading about it...and if I remember correctly...what you're saying is that the fact that there is no angular displacement at 60° then it would = 0? Is that correct?

 

[SammyS]

How do you get that mgcos(θ) is equal to the centripetal force? It's not.


mgcos(θ) = m v2/r can't possibly work, because for instance, at the instant that the mass is released, the velocity is zero, but θ = 60° .

The angle θ = 60 for the pendulum diagram corresponds to (ωt) = 90 (the release point) in the equivalent circular motion and the amplitude A = LSinθ, ie A=LSin(ωt)

mgCosθ = mv^2/r = centripetal force is correct.
mgCosθ = ma where a = centripetal acceleration (v^2/r)...r = length of pendulum

Are you replying to me, since you quoted from my post ? (There is a QUOTE feature that is provided.)