How to calculate velocity in this question

[Gardalay]

Homework Statement

a pendulum of mass 1.0kg and length 1.0m is swung from a height of 0.110m above its equilibrium position. Find the tension of the pendulum string at the bottom of its swing.

Formula:
T=mv^2/r+mg

How do i get the V^2?

 

[Gardalay]

I'm trying to find the tension...I don't have the # for T. do i just use v^2=Vi^2+2ad? and make velocity initial 0?

 

[pat666]

no its periodic motion not projectile

 

[Gardalay]

what is the # for T and u?

 

[pat666]

do you mean coefficient by #?
the velocity at the bottem is given by -wAsin(wt) and from x=Acos(wt) you know that wt =1 since velocity is max at the bottem

 

[Gardalay]

yeah...what's the coefficient for T and u

 

[pat666]

dont worry about what i said before it was for something else,,,, I get v at the bottom as 0.3445m/s, is the T in the formula up the top period or tension just asking because I've never seen it before.

 

[Gardalay]

it's for tension

 

[pat666]

do you get the right answer when you sub the v i found into your equation? answer 9.93N

 

[Gardalay]

yeah i got that answer (I don't know if it was right though since I don't have the answer key). How did you find V again? I have no idea what '-wAsin(wt) and from x=Acos(wt)' was.

 

[graphene]

use energy conversion to get mgh = (1/2) mv^2. v is the speed at the bottom.
and then T = mg + mv^2 / r

 

[Gardalay]

that equation doesn't give me a velocity of 0.3445m/s.
mgh=1/2mv^2
(1kg)(9.8m/s^2)(1.1m)=1/2(1kg)(v^2)
=4.64m/s

Is it 4.64 m/s?

 

[pat666]

t my procedure should yield exactly the same answer as using energy methods, T=2pisqrt(L/G) T=2.0 f= 0.498 w=3.13 v=wA v=0.244m/s , anyone know why that doesn't work?

 

[pat666]

from energy methods v=1.469 but I am not sure i agree with it since your acceleration (9.81) is for free fall but this is not the case for pendulums which is why our answer don't line up.

there is a component of that acceleration in the x direction, I am quietly confident that you can't use energy methods to solve this now,,,, more proof there is a horizontal displacement that looses energy. in conclusion you can't use energy methods!

 

[Gardalay]

I have no idea, can someone clarify this please?

 

[rl.bhat]

that equation doesn't give me a velocity of 0.3445m/s.
mgh=1/2mv^2
(1kg)(9.8m/s^2)(1.1m)=1/2(1kg)(v^2)
=4.64m/s

Is it 4.64 m/s?

The above equation should be

(1kg)(9.8m/s^2)(0.11m)=1/2(1kg)(v^2)

 

[pat666]

Equations you need: note w is onega or angular velocity
X(t)=Acos(wt)
now you know that at equilibrium velocity is a maximum. so at this point A=x
cos 0 =1 therefore wt=1
T(period)=2pi*sqrt(L/g)
T=2 , f=t^-1 f=0.498Hz , w =2pif, w=3.13
v(t)=wAsin(wt) as said before, wt is 1
vmax=wa
vmax=3.13*0.11
v=0.344m/s

need more clarification? ps as i said before you can't use energy methods

 

[pat666]

rl.bhat do you think that that will give the correct velocity at the bottom? ie do you think you can use energy methods for this?

 

[rl.bhat]

rl.bhat do you think that that will give the correct velocity at the bottom? ie do you think you can use energy methods for this?

In the simple pendulum, for small oscillations, the total energy remains the constant.

 

[pat666]

The energy in the system would certainly remain constant. I don't understand how mgh=1/2mv^2 would possibly work though. the velocity that gives would be the velocity if it was dropped straight down with nothing attached. the velocity i calculated is significantly different than the one using energy and this oscillation would be called a "small oscillation" can you see anything wrong with what i did??

 

[graphene]

lemme elaborate the enrgy method

during the pendulum's motion, there is conversion of potential energy into kinetic energy. the difference in potential energies between the end point(top) & the centre(bottom) gets converted to kinetic enrgy that the pendulum has at the bottom.

potential energy = mgh1 at the top & mgh2 at the bottom.

energy at the extreme position = mgh1
energy at the bottom position = mgh2 + (1/2)mv^2.

equate & u get mg*0.11 = (1/2)mv^2.


the other method a few are speaking about, is really tedious. the energy method is v.simple & straight-forward. it just requires the energies at the two paths & does not bother about the motion between the top paths, unlike the other method

 

[pat666]

i f know the energy method, my problem is that there is a component of g in the x direction unless it is released perpendicularly. also if the energy method could be used accurately here why don't the two answers line up, either i screwed something up in my calculations (entirely possible) or the energy method (the way you are using it) doesn't hold true for pendulums.

 

[graphene]

so u want to compose g into - the radial component (pointing outward along the sting) & the theta component (pointing perpendicular to the string in the direction of motion).

the theta component is g sin(theta).
this is the acceleration experienced by the pendulum. u can very well use it to find the velocity at the bottom. but clearly this is much more complicated.

 

[pat666]

but much more accurate and true. i think saying that g acts straight down on the mass is inaccurate enough to say the answer is just wrong! again unless I am missing some fundamental principle.