Pendulum Speed at Different Heights

AI Thread Summary
The discussion focuses on calculating the speed of a pendulum at different heights using energy conservation principles. The potential energy (PE) at the highest point converts to kinetic energy (KE) at the lowest point, allowing the derivation of speed using the equation v^2 = 2gh. The initial calculation for the pendulum's speed at the lowest point, 1.4 m/s, is confirmed correct, and the speed at 1 cm above the lowest point is calculated to be 1.33 m/s. The importance of understanding the relationship between potential and kinetic energy is emphasized, along with the benefit of performing algebraic manipulations before substituting numerical values. Overall, the calculations align with the expected physics outcomes, reinforcing the concepts of energy conservation.
xasuma
Messages
8
Reaction score
0

Homework Statement


The pendulum shown (http://puu.sh/chwMZ/90f4b96fb1.png ) swings freely, turning around at a height of 10 cm above the lowest point in its swing. It has an unknown mass, but air resistance can be ignored.

a) Use energy conservation to find its speed at the lowest point of its swing.
b)What is the pendulum speed when it is at a height of only 1 cm above its lowest point?

Homework Equations


PE=mgh
V^2 = (2PE)/m

The Attempt at a Solution


I know I need to use the two equations above. But I don't know how. I can't understand how to do this problem. Therefore there is no attempted solution. I am not asking for a solution, I want to learn, but can someone explain to me what to do here?
 
Last edited by a moderator:
Physics news on Phys.org
Your second equation has basically already done most of the work for you, but you have to understand how to use it. In fact, I think that this equation will be more confusing to you than it will be helpful, so let's start with some simpler concepts.

Do you understand the concepts of Potential Energy and Kinetic Energy, and what IS this concept of "Energy conservation"?
 
This is my understanding of it.
Potential energy is stored energy which has the potential to be released.
Kinetic energy is the energy an object has while in motion.
And energy conservation means that energy can change from potential to kinetic (or others) , since it can't be destroyed nor created.

I actually came up with the second equation by myself (I mean it wasn't given to me) .The problem I have is the mass. If the mass of the pendulum ball was given to me this would be no problem, but I can't get my head around it being unknown. I know that if air resistance is ignored, the acceleration would be 9.8m/s (if it were going straight down) but with the pendulum motion I get confused. I thought of the tension of the pendulum as well, drew a body diagram for the ball , but I just keep getting more confused.

Maybe the mass =1 , so it won't be important in this equation?
In that case I would say:
PE=9.8*0.1= 0.98N
v^2=(2*0.98)/1 = 1.96
v=1.4 m/sI don't really know though.
 
PE=9.8*0.1= 0.98N
PE has units of energy: "Joules" (J); "Newtons" (N) are units of force.
You seem to have written PE = gh ... gh has units of m^2/s^2 - not N or J.
This is a clue and should clear up your confusion.

This is one place where doing all the algebra before you put numbers into the equations really pays off.
You've already figured out that all the gravitational PE lost at the top of the arc goes into kinetic energy at the bottom.
So just write out <the formula for kinetic energy> = <the formula for potential energy> and then cancel terms.
 
I see,
so I got:
1/2mv^2 = mgh , solve for v (speed)

v^2 = 2gh
v^2 = 2*9.8*0.1
v= 1.4 m/s

---------------------

Now for B. The speed of the pendulum 1 cm above the lowest point.
I just changed 'h' from 0.1 to 0.09, did the same calculation as above and got:

v= 1.33 m/s

Did I do it right? I believe I get it :)
 
It came out with the same answer - just for better reasons.
Well done - the thing about laying it out clearly like that is you end up with more confidence in your results.
Also notice that doing the algebra first gives you an equation that you can use in the next question with no extra work.

You can reality check your answers by thinking about the physics ... the v you calculated before was at the bottom of the swing - do you expect that the speed is faster or slower higher up? Is what you got consistent with that?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top