MCAT Physics: Find Rod Length After 20% Decrease in Pendulum Period

  • Thread starter Thread starter caviar
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
The discussion revolves around calculating the decrease in rod length of a pendulum after a 20% reduction in its period. The relevant equation for the period is T = 2π√(l/g), and participants clarify how to rearrange this equation to express length (l) in terms of period (T). The confusion arises over the correct algebraic manipulation, with one participant asserting that the correct formula is l = 4π²gT². The conclusion drawn is that a 20% decrease in period results in a 36% decrease in rod length, leading to some skepticism about the accuracy of this calculation. Overall, the thread highlights the importance of understanding algebraic transformations in solving physics problems.
caviar
Messages
2
Reaction score
0

Homework Statement


The length of the rod of a certain pendulum is decreased, and the period then decreases by 20%. By how much was the rod length decreased?


Homework Equations


T=2π√(l/g)
T=period
l=length of the rod
g=acceleration due to gravity


The Attempt at a Solution


I know how to solve the problem, I just don't know how to do the math to solve for l (length).
 
Physics news on Phys.org
If you know how to solve the problem, what is then the difficulty that you have?

you have the relevant equation that gives you the period as function of length. But you want to have an expression for the length as function of the period. And then you want an expression that gives you the length for a 20% longer period.
 
I don't understand what you are asking.

Are you saying you don't know how to solve T = 2pi*sqrt(l/g) for l?
Because this is basic algebra: l = g*(T/2pi)^2
 
Sorry for the confusion, yeah I forgot how to do basic algebra and don't understand why this is the answer: l=4π^2gT^2, and not this l=gT^2/4π^2 (what you said and what I think it is too)?

Either way you can still find the answer because when you multiply T by 0.80, you multiply l by 0.64 (0.8^2). So, the length (l) decreases by 36% (1-0.64).
 
Mm, I wouldn't trust that. Must be some sort of misprint or mistake...unless my 6th grade teacher was wrong all this time :o
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top