Assume that you proved that the period of the pendulum does not depend on its ma
ID: 1460352 • Letter: A
Question
Assume that you proved that the period of the pendulum does not depend on its mass. Which of the following parameters do you think you can account for the prediction of the period of the pendulum using dimensional analysis techniques?
L - pendulum length
I - moment of inertia of the pendulum bob
2 - correct numerical constant for the pendulum period
g - acceleration of free fall
- launching angle
F - force of friction in the air
Write down the resulting expression for the period of the pendulum in terms of whichever parameters,L,g,0,I,2 you chose.
T ~ _______________
Explanation / Answer
here,
We are looking for an equation that gives T the period of a pendulum that uses all or some of the following properties, with their units / dimensions in []:
(1) length L in [m]
(2) mass m in [kg]
(3) gravitational field g in [m / s^2]
(4) theta has unit radians , but this can not be used because it can not be cancel out
(5) T also does not contain kg , so the force can also not be used
In other words, we are trying to figure out how T might depend on these 3 quantities just by looking at the dimensions.
T is in [s]
The only factor that has [s] in its dimensions is g [m / s^2]. There is no other factor that has time in it, therefore g must be involved in the equation for T.
You can see that g must be in the denominator to get the [s] into the numerator for T; you would also have to take the square root of 1/g to turn the s^2 into an [s}.
Therefore, we know that
T involves sqrt(1 / g) = sqrt ( [s^2] / [m])
Now, we have the seconds in the numerator, but also a sqrt(1/[m]) that needs to be cancelled out.
So we need to take the square root of some factor that has a dimension of [m]; fortunately, we have that in L so we can say that:
T = k * sqrt( L / g), where k is a dimensionless constant.
The expression on the right side of , when calculated out, gives the proper dimension for the left side T. So we've found the answer to the question
therefore 2*pi is correct numerical constant for the pendulum period
so , T = 2*pi * sqrt( L/g)