Question

A pendulum consists of a massless rigid rod with a mass at one end. The other end is pivoted on a frictionless pivot so that it can turn through a complete circle. The pendulum is inverted, so the mass is directly above the pivot point, and then released. The speed of the mass as it passes through the lowest point is 6.0 m/s. If the pendulum undergoes small amplitude oscillations at the bottom of the arc, what will be the frequency of the oscillations?

Answer 1

Not knowing the center of gravity, lets look at energy.

Assume the distance from the piviot point to the center of mass is r.

Then the distance the mass fell is 2r, and the PE lost is 2rmg.the KE at the bottom is 1/2 m v^2, and they are equal

1/2 m v^2=2rmg
r= 1/4 v^2/g

so, period of pendulum= 2PI sqrt(r/g)
and frequncey is 1/Period

Answer 2

Is the formula r = (1/4)(v^2/g)? or (1/4 v^2)/g ?

Answer 3

period= 2PI sqrt (1/4 v^2/g^2)=2PI v/2g=PI v/g

frequency= 1/period= g/(PIv)

check that.

Answer 4

I am just confused on the way some of these are written out like in the period equation you just gave, is pi multiplied by v and then divided by g or is the whole quantity given by v/g multiplied by pi?

Answer 5

period= 2PI * sqrt(v/g)

Answer 6

So the first formula period=2pi*sqrt(r/g) is incorrect? I am just confused I am sorry

Answer 7

sorry about that, it is r/g, I am trying to clear this board too quickly.

Answer 8

So is the formula for r: r = (1/4)(v^2/g)? or (1/4 v^2)/g ? sorry I am asking so much

Answer 9

There is no difference.

a/b * c/d= ac/bd= ac/b * 1/d
1/4 v^2/g= 1/4 v^2/g= 1/(4g) * v^2