How Is the Angular Momentum of a Clock's Second Hand Calculated?

[Heat]

[SOLVED] Angular Momentum

Homework Statement

Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0cm and a mass of 6.00g. Take the second hand to be a slender rod rotating with constant angular velocity about one end.

The Attempt at a Solution

Clock hand: 15.0 cm --> .15m
Mass: 6.00 g --> .006kg

L=I(omega)

To get omega:
The second hand does one revolution in 60seconds.

so 1/60 = .01667 rev/s --> .10472 rad/s

To get I:

I = 1/3 ML^2

I = (.006)(.15)^2 = .000135

Now that we got I and omega,

L = (.000135)(.10472)
L = .0000141372

I would have thought that this procedure would have been right, but I got it incorrect.

Final Answer: L= _________ kg m^2/s

Please and thank you.

 

[Doc Al]

To get I:

I = 1/3 ML^2

I = (.006)(.15)^2 = .000135

You forgot to multiply by 1/3.

 

[Heat]

you have good eye. Thank you for your help. :)