Total angular momentum of hands of Big Ben

[DarkerStorm]

Hello I just can't seem to get this problem right

Homework Statement

The hour and minute hands of Big Ben in London are 2.6 m and 4.58 m long and have masses of 68.6 kg and 97 kg respectively. Calculate the total angular momentum of the minute and hour hand about the center point. Treat the hand as long, thin rod. Treat “into the clock” as the positive direction. Answer in units of kg · m2/s.

Homework Equations

$$\omega = \frac{ 2\pi }{T}$$

$$I = \frac{1}{3} M R^2$$
$$I = \frac{1}{12} M R^2$$?

$$L = I \omega$$

The Attempt at a Solution

Mmin= 68.6
Mhour = 97

lmin = 2.6
lhour = 4.58

$$\omega min = \frac{2\pi}{3600}$$
$$\omega hour = \frac{2\pi}{86400}$$

$$Imin = \frac{1}{3} Mmin lmin^2$$ = 154.579
$$Ihour = \frac{1}{3} Mhour lhour^2$$ = 678.237

$$Lmin = Imin \omega min$$ = 0.269791
$$Lhour = Ihour \omega hour$$ = 0.0493228

 

[CompuChip]

Your calculation looks all right, so probably you've just made a little mistake somewhere.

Are you sure you aren't switching the minute and hour hands somewhere halfway? The hour hand is 2.6 m and 68.6 kg and the minute hand is 4.58 m and 97 kg. Are you using that consistently?

You are giving two formulas for the moment of inertia, with 1/3 and 1/12 in front. Can you explain the difference?

Also, don't forget your units, especially in the final answer (what is the unit of angular momentum?)

 

[DarkerStorm]

Thanks I found out where I was wrong now, I was switching the mass numbers of minutes and hour hands.