Question

The Parliament Building tower clock in London, has hour and minute hands with lengths of 2.70 m and 4.50 m and masses of 60.0 kg and 100 kg, respectively. Calculate the total angular momentum of these hands about the center point. Treat the hands as long, thin uniform rods.

Answer 1

Answer:

The total angular momentum of these hands is 1.20 kg m²/s.

Explanation:

Given that,

Length of hour hand = 2.70 m

Length of minute hand = 4.50 m

Mass of hour hand = 60.0 kg

Mass of minute hand = 100 kg

We need to calculate the total angular momentum

Using formula of angular momentum

$L=I_{k}\omega_{k}+I_{m}\omega_{m}$

$L=(\dfrac{ml_{h}^2}{3})\times\dfrac{2\pi}{T}+\dfrac{ml_{m}^2}{3}\times\dfrac{2\pi}{T}$

Where, $l_{k}$ = length of hour hand

$l_{m}$=length of minute hand

Put the value into the formula

$L=\dfrac{60.0\times(2.70)^2}{3}\times\dfrac{2\pi}{12\times3600}+\dfrac{100\times(4.50)^2}{3}\times\dfrac{2\pi}{3600}$

$L=1.199 = 1.20\ kg m^2/s$

Hence, The total angular momentum of these hands is 1.20 kg m²/s.

Answer 2

Answer:

Explanation:$1.1993 kg-m^2/s$

Given

length of hour hand is 2.7m

length if minute hand is 4.5m

And Moment of inertia of a rod about its one of a end $I=\frac{ml^2}{3}$

$\omega$for hour hand is

$\omega =\frac{2\pi }{12\cdot 3600} rad/s$

Angular speed for minute hand $=frac{2\pi }{3600}$

Angular moment(L) is given $I\omega$

$L_{minute}=\frac{100\times 4.5^2}{3}\times frac{2\pi }{3600}$

$L_{minute}=1.178$

For hour hand

$L_{hour}=\frac{ml^2}{3}\omega$

$L_{hour}=\frac{60\times 2.7^2}{3}\times \frac{2\pi }{12\cdot 3600}$

$L_{hour}=0.0212 kg-m^2/s$

$L_{total}=1.1993 kg-m^2/s$