# Finding Rotational Kinetic Energy Of A Clock [closed]

The problem I am working on is:

Big Ben, the Parliament tower clock in London, has an hour hand 2.70 m long with a mass of 300 kg, and a minute hand 4.20 m long with a mass of 100 kg (see figure below). Calculate the total rotational kinetic energy of the two hands about the axis of rotation. (You may model the hands as long, thin rods rotated about one end. Assume the hour and minute hands are rotating at a constant rate of one revolution per 12 hours and 60 minutes, respectively.)

(Converted) Angular Speed of Clock Hands:

Hour Hand $1.45\cdot10^{-4}~rad/s$

Minute Hand $1.75\cdot10^{-3}~rad/s$

Rotational Moment of Inertia:

Hour Hand $I=1/3(300~kg)(2.70~m)^2=729~kg\cdot m^2$

Minute Hand $I=1/3(100~kg)(4.20~m)^2=243~kg\cdot m^2$

Rotational Kinetic Energy:

$K_{rot}=1/2(729~kg\cdot m^2)(1.45\cdot10^{-4}~rad/s)^2+1/2(243~kg\cdot m^2)(1.75\cdot10^{-3}~rad/s)^2$

When I calculate this, it comes out incorrect, what has happened?

## closed as off-topic by ACuriousMind♦Feb 9 '17 at 0:11

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• It looks like your angular speed of the minute hand is an order of magnitude off. The rest of it looks good. – user16454 Dec 1 '12 at 19:56

## 2 Answers

The angular speed of the minute hand is actually $1.75\times 10^{-3} rad/s$. Always do a quick consistency check on your calculations. The hour hand is moving at roughly 1/10 angular speed so the orders of magnitude should differ by ~1.

• Even with the corrected values, I still get a wrong answer. The answer yielded from the calculations above is $3.80\cdot 10^{-4} J$, which is wrong. – Mack Dec 2 '12 at 13:31
• Instead of using 4.20(m) for the minute hand you used 2.70(m) – descheleschilder Feb 9 '17 at 0:16

When you multiplied you minute hand I, it should have come out to be 588kgm^2 not 243 kgm^2