# Clock hand's rotation

1. Feb 22, 2005

### UrbanXrisis

The hour hand and the minute hand of a clock are 2.7m long and 4.5m long and have masses of 60kg and 100kg respectively. Calculate the total rotational kinetic energy of the two hands about the axis of rotation.

$$w=circumfrence/time=2rpi/t=2(2.7m)pi/86400s$$
$$w=1.96x10^{-4}rad/s$$
$$w=circumfrence/time=2rpi/t=2(4.5m)pi/3600s$$
$$w=0.00785rad/s$$
$$energy=[m_1r_1^2w_1^2]+[m_2r_2^2w_2^2]$$
$$energy=[60kg*(2.7m)^2*(1.96x10^{-4}rad/s)^2]+[100kg*(4.5m)^2*(0.00785rad/s)^2]$$
$$energy=0.1248J$$

is this the total energy of the two hands?

Last edited: Feb 22, 2005
2. Feb 23, 2005

### scholar

Firstly, your angular velocity $$\omega$$ is a measure of radians per second. This means that it is independent of the radius. So your forumula for each angular velocity should be $$\omega=2\pi/T$$.

Also, the rotational inertia of a thin rod rotating about an axis through one end and perpendicular to length (IIRC) should be $$I=1/3MR^2$$.

I think that's about right, check it out and tell me if you see any mistakes. :)

3. Feb 23, 2005

### UrbanXrisis

$$w=2pi/t=2pi/86400s$$
$$w=7.27x10^{-5}rad/s$$
$$w=2pi/t=2pi/3600s$$
$$w=0.001745rad/s$$
$$energy=[(1/3)m_1r_1^2w_1^2]+[(1/3)m_2r_2^2w_2^2]$$
$$energy=[(1/3)*60kg*(2.7m)^2*(7.27x10^{-5}rad/s)^2]+[(1/3)*100kg*(4.5m)^2*(0.001745rad/s)^2]$$
$$energy=0.0020569J$$

is this correct? and why is it so small?

4. Feb 23, 2005

### dextercioby

Because you used the WRONG periods:the one for a (mean solar) DAY and an hour,instead of an hour and one minute,respectively.:tongue:

Daniel.

5. Feb 23, 2005

### UrbanXrisis

it takes 12 hrs for the hour hand to rotate 360 degrees...(43200s)
it takes 1 hr for the minute hand to rotate 360 degrees...(3600s)

$$w=2pi/t=2pi/43200s$$
$$w=1.45x10^{-4}rad/s$$
$$w=2pi/t=2pi/3600s$$
$$w=0.001745rad/s$$
$$energy=[(1/3)m_1r_1^2w_1^2]+[(1/3)m_2r_2^2w_2^2]$$
$$energy=[(1/3)*60kg*(2.7m)^2*(1.45x10^{-4}rad/s)^2]+[(1/3)*100kg*(4.5m)^2*(0.001745rad/s)^2]$$
$$energy=0.002058J$$

is this correct?

6. Feb 23, 2005

### dextercioby

Yes,now it looks good.

Daniel.

7. Feb 23, 2005

### ramollari

UrbanXrisis, you are simplifying the clock hands as thin rods, right? Then where did you derive the formula for rotational KE: $$KE_{rot} = 1/3mr^2\omega^2$$? If the moment of inertia for a thin rod is 1/12 then $$KE_{rot} = 1/24mr^2\omega^2$$.

Last edited: Feb 23, 2005
8. Feb 23, 2005

### dextercioby

Nope,the moment of inertia is $I=\frac{1}{3}ml^{2}$...Don't believe me,see here:http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Scroll down till u hit a link with moment of inertia...

Daniel.

P.S.I hope u see where your assumption went wrong... :uhh:

Last edited: Feb 23, 2005
9. Feb 23, 2005

### ramollari

Aaaa, I see, then the coefficient is 1/2*1/3 = 1/6.

10. Feb 23, 2005

### dextercioby

No,it is $$\frac{1}{3}$$

Daniel.

11. Feb 23, 2005

### ramollari

Are you kidding dude? I'm now talking about the rotational KE not about the moment of inertia. If $$I = 1/3mr^2$$, then
$$KE_{rot} = 1/2I\omega^2 = 1/2(1/3mr^2)\omega^2 = 1/6mr^2\omega^2$$, and not the result from UrbanXrisis:
$$KE_{rot} = 1/3mr^2\omega^2$$

12. Feb 23, 2005

### dextercioby

Yes,then i agree...Didn't figure out what u meant.

Daniel.