Homework Help: 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.