What is the rotational kinetic energy of a clock's hour and minute hands?

[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?

 

[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. :)

 

[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?

 

[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.

Daniel.

 

[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?

 

[dextercioby]

Yes,now it looks good.

Daniel.

 

[ramollari]

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?

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.

 

[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...

 

[ramollari]

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...

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

 

[dextercioby]

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

Daniel.

 

[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

 

[dextercioby]

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


Daniel.