What is the kinetic energy of an hour hand? Solution to this problem needs to ad

Question

What is the kinetic energy of an hour hand? Solution to this problem needs to address the following:

Consider both a wristwatch and Big Ben (the clock on the tower at the Palace of Westminster in London)

How did you determine the kinetic energy? What equations did you need? What values did you use for the variables? How did you determine those values?

How do the two kinetic energies compare to each other?

Is this kinetic energy responsible for powering kinetic watches?

Draw a cheap graph to illustrate. Thanks!

Explanation / Answer

The rotational kinetic energy for a rod of length l pivoting around its end is

  KE 0.5I^2 = 0.5(ml^2/3)*^2

The rotational velocity, , will be the same for all hour hands. The hour hand makes two rotations per day so that

= 2/(12*60*60) = 1.5*10^-4 rad/s

Now we need to estimate the masses and sizes of our hour hands. The hour hand on my wrist watch is about 7 mm long and 1 mm wide. It is probably 0.1 to 0.2 mm thick and made of aluminum or steel. Using an intermediate density we get

  m = V = (5 *10^3 kg/m^3 )(7 *10^-3 m)

  (10^-3 m)(1.5 *10^-4 m) = 5 *10^-6 kg

and I = ml^2/3 = (5 *10^-6 kg)(7 *10^-3 m)^2 = 8*10^-11 kgm^2

so that KE = 0.5*(8*10^-11 kgm^2)*(1.5*10^-4 rad/s)^2 = 10^-18 J

That is definitely not a lot. You were right not to worry too much about the precise factor in the moment of inertia.

Maybe Big Ben will have more energy. The face of Big Ben is more than two stories tall so that the minute hand appears to be about 4 m long and the hour hand a mere 2 m. The hands need to be about 0.3 m wide to be visible and probably need to be about 0.05 m thick to not bend. The clock is old enough that the original hands would have been made from iron or wood (aluminum being extremely expensive throughout most of the 19th century). Wooden hands would probably have been solid; iron (or steel) could have been hollow. A hollow steel hour hand with 5 mm thick walls would have 1/5 the mass of a solid steel hour hand and about twice the mass of a wooden one. Now let’s calculate:

  m = V = (8 *10^3 kg/m^3 )(2 m)(0.3 m)(10^-2 m) = 50 kg

and I = ml^2/3 = [(50 kg)(2 m)^2]/3 = 70 kg-m^2

so that KEbigben = 0.5*(70 kg-m^2 )(1.5 * 10^-4 rad/s)^2 = 10^-6 J

OK. So even the Big Ben hour hands have a negligible amount of kinetic energy

In fact, even the minute hands do not have much kinetic energy. The minute hand will have twice the length and thus twice the mass, giving eight times the rotational inertia. It will also have 12 times the angular velocity, giving 10^3 times as much kinetic energy. Thus, even Big Ben’s mighty minute hand only has a milliJoule of kinetic energy.   

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.