The hour and minute hands of a tower clock like Big Ben in London are 2.79 m and 4.44 m long and have masses of 58.2 kg and 90 kg, respectively. Calculate the total rotational kinetic energy of the two hands about the axis of rotation. Model the hands as long thin rods. Answer in units of J.

Answer: 895.85 x 10^-6 J or 8.96 x 10^-4 J

Explanation:

Angular kinetic energy E in Joules

E = ½Iw^2

W is angular velocity in radians/sec

1 radian/sec = 9.55 rev/min

I is moment of inertia in kgm^2

I = cMR^2

M is mass (kg), R is radius (meters)

c = 1/3 for a rod around its end, R = length

For minute hand

I = (1/3) (90) (4.44) ^2 = 0.33 x 90 x 19.7136 = 585.49

w = 1 rev/hour = 1 rev/3600sec = 2pi / 3600 = pi/1800 rad/s

KE = 1/2 (1/3) (90) (4.44) ^2 (pi/1800) ^2 = 0.5 x 0.33 x 90 x 19.7136 x 3.05 x 10^-6

KE = 0.00089 J

For hour hand

I = (1/3) (58.2) (2.79) ^2 = 0.33 x 58.2 x 2.79^2 = 149.5

w = 1 rev/12hour = 1 rev / (12x3600sec) = 2pi / 12x3600 = pi/21600 rad/s

KE = 1/2 (1/3) (90) (4.44) ^2 (pi/21600) ^2 = 0.5 x 0.33 x 90 x 19.7136 x 2.12 x 10^-8

KE = 5.85 x 10^-6 J

Therefore total kinetic energy = 895.85 x 10^-6 J