The hour and minute hands of a tower clock like Big Ben in London are 2.62 m and 4.5 m long and have masses of 66.6 kg and 99 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
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Angular kinetic energy E in Joules
E = ½Iω²
ω is angular velocity in radians/sec
1 radian/sec = 9.55 rev/min
I is moment of inertia in kg•m²
I = cMR²
M is mass (kg), R is radius (meters)
c = ⅓ for a rod around its end, R = length
For minute hand
I = (⅓)(99)(4.5)²
ω = 1 rev/hour = 1 rev/3600sec = 2π / 3600 = π/1800 rad/s
KE = ½(⅓)(99)(4.5)²(π/1800)²
For hour hand
I = (⅓)(66.6)(2.62)²
ω = 1 rev/12hour = 1 rev/(12•3600sec) = 2π / 12•3600 = π/21600 rad/s
KE = ½(⅓)(99)(4.5)²(π/21600)²
Do the calc and add the two.
E = ½Iω²
ω is angular velocity in radians/sec
1 radian/sec = 9.55 rev/min
I is moment of inertia in kg•m²
I = cMR²
M is mass (kg), R is radius (meters)
c = ⅓ for a rod around its end, R = length
For minute hand
I = (⅓)(99)(4.5)²
ω = 1 rev/hour = 1 rev/3600sec = 2π / 3600 = π/1800 rad/s
KE = ½(⅓)(99)(4.5)²(π/1800)²
For hour hand
I = (⅓)(66.6)(2.62)²
ω = 1 rev/12hour = 1 rev/(12•3600sec) = 2π / 12•3600 = π/21600 rad/s
KE = ½(⅓)(99)(4.5)²(π/21600)²
Do the calc and add the two.