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Answer :
Given the lengths and masses of the clock hands, we calculated the moment of inertia for each. Using their respective rotational speeds, we computed the rotational kinetic energy. The total rotational kinetic energy of the clock hands is approximately 1.08 J.
Calculating the Rotational Kinetic Energy of Clock Hands
To calculate the total rotational kinetic energy of the hour and minute hands of a tower clock. Given the lengths and masses of the hands, we can model them as long thin rods rotating about one end.
1. Moment of Inertia (I) for a thin rod rotating about one end: I = (1/3) * m * L².
2. For the hour hand: mh = 53 kg, Lh = 2.69 m
Ih = (1/3) * 53 kg * (2.69 m)² = 127.18 kg*m².
3. For the minute hand: mm = 103 kg, Lm = 4.55 m
Im = (1/3) * 103 kg * (4.55 m)² = 715.59 kg*m².
4. Rotational speed:
hour hand: ωh = 2π/(12*3600) rad/s = 1.45 × 10-4 rad/s
minute hand: ωm = 2π/3600 rad/s = 1.74 × 10-3 rad/s
5. Rotational kinetic energy (KErot) formula: KErot = (1/2) * I * ω².
6. For the hour hand:
KErot,h = (1/2) * 127.18 kg*m² * (1.45 × 10-4 rad/s)² = 1.34 × 10-3 J
7. For the minute hand:
KErot,m = (1/2) * 715.59 kg*m² * (1.74 × 10-3 rad/s)² = 1.08 J
8. Total rotational kinetic energy:
KE = KErot,h + KErot,m = 1.34 × 10-3 J + 1.08 J ≈ 1.08 J
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