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Chapter 8: Problem 103

A large clock has a second hand with a mass of \(0.10 \mathrm{kg}\) concentrated at the tip of the pointer. (a) If the length of the second hand is $30.0 \mathrm{cm},$ what is its angular momentum? (b) The same clock has an hour hand with a mass of \(0.20 \mathrm{kg}\) concentrated at the tip of the pointer. If the hour hand has a length of \(20.0 \mathrm{cm},\) what is its angular momentum?

Short Answer

Expert verified
Answer: The angular momentum of the second hand is 9.45 × 10^{-4} kg * m^2/s, and the hour hand is 1.16 × 10^{-6} kg * m^2/s.

Step by step solution

01

Find the moment of inertia of the second hand and the hour hand

To find the moment of inertia, use the formula I = m * r^2. For the second hand: I = 0.10 kg * (0.30 m)^2 = 0.009 kg * m^2. For the hour hand: I = 0.20 kg * (0.20 m)^2 = 0.008 kg * m^2.
02

Calculate the angular speed of the second hand and the hour hand

To find the angular speed, use the formula ω = 2π / T. The second hand takes 60 seconds to complete one rotation, so T = 60 seconds for the second hand. For the hour hand, it takes 12 hours (43200 seconds) to complete one rotation, so T = 43200 seconds for the hour hand. For the second hand: ω = 2π / 60 = 0.105 rad/s. For the hour hand: ω = 2π / 43200 = 1.45 × 10^{-4} rad/s.
03

Find the angular momentum of the second hand and the hour hand

Now that we have the moment of inertia (I) and angular speed (ω) for both hands, we can find the angular momentum (L) using the formula L = I * ω. For the second hand: L = 0.009 kg * m^2 * 0.105 rad/s = 9.45 × 10^{-4} kg * m^2/s. For the hour hand: L = 0.008 kg * m^2 * 1.45 × 10^{-4} rad/s = 1.16 × 10^{-6} kg * m^2/s. The angular momentum of the second hand is 9.45 × 10^{-4} kg * m^2/s, and the hour hand is 1.16 × 10^{-6} kg * m^2/s.

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