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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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Most popular questions from this chapter

A bicycle has wheels of radius \(0.32 \mathrm{m}\). Each wheel has a rotational inertia of \(0.080 \mathrm{kg} \cdot \mathrm{m}^{2}\) about its axle. The total mass of the bicycle including the wheels and the rider is 79 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?
A crustacean (Hemisquilla ensigera) rotates its anterior limb to strike a mollusk, intending to break it open. The limb reaches an angular velocity of 175 rad/s in \(1.50 \mathrm{ms} .\) We can approximate the limb as a thin rod rotating about an axis perpendicular to one end (the joint where the limb attaches to the crustacean). (a) If the mass of the limb is \(28.0 \mathrm{g}\) and the length is \(3.80 \mathrm{cm}\) what is the rotational inertia of the limb about that axis? (b) If the ex-tensor muscle is 3.00 mm from the joint and acts perpendicular to the limb, what is the muscular force required to achieve the blow?
Verify that the units of the rotational form of Newton's second law [Eq. (8-9)] are consistent. In other words, show that the product of a rotational inertia expressed in \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular acceleration expressed in \(\mathrm{rad} / \mathrm{s}^{2}\) is a torque expressed in \(\mathrm{N} \cdot \mathrm{m}\).
A mechanic turns a wrench using a force of \(25 \mathrm{N}\) at a distance of \(16 \mathrm{cm}\) from the rotation axis. The force is perpendicular to the wrench handle. What magnitude torque does she apply to the wrench?
The pull cord of a lawnmower engine is wound around a drum of radius $6.00 \mathrm{cm} .\( While the cord is pulled with a force of \)75 \mathrm{N}$ to start the engine, what magnitude torque does the cord apply to the drum?
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