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Chapter 9: Problem 50

Two thin rods of length \(L\) are rotating with the same angular speed \(\omega(\) in \(\mathrm{rad} / \mathrm{s})\) about axes that pass perpendicularly through one end. Rod \(\mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{kg}\) attached to its free end. Rod B has a mass of \(0.66 \mathrm{kg}\), which is distributed uniformly along its length. The length of each rod is \(0.75 \mathrm{m},\) and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod \(A\) with its attached particle and of rod \(B\).

Short Answer

Expert verified
Rod A's kinetic energy is 3.273 J; Rod B's is 1.091 J.

Step by step solution

01

Understanding the System

We have two rods of length \(L = 0.75\,\mathrm{m}\) rotating with angular speed \(\omega = 4.2\,\mathrm{rad/s}\). Rod A is massless with a particle of mass \(m = 0.66\,\mathrm{kg}\) at the end, while Rod B has a uniform mass distribution of \(0.66\,\mathrm{kg}\). We are tasked with finding their kinetic energies.
02

Calculate Kinetic Energy for Rod A

For Rod A, all the mass is at the end where the particle is. The kinetic energy \(K_A\) can be calculated using the formula for rotational kinetic energy \(K = \frac{1}{2} I \omega^2\), where the moment of inertia \(I\) is \(mL^2\) for a point mass at a distance \(L\) from the axis. So, \(I = 0.66 \times (0.75)^2\,\mathrm{kg \cdot m^2}\). Substitute \(I\) and \(\omega\) to find \(K_A\).
03

Plug in Values for Rod A

Substitute \(I = 0.66 \times (0.75)^2 = 0.37125\,\mathrm{kg \cdot m^2}\) and \(\omega = 4.2\,\mathrm{rad/s}\) into the kinetic energy formula: \(K_A = \frac{1}{2} \times 0.37125 \times (4.2)^2\). Calculate to get \(K_A\).
04

Calculate Kinetic Energy for Rod B

For Rod B, the mass is uniformly distributed, so we use the formula \(I = \frac{1}{3} m L^2\) for a rod about an end. Thus, \(I = \frac{1}{3} \times 0.66 \times (0.75)^2\,\mathrm{kg \cdot m^2}\). Substitute \(I\) and \(\omega\) into the kinetic energy formula to find \(K_B\).
05

Plug in Values for Rod B

Substitute \(I = \frac{1}{3} \times 0.66 \times (0.75)^2 = 0.12375\,\mathrm{kg \cdot m^2}\) and \(\omega = 4.2\,\mathrm{rad/s}\) into the kinetic energy formula: \(K_B = \frac{1}{2} \times 0.12375 \times (4.2)^2\). Calculate to get \(K_B\).
06

Computation of Kinetic Energies

Calculate \(K_A = \frac{1}{2} \times 0.37125 \times (4.2)^2 = 3.273\,\mathrm{J}\). Then, calculate \(K_B = \frac{1}{2} \times 0.12375 \times (4.2)^2 = 1.091\,\mathrm{J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
Imagine trying to spin a rod that's held at one end. Now, depending on how the mass is spread along the rod, it could be easier or harder to spin. This is exactly what the "Moment of Inertia" helps us understand. It essentially measures how difficult it is to change the rotational motion of an object.

For different shapes and mass distributions, the moment of inertia is calculated using different formulas. For Rod A, all the mass (0.66 kg) is at the end, treated like a point mass, and the formula is simple: \[ I = mL^2 \]where \(m\) is mass and \(L\) is the length from the rotation point. However, for Rod B, the mass is spread uniformly. Here, the formula is slightly different to account for this even spread:\[ I = \frac{1}{3}mL^2 \]Remember, larger values of moment of inertia mean it's harder to spin or stop spinning, which is crucial when dealing with objects in motion.
Angular Speed
Think of angular speed as how fast something spins around a central point. It's like how quickly the hands of a clock move, measured in radians per second (rad/s). In this exercise, both rods rotate with the same angular speed \(\omega = 4.2 \,\mathrm{rad/s}\).

Angular speed is important because it tells us how quickly an object is rotating. When combined with the moment of inertia, it helps us calculate rotational kinetic energy. The formula used is:\[ K = \frac{1}{2} I \omega^2 \]where \(I\) represents the moment of inertia, and \(\omega\) is the angular speed. Notice how angular speed is squared in the formula, showing that even small changes in \(\omega\) have a big impact on kinetic energy.
Uniform Mass Distribution
Uniform mass distribution means that the mass of an object is spread out evenly throughout its volume. For Rod B, this means that every segment of the rod has the same mass. This distribution affects how the rod rotates and its moment of inertia.

The concept drastically changes the approach we take in physics calculations. With a uniform distribution, we use specific formulas to find properties like moment of inertia because the mass isn't concentrated at a single point but rather spread across the object's length.
  • Uniform distribution requires the use of \(I = \frac{1}{3} m L^2\).
  • It results in lower moment of inertia compared to a mass at the end, hence less resistance to change in rotational speed.
For Rod B, having 0.66 kg uniformly along 0.75 m gives it smaller rotational resistance compared to Rod A, where the mass was concentrated at the end.

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Most popular questions from this chapter

Two disks are rotating about the same axis. Disk A has a moment of inertia of \(3.4 \mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular velocity of \(+7.2 \mathrm{rad} / \mathrm{s} .\) Disk \(\mathrm{B}\) is rotating with an angular velocity of \(-9.8 \mathrm{rad} / \mathrm{s}\). The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of \(-2.4 \mathrm{rad} / \mathrm{s}\). The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?

A block (mass \(=2.0 \mathrm{kg}\) ) is hanging from a massless cord that is wrapped around a pulley (moment of inertia \(=1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of \(0.040 \mathrm{m}\) during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a $$30.0^{\circ}$$ angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

Multiple-Concept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{m},\) an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{kg} \cdot \mathrm{m}^{2} .\) A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of \(6.40 \mathrm{s} .\) (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{c m}+M h^{2},\) where \(I_{c m}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.
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