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Chapter 10: Problem 19

A \(2.20-\) kg hoop 1.20 \(\mathrm{m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00 \(\mathrm{rad} / \mathrm{s}\) . (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

Short Answer

Expert verified
(a) 1.80 m/s. (b) 7.128 J. (c-i) 3.60 m/s; (c-ii) 0 m/s; (c-iii) 1.80 m/s. (d-i) 1.80 m/s up; (d-ii) -1.80 m/s down; (d-iii) 1.80 m/s left.

Step by step solution

01

Calculate Linear Velocity of the Center

To find the linear velocity of the hoop's center, use the relation between angular velocity and linear velocity: \[ v = r \cdot \omega \] where \( r \) is the radius of the hoop, which is half of the diameter. The given diameter is 1.20 m, so \( r = \frac{1.20}{2} = 0.60 \) m. The angular velocity \( \omega \) is 3.00 rad/s. Hence, \[ v = 0.60 \times 3.00 = 1.80 \text{ m/s} \] Thus, the linear velocity of the center of the hoop is 1.80 m/s.
02

Calculate Total Kinetic Energy of the Hoop

A hoop rolling without slipping possesses both translational and rotational kinetic energy. The total kinetic energy \( K \) is given by: \[ K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] For a hoop, the moment of inertia \( I = mr^2 \). Using the mass \( m = 2.20 \) kg, radius \( r = 0.60 \) m, and previously found \( v = 1.80 \) m/s: \[ K = \frac{1}{2}(2.20)(1.80)^2 + \frac{1}{2}(2.20)(0.60)^2(3.00)^2 \] \[ K = \frac{1}{2}(2.20)(3.24) + \frac{1}{2}(2.20)(0.36)(9) \] \[ K = 3.564 + 3.564 = 7.128 \text{ J} \] Thus, the total kinetic energy is 7.128 Joules.
03

Calculate Velocity of the Highest Point (Ground Observer)

The velocity of the highest point relative to the ground is the sum of the linear velocity of the center and the rotational component at a distance \( r \) from the axis of rotation. Here, \[ v_\text{highest} = v + r\omega = 1.80 + 0.60 \times 3.00 = 3.60 \text{ m/s} \]Thus, velocity of the highest point is 3.60 m/s to the right.
04

Calculate Velocity of the Lowest Point (Ground Observer)

The velocity of the lowest point is the linear velocity minus the rotational component because the rotational motion is opposite the linear. It's \[ v_\text{lowest} = v - r\omega = 1.80 - 0.60 \times 3.00 = 0 \text{ m/s} \]
05

Velocity of the Right Midway Point (Ground Observer)

For the point midway on the right side, the velocity only has horizontal component of the center's linear velocity, as its radial velocity cancels due to symmetry: \[ v_\text{right} = 1.80 \text{ m/s} \]Thus, this point's velocity is 1.80 m/s to the right.
06

Recalculate Velocities for Moving Observer

When observed from a frame moving at the same speed as the hoop's center, the linear component cancels out. (i) Highest point: Observed velocity is \[ v_\text{highest} = r\omega = 0.60 \times 3.00 = 1.80 \text{ m/s} \] upwards.(ii) Lowest point: Observed velocity \[ v_\text{lowest} = -r\omega = -0.60 \times 3.00 = -1.80 \text{ m/s} \] downwards.(iii) Right Midway Point: Velocity is \[ v_\text{right} = r\omega \] perpendicular and no net component horizontally, hence, 1.80 m/s vertically to the left.

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Key Concepts

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

Linear Velocity
Linear velocity is the rate at which an object moves in a straight line. In the context of a hoop rolling without slipping, it refers to the speed of its center. The center's linear velocity is directly related to its angular velocity, which is the rate of rotation.
  • The formula connecting the two is: \( v = r \cdot \omega \), where \( v \) is the linear velocity, \( r \) is the radius of the hoop, and \( \omega \) is the angular velocity.
  • For example, with a hoop of 1.20 m diameter (0.60 m radius) rotating at 3.00 rad/s, the linear velocity will be \( 0.60 \times 3.00 = 1.80 \) m/s.
Understanding this relation is crucial because it connects the linear motion of the hoop's body and the rotational motion it undergoes as it rolls.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For rolling objects like hoops, kinetic energy has two components: translational (due to linear motion) and rotational (due to spinning).
  • Translational kinetic energy is calculated with \( \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the linear velocity.
  • Rotational kinetic energy uses \( \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
  • For a hoop, the moment of inertia \( I \) is \( mr^2 \), making the total kinetic energy: \( \frac{1}{2}mv^2 + \frac{1}{2}mr^2\omega^2 \).
For our hoop example, the total kinetic energy is the sum of both energy forms, reflecting its complete movement through both space and spin. Calculating this helps understand how the hoop's energy is distributed between different types of motion.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates. Specifically, it tells how much the angle changes per unit of time as a wheel or hoop spins.
  • It is denoted by \( \omega \) and measured in radians per second.
  • Angular velocity for a rolling object is directly tied with linear velocity through the radius. The relation is \( v = r \cdot \omega \).
  • For our hoop, with an angular velocity of 3.00 rad/s and a radius of 0.60 m, the associated linear speed is 1.80 m/s.
In rolling motion, angular velocity serves as a link between how fast the object rotates and how fast it translates (or moves linearly) along a path. Understanding this helps in analyzing the dynamics of rotating objects.
Moment of Inertia
The moment of inertia is a property that measures how difficult it is to change the rotational motion of an object. It depends on the mass distribution relative to the axis of rotation.
  • For objects like hoops, the moment of inertia \( I \) is calculated as \( mr^2 \), where \( m \) is mass and \( r \) is the radius.
  • For our 2.20 kg hoop with a radius of 0.60 m, \( I = 2.20 \times (0.60)^2 \), reflecting its resistance to changes in spin.
The moment of inertia plays a vital role in determining the hoop's rotational kinetic energy. It's crucial for understanding how an object's mass distribution affects its rotational movement. Without knowing \( I \), predicting how a hoop or any object will rotate under given forces would be challenging.

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Most popular questions from this chapter

A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 \(\mathrm{m}\) and a total mass of 120 \(\mathrm{kg}\) . The turntable is initially rotating at 3.00 \(\mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a 70.0 -kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

A woman with mass 50 \(\mathrm{kg}\) is standing on the rim of a large disk that is rotating at 0.50 \(\mathrm{rev} / \mathrm{s}\) about an axis through its center. The disk has mass 110 \(\mathrm{kg}\) and radius 4.0 \(\mathrm{m} .\) Calculate the magnitude of the total angular momentum of the woman-disk system. (Assume that you can treat the woman as a point.)

While exploring a castle, Exena the Exterminator is spotted by a dragon that chases her down a hallway. Exena runs into a room and attempts to swing the heavy door shut before the dragon gets her. The door is initially perpendicular to the wall, so it must be turned through \(90^{\circ}\) to close. The door is 3.00 \(\mathrm{m}\) tall and 1.25 \(\mathrm{m}\) wide, and it weighs 750 \(\mathrm{N} .\) You can ignore the friction at the hinges. If Exena applies a force of 220 \(\mathrm{N}\) at the edge of the door and perpendicular to it, how much time does it take her to close the door?

A Gyroscope on the Moon. A certain gyroscope precesses at a rate of 0.50 \(\mathrm{rad} / \mathrm{s}\) when used on earth. If it were taken to a lunar base, where the acceleration due to gravity is \(0.165 g,\) what would be its precession rate?

A small block on a frictionless, horizontal surface has a mass of 0.0250 \(\mathrm{kg} .\) It is attached to a massless cord passing through a hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 1.75 \(\mathrm{rad} / \mathrm{s}\) . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 \(\mathrm{m} .\) Model the block as a particle. (a) Is the angular momentum of the block conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block. (d) How much work was done in pulling the cord?
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