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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), except as viewed by someone moving along with same velocity as the hoop.

Short Answer

Expert verified
(a) 1.80 m/s; (b) 14.256 J; (c)(i) 3.60 m/s, (ii) 0 m/s, (iii) 2.70 m/s at 45掳; (d)(i) 1.80 m/s, (ii) -1.80 m/s, (iii) 0.90 m/s at 45掳.

Step by step solution

01

Understand the Hoop Parameters

The hoop has a mass of 2.20 kg and a diameter of 1.20 m, giving it a radius of 0.60 m \((\frac{1.20}{2})\). It's rolling at an angular velocity of 3.00 rad/s. The condition that it's rolling without slipping means the linear velocity \(v\) of the center of the hoop is related to the angular velocity \(\omega\) by \(v = \omega r\).
02

Calculate Linear Velocity

Using the relation \(v = \omega r\), the linear velocity \(v\) of the hoop's center is \(3.00 \; \mathrm{rad/s} \times 0.60 \; \mathrm{m} = 1.80 \; \mathrm{m/s}\). Therefore, the center of the hoop is moving at 1.80 m/s.
03

Calculate Total Kinetic Energy

The total kinetic energy of the hoop includes translational and rotational kinetic energy. Translational kinetic energy is given by \(\frac{1}{2}mv^2\) and rotational kinetic energy by \(\frac{1}{2}I\omega^2\). The moment of inertia \(I\) for a hoop about its center is \(mr^2\). Using \(m = 2.20 \; \mathrm{kg}\), \(v = 1.80 \; \mathrm{m/s}\), \(\omega = 3.00 \; \mathrm{rad/s}\), \(I = 2.20 \times 0.60^2\), we find the total kinetic energy \(K = \frac{1}{2}(2.20)(1.80)^2 + \frac{1}{2}(2.20 \times 0.60^2)(3.00)^2\) gives \(7.128 + 7.128 = 14.256 \; \mathrm{J}\).
04

Velocity of Highest Point Relative to Ground

For the highest point, velocity relative to the center's linear velocity is \(v + r\omega\). Thus, the velocity is \(1.80 + 1.80 = 3.60\; \mathrm{m/s}\) to the right.
05

Velocity of Lowest Point Relative to Ground

For the lowest point, velocity relative to the center's linear velocity is \(v - r\omega\). Thus, the velocity is \(1.80 - 1.80 = 0\; \mathrm{m/s}\).
06

Velocity of Right-side Point Midway Relative to Ground

For a point on the right side, its velocity is perpendicular to the radial line and has a magnitude \(\frac{1}{2}r\omega\) because this location is 45掳 below the topmost point. The velocity is \(1.80 + (\frac{1}{2} \times 1.80) = 2.70 \; \mathrm{m/s}\) directed at 45掳 to the horizontal.
07

Velocity of Points in Part (d) Relative to Moving Observer

For part (d), since the observer is moving with the same velocity as the hoop's center, the velocities of each point will be reduced by the center's velocity of the hoop. Therefore: (i) highest point's velocity is \(1.80 \; \mathrm{m/s}\); (ii) lowest point's velocity is \( -1.80 \; \mathrm{m/s}\); (iii) right-side point's velocity is \(0.90 \; \mathrm{m/s}\) at a 45掳 angle.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fundamental concept that explains the energy possessed by an object in motion. There are two forms of kinetic energy we consider for a rolling hoop 鈥� translational kinetic energy and rotational kinetic energy.

Translational kinetic energy applies to the movement of the hoop's center of mass across a surface. It is given by the formula:
  • \[KE_{translational} = \frac{1}{2}mv^2\]
where "m" is the mass of the hoop, and "v" is the linear velocity of its center.

Rotational kinetic energy is associated with the hoop's spinning motion around its axis. The formula for rotational kinetic energy is:
  • \[KE_{rotational} = \frac{1}{2}I\omega^2\]
where "I" is the moment of inertia, and "\omega" is the angular velocity.

For a complete description of the hoop's kinetic energy, you sum both translational and rotational kinetic energies. In this example, each contributes equally to the total kinetic energy, making both essential to understand in rolling motion.
Angular Velocity
Angular velocity \(\omega\) refers to how fast an object rotates or spins around a particular axis. It is a crucial concept for objects undergoing rotational motion, like a rolling hoop. Angular velocity is measured in radians per second (rad/s), indicating the angle an object sweeps in a given time.

In a rolling situation without slipping, angular velocity directly relates to linear velocity. The equation that connects these two is:
  • \[v = \omega r\]
where "v" is the linear velocity, and "r" is the radius of the hoop.

For the hoop rolling on a floor, it maintains an angular velocity of 3.00 rad/s. This consistency is essential because any slipping would change the dynamics significantly, affecting both angular and linear velocities.
Linear Velocity
Linear velocity is a measure of the rate of change of an object's position along a path, essentially how fast something moves in a straight line. For rolling objects like hoops, linear velocity refers to the speed of the center of mass.

In rolling motion, linear velocity is linked to angular velocity by the equation
  • \[v = \omega r\]
where "v" is linear velocity, "\omega" is angular velocity, and "r" is the radius.

For the given problem, the hoop's center moves with a linear velocity of 1.80 m/s. This velocity dictates the overall movement of the hoop while maintaining no slippage.
Moment of Inertia
Moment of inertia is a property that indicates an object's resistance to changes in its rotation. It is crucial for understanding and calculating rotational dynamics. For a hoop or ring, the moment of inertia about its center is calculated as:
  • \[I = mr^2\]
where "m" is mass and "r" is radius.

In this exercise, the hoop has a mass of 2.20 kg, giving it a moment of inertia based on the distance from its center to the edge. With rotations, the moment of inertia influences the hoop's rotational kinetic energy.

The higher the moment of inertia, the more effort needed to change the hoop's rotational speed. This concept helps in understanding why different objects with various structures have unique rotational features.

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Most popular questions from this chapter

In a physics laboratory you do the following ballistic pen- dulum experiment: You shoot a ball of mass \(m\) horizontally from a spring gun with a speed \(v\) . The ball is immediately caught a distance \(r\) below a frictionless pivot by a pivoted catcher assembly of mass \(M\) . The moment of inertia of this assembly about its rotation axis through the pivot is \(L\) . The distance \(r\) is much greater than the radius of the ball. (a) Use conservation of angular momentum to show that the angular speed of the ball and catcher just after the ball is caught is \(\omega=\operatorname{mor} /\left(m r^{2}+I\right) .\) (b) After the ball is caught, the center of mass of the ball-catcher assembly system swings up with a maximum height increase \(h\) . Use conservation of energy to show that \(\omega=\sqrt{2(M+m) g h /\left(m r^{2}+1\right)} .\) (c) Your lab partner says that linear momentum is conserved in the collision and derives the expression \(m v=(m+M) V,\) where \(V\) is the speed of the ball immediately after the collision. She then uses conservation of energy to derive that \(V=\sqrt{2 g h},\) so that \(m v=(m+M) \sqrt{2 g h} .\) Use the results of parts (a) and (b) to show that this equation is satisfied only for the special case when \(r\) is given by \(I=M r^{2} .\)

A child rolls a 0.600 -kg basketball up a long ramp. The bas- ketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 \(\mathrm{m} / \mathrm{s}\) . When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\). Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.

A thin, uniform metal bar, 2.00 \(\mathrm{m}\) long and weighing \(90.0 \mathrm{N},\) is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 \(\mathrm{m}\) below the ceiling by a small \(3.00-\mathrm{kg}\) ball, initially traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s}\) . The ball rebounds in the opposite direction with a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) . (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

An electric motor consumes 9.00 \(\mathrm{kJ}\) of electrical energy in 1.00 \(\mathrm{min}\) . If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500 \(\mathrm{rpm} ?\)

The Spinning Figure Skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center (Fig. 10.49 ) When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled, hollow cylinder. His hands and arms have a combined mass 8.0 \(\mathrm{kg}\) . When outstretched, they span 1.8 \(\mathrm{m}\) ; when wrapped, they form a cylinder of radius 25 \(\mathrm{cm}\) . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . If his original angular speed is 0.40 \(\mathrm{rev} / \mathrm{s}\) , what is his final angular speed?
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