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Chapter 10: Problem 8

A basketball of mass \(610 \mathrm{~g}\) and circumference \(76 \mathrm{~cm}\) is rolling without slipping across a gymnasium floor. Treating the ball as a hollow sphere, what fraction of its total kinetic energy is associated with its rotational motion? a) 0.14 b) 0.19 c) 0.29 d) 0.40 e) 0.67

Short Answer

Expert verified
Answer: The fraction of the total kinetic energy associated with the rotational motion of the basketball is approximately 0.14.

Step by step solution

01

Calculate the moment of inertia of a hollow sphere

The formula for the moment of inertia I of a hollow sphere with mass m and radius R is given by: I = \frac{2}{3}mR^2 We know the mass m = 610 g and the circumference C = 76 cm. We can find the radius R using the relation: R = \frac{C}{2\pi}
02

Find radius of the basketball

Using the circumference, calculate the radius of the basketball: R = \frac{76}{2\pi} = \frac{76}{6.28} \approx 12.1 \mathrm{~cm} Now, we can calculate the moment of inertia I.
03

Calculate the moment of inertia

Substitute the values for mass and radius in the formula for the moment of inertia: I = \frac{2}{3}(610)(12.1)^2 \approx 14926 \mathrm{~g~ cm^2}
04

Determine the relationship between linear and rotational kinetic energies

The total kinetic energy is the sum of the linear kinetic energy (K_l) and the rotational kinetic energy (K_r). The linear kinetic energy is given by: K_l = \frac{1}{2}mv^2 The rotational kinetic energy is given by: K_r = \frac{1}{2}Iω^2 Since the basketball is rolling without slipping, the relationship between linear velocity v and angular velocity ω is: v = Rω Now, we can express the rotational kinetic energy in terms of linear velocity as follows: K_r = \frac{1}{2}I\frac{v^2}{R^2}
05

Calculate the fraction of the total kinetic energy due to rotation

Find the fraction of the total kinetic energy associated with rotational motion: Fraction = \frac{K_r}{K_r + K_l} = \frac{\frac{1}{2}I\frac{v^2}{R^2}}{\frac{1}{2}I\frac{v^2}{R^2} + \frac{1}{2}mv^2} Upon simplification, we get: Fraction = \frac{I}{I + mR^2} Now, substitute the values for the moment of inertia I, mass m, and radius R into the equation: Fraction = \frac{14926}{14926 + 610(12.1)^2} = \frac{14926}{14926+91142} = \frac{14926}{106068} \approx 0.14 The fraction of the total kinetic energy associated with the rotational motion is a) 0.14.

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Most popular questions from this chapter

An oxygen molecule \(\left(\mathrm{O}_{2}\right)\) rotates in the \(x y\) -plane about the \(z\) -axis. The axis of rotation passes through the center of the molecule, perpendicular to its length. The mass of each oxygen atom is \(2.66 \cdot 10^{-26} \mathrm{~kg},\) and the average separation between the two atoms is \(d=1.21 \cdot 10^{-10} \mathrm{~m}\) a) Calculate the moment of inertia of the molecule about the \(z\) -axis. b) If the angular speed of the molecule about the \(z\) -axis is \(4.60 \cdot 10^{12} \mathrm{rad} / \mathrm{s},\) what is its rotational kinetic energy?

A wagon wheel is made entirely of wood. Its components consist of a rim, 12 spokes, and a hub. The rim has mass \(5.2 \mathrm{~kg}\), outer radius \(0.90 \mathrm{~m}\), and inner radius \(0.86 \mathrm{~m}\). The hub is a solid cylinder with mass \(3.4 \mathrm{~kg}\) and radius \(0.12 \mathrm{~m} .\) The spokes are thin rods of mass \(1.1 \mathrm{~kg}\) that extend from the hub to the inner side of the rim. Determine the constant \(c=I / M R^{2}\) for this wagon wheel.

An ice skater rotating on frictionless ice brings her hands into her body so that she rotates faster. Which, if any, of the conservation laws hold? a) conservation of mechanical energy and conservation of angular momentum b) conservation of mechanical energy only c) conservation of angular momentum only d) neither conservation of mechanical energy nor conservation of angular momentum

A student of mass \(52 \mathrm{~kg}\) wants to measure the mass of a playground merry-go-round, which consists of a solid metal disk of radius \(R=1.5 \mathrm{~m}\) that is mounted in a horizontal position on a low-friction axle. She tries an experiment: She runs with speed \(v=6.8 \mathrm{~m} / \mathrm{s}\) toward the outer rim of the merry-go-round and jumps on to the outer rim, as shown in the figure. The merry-go-round is initially at rest before the student jumps on and rotates at \(1.3 \mathrm{rad} / \mathrm{s}\) immediately after she jumps on. You may assume that the student's mass is concentrated at a point. a) What is the mass of the merry-go-round? b) If it takes 35 s for the merry-go-round to come to a stop after the student has jumped on, what is the average torque due to friction in the axle? c) How many times does the merry-go-round rotate before it stops, assuming that the torque due to friction is constant?

A uniform rod of mass \(M=250.0 \mathrm{~g}\) and length \(L=50.0 \mathrm{~cm}\) stands vertically on a horizontal table. It is released from rest to fall. a) What forces are acting on the rod? b) Calculate the angular speed of the rod, the vertical acceleration of the moving end of the rod, and the normal force exerted by the table on the rod as it makes an angle \(\theta=45.0^{\circ}\) with respect to the vertical. c) If the rod falls onto the table without slipping, find the linear acceleration of the end point of the rod when it hits the table and compare it with \(g\).
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