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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

A force, \(\vec{F}=(2 \hat{x}+3 \hat{y}) \mathrm{N},\) is applied to an object at a point whose position vector with respect to the pivot point is \(\vec{r}=(4 \hat{x}+4 \hat{y}+4 \hat{z}) \mathrm{m} .\) Calculate the torque created by the force about that pivot point.

A disk of clay is rotating with angular velocity \(\omega .\) A blob of clay is stuck to the outer rim of the disk, and it has a mass \(\frac{1}{10}\) of that of the disk. If the blob detaches and flies off tangent to the outer rim of the disk, what is the angular velocity of the disk after the blob separates? a) \(\frac{5}{6} \omega\) b) \(\frac{10}{11} \omega\) c) \(\omega\) d) \(\frac{11}{10} \omega\) e) \(\frac{6}{5} \omega\)

The turbine and associated rotating parts of a jet engine have a total moment of inertia of \(25 \mathrm{~kg} \mathrm{~m}^{2}\). The turbine is accelerated uniformly from rest to an angular speed of \(150 \mathrm{rad} / \mathrm{s}\) in a time of \(25 \mathrm{~s}\). Find a) the angular acceleration, b) the net torque required, c) the angle turned through in \(25 \mathrm{~s}\) d) the work done by the net torque, and e) the kinetic energy of the turbine at the end of the \(25 \mathrm{~s}\).

A professor doing a lecture demonstration stands at the center of a frictionless turntable, holding 5.00 -kg masses in each hand with arms extended so that each mass is \(1.20 \mathrm{~m}\) from his centerline. A (carefully selected!) student spins the professor up to a rotational speed of \(1.00 \mathrm{rpm} .\) If he then pulls his arms in by his sides so that each mass is \(0.300 \mathrm{~m}\) from his centerline, what is his new rotation rate? Assume that his rotational inertia without the masses is \(2.80 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\), and neglect the effect on the rotational inertia of the position of his arms, since their mass is small compared to the mass of the body.

A thin uniform rod (length \(=1.00 \mathrm{~m},\) mass \(=2.00 \mathrm{~kg})\) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is \(\frac{1}{3} m L^{2} .\) The rod is released when it is \(60.0^{\circ}\) below the horizontal. What is the angular acceleration of the rod at the instant it is released?
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