Problem 8 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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}

Step 2: 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.

Step 3: 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}

Step 4: 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}

Step 5: 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.