Question

A basketball of circumference \(75.5 \mathrm{~cm}\) and mass \(598 \mathrm{~g}\) is forced to the bottom of a swimming pool and then released. After initially accelerating upward, it rises at a constant velocity. a) Calculate the buoyant force on the basketball. b) Calculate the drag force the basketball experiences while it is moving upward at constant velocity.

Short answer

Solution: 1. Calculate the radius of the basketball: $$ r = \frac{75.5}{2 \pi} $$ 2. Calculate the volume of the basketball: $$ V = \frac{4}{3} \pi \left(\frac{75.5}{2 \pi}\right)^3 $$ 3. Convert the basketball's volume from cm³ to m³ and find the buoyant force acting on the basketball: $$ F_B = 1000\frac{\mathrm{kg}}{\mathrm{m}^3} \cdot (V \cdot 10^{-6}\,\mathrm{m}^3) \cdot 9.81\frac{\mathrm{m}}{{s}^2} $$ 4. Compute the force of gravity acting on the basketball and find the drag force: $$ F_G = 0.598\,\mathrm{kg} \cdot 9.81\frac{\mathrm{m}}{{s}^2} $$ $$ F_D = F_B - F_G $$ Find the values for the buoyant force (F_B) and the drag force (F_D) experienced by the basketball.

Step-by-step solution

Find the volume of the basketball

To find the volume \((V)\) of the basketball, we'll use the formula for the volume of a sphere, which is given by: $$ V = \frac{4}{3} \pi r^3 $$ First, we need to find the radius \((r)\) of the basketball using the circumference \((C)\). The formula for the circumference of a sphere is given by: $$ C = 2 \pi r $$ We are given the circumference as \(75.5\mathrm{~cm}\). So, we can find the radius by rearranging the formula: $$ r = \frac{C}{2 \pi} = \frac{75.5}{2 \pi} $$ Now, we can find the volume of the basketball: $$ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{75.5}{2 \pi}\right)^3 $$

Calculate the buoyant force

The buoyant force \((F_B)\) on the basketball is equal to the weight of the water displaced by the basketball. This can be calculated using the Archimedes' principle: $$ F_B = \rho_{water} \cdot V \cdot g $$ where \(\rho_{water}\) is the density of water, which is typically \(1000\frac{\mathrm{kg}}{\mathrm{m}^3}\), and \(g\) is the acceleration due to gravity (approximately \(9.81\frac{\mathrm{m}}{{s}^2}\)). Make sure to convert the volume of the basketball from \(cm^3\) to \(m^3\), by multiplying the volume by \(10^{-6}\). Then, calculate the buoyant force: $$ F_B = 1000\frac{\mathrm{kg}}{\mathrm{m}^3} \cdot (V \cdot 10^{-6}\,\mathrm{m}^3) \cdot 9.81\frac{\mathrm{m}}{{s}^2} $$

Calculate the drag force

The drag force \((F_D)\) acting on the basketball while it is moving upward at constant velocity can be determined using the following equation: $$ F_D = F_B - F_G $$ where \(F_G\) is the force of gravity acting on the basketball. We can calculate the force of gravity by using the basketball's mass, converting it from grams to kilograms, and multiplying it by the acceleration due to gravity: $$ F_G = m \cdot g = (0.598\,\mathrm{kg}) \cdot 9.81\frac{\mathrm{m}}{{s}^2} $$ Finally, we can calculate the drag force: $$ F_D = F_B - F_G $$