Question

A bowling ball made for a child has half the radius of an adult bowling ball. They are made of the same material (and therefore have the same mass per unit volume). By what factor is the (a) mass and (b) rotational inertia of the child's ball reduced compared with the adult ball?

Short answer

Answer: The mass of the child's ball is reduced by a factor of 8 and the rotational inertia is reduced by a factor of 32 compared to the adult's ball.

Step-by-step solution

Define variables and given information

Let's define the variables: - Radius of child's ball: r - Radius of adult's ball: R - Density of the material: ρ (mass/volume) Given information: - r = 0.5 · R (child's ball has half the radius of the adult's ball)

Calculate the volumes of the balls

To find the mass, we first need to calculate the volume of the child's and adult's ball using the formula for the volume of a sphere: $$V = \frac{4}{3} \pi R^3$$ Child's ball volume: $$V_c = \frac{4}{3} \pi r^3$$ Adult's ball volume: $$V_a = \frac{4}{3} \pi R^3$$

Calculate the masses of the balls

To find the mass, we can use the formula for mass: $$m = \rho \cdot V$$ Child's ball mass: $$m_c = \rho \cdot \frac{4}{3} \pi r^3$$ Adult's ball mass: $$m_a = \rho\cdot\frac{4}{3} \pi R^3$$

Find the ratio of the masses

Now we can calculate the ratio of the mass of the child's ball to the adult's ball. Since r = 0.5 · R, we can use this relation: $$\frac{m_c}{m_a} = \frac{\rho\cdot\frac{4}{3} \pi r^3}{\rho\cdot\frac{4}{3} \pi R^3} = \frac{r^3}{R^3} = \frac{(0.5R)^3}{R^3} = \frac{1}{8}$$ So the mass of the child's ball is reduced by a factor of 8 compared to the adult ball. (Answer a)

Calculate the moment of inertia of the balls

To find the rotational inertia, we can use the formula for the moment of inertia of a solid sphere: $$I = \frac{2}{5} mR^2$$ Child's ball inertia: $$I_c = \frac{2}{5} m_cr^2$$ Adult's ball inertia: $$I_a = \frac{2}{5} m_aR^2$$

Find the ratio of the moments of inertia

Now we can calculate the ratio of the moment of inertia of the child's ball to the adult's ball: $$\frac{I_c}{I_a} = \frac{\frac{2}{5} m_cr^2}{\frac{2}{5} m_aR^2} = \frac{m_cr^2}{m_aR^2} = \frac{m_c}{m_a}\cdot\frac{r^2}{R^2} = \frac{1}{8}\cdot\frac{(0.5R)^2}{R^2} = \frac{1}{32}$$ So the rotational inertia of the child's ball is reduced by a factor of 32 compared to the adult ball. (Answer b)