Final answer:
The ratio of rotational to translational kinetic energies of a baseball with a mass of 0.14 kg, radius of 3.7 cm, spinning with an angular velocity of 52 rad/s, and a linear speed of 42 m/s is approximately 3.98 × 10^-4.
Step-by-step explanation:
The question asks us to calculate the ratio of rotational and translational kinetic energies of a spinning baseball with a given mass, radius, angular velocity, and linear speed.
The rotational kinetic energy (Krot) can be calculated using the formula
Krot = 1/2 I ω^2,
where I is the moment of inertia and ω is the angular velocity.
The translational kinetic energy (Ktrans) is given by
Ktrans = 1/2 m v^2,
where m is the mass and v is the linear speed.
The moment of inertia for a sphere is
I = 2/5 m r^2,
where r is the radius.
First, we calculate the moment of inertia of the baseball:
I = 2/5 × 0.14 kg × (0.037 m)^2 ≈ 3.64 × 10^-4 kg·m^2.
Next, we find the rotational kinetic energy:
Krot = 1/2 × 3.64 × 10^-4 kg·m^2 × (52 rad/s)^2 ≈ 4.92 × 10^-2 J.
Then, the translational kinetic energy:
Ktrans = 1/2 × 0.14 kg × (42 m/s)^2 ≈ 123.48 J.
The ratio of rotational to translational kinetic energy is:
Krot/Ktrans = 4.92 × 10^-2 J / 123.48 J ≈ 3.98 × 10^-4.
Therefore, the ratio of rotational to translational kinetic energy is approximately 3.98 × 10^-4.