When a pitcher throws a curve ball, the ball is given a fairly rapid spin. A 0.15 kg baseball with a radius of 3.7 cm is thrown with a linear speed of 41 m/s and an angular speed of 37 rad/s. Assume that the ball is a uniform, solid sphere.
Calculate how much of its kinetic energy is rotational.
You need the moment of inertia for a solid sphere, look it up.
rotationalKE=1/2 I w^2
translation KE=1/2 m v^2
To calculate the rotational kinetic energy of the baseball, you can use the formula:
Rotational Kinetic Energy = (1/2) * I * ω²
Where:
I = Moment of inertia
ω = Angular speed
The moment of inertia (I) for a solid sphere is given by the formula:
I = (2/5) * m * r²
Where:
m = Mass of the sphere
r = Radius of the sphere
Let's calculate the moment of inertia first:
m = 0.15 kg (given)
r = 3.7 cm = 0.037 m (converted to meters)
I = (2/5) * m * r²
I = (2/5) * 0.15 kg * (0.037 m)²
I ≈ 0.000463 kg * m²
Now, substitute the values of I and ω into the rotational kinetic energy formula:
Rotational Kinetic Energy = (1/2) * I * ω²
Rotational Kinetic Energy = (1/2) * 0.000463 kg * m² * (37 rad/s)²
Simplifying the equation:
Rotational Kinetic Energy ≈ 0.000463 kg * m² * 37² rad²/s²
Rotational Kinetic Energy ≈ 0.000463 kg * m² * 1369 rad²/s²
Multiplying the values:
Rotational Kinetic Energy ≈ 0.6339 kg * m² * rad²/s²
Therefore, approximately 0.6339 kg * m² * rad²/s² of the ball's kinetic energy is rotational.
To calculate the amount of kinetic energy that is due to rotation, we need to determine the rotational kinetic energy of the baseball.
The formula for rotational kinetic energy is:
K_rot = (1/2) * I * ω^2
Where:
K_rot is the rotational kinetic energy
I is the moment of inertia
ω (omega) is the angular speed
To find the moment of inertia (I) of the baseball, we need to use the formula for a solid sphere:
I = (2/5) * m * r^2
Where:
m is the mass of the baseball
r is the radius of the baseball
Given:
m = 0.15 kg
r = 0.037 m
First, we'll calculate the moment of inertia (I):
I = (2/5) * m * r^2
I = (2/5) * 0.15 kg * (0.037 m)^2
I ≈ 0.0003426 kg * m^2
Next, we'll calculate the rotational kinetic energy (K_rot):
K_rot = (1/2) * I * ω^2
K_rot = (1/2) * 0.0003426 kg * m^2 * (37 rad/s)^2
K_rot ≈ 0.2304 joules
Therefore, approximately 0.2304 joules of the kinetic energy of the baseball is due to its rotation.