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.
(b) Calculate how much of its kinetic energy is rotational.
rotational is 1/2 I w^2 yes?
I of sphere is 2/5mr^2 yes?
so would it not be 1/2[(2/5 X .15 X .037^2) X (41/.037)^2]
please help!
"Yes" to your first two questions.
You used a wrong number for the angular speed w. It is 37 rad/s, as stated.
You only use the relation w = V/R for rolling objects (cylinders and spheres)that are not slipping
The speed of the ball, V = 41 m/s, has nothing to do with the rotational kinetic energy
thank you so much! i figured it out!
My pleasure and congratulations! It is great to help students who show their work
To calculate the rotational kinetic energy of the baseball, you're on the right track using the formula:
Rotational Kinetic Energy = (1/2) * I * ω^2
Where:
I is the moment of inertia of the baseball
ω is the angular speed of the baseball
The moment of inertia of a solid sphere can be calculated using the formula:
I = (2/5) * m * r^2
Where:
m is the mass of the baseball
r is the radius of the baseball
In this case, you have the following values:
m = 0.15 kg
r = 0.037 m
ω = 37 rad/s
Now we can plug these values into the formulas:
I = (2/5) * (0.15 kg) * (0.037 m)^2
I = 0.00061725 kg⋅m^2
Rotational Kinetic Energy = (1/2) * (0.00061725 kg⋅m^2) * (37 rad/s)^2
Rotational Kinetic Energy ≈ 0.433 J
So, approximately 0.433 J of the kinetic energy of the baseball is due to its rotation.