A baseball pitcher loosens up his pitching arm. He tosses a 0.20-kg ball using only the rotation of his forearm, 0.28 m in length, to accelerate the ball. What is the moment of inertia of the ball alone as it moves in a circular arc with a radius of 0.28 m?

Moment of Inertia, I = 0.016 kgm²

Explanation:

Mass of the ball, m = 0.20 kg

Length of the pitcher's arm, l = 0.28

Radius of the circular arc, r = 0.28 m

Moment of Inertia is given by the formula:

I = mr²

I = 0.20 * 0.28²

I = 0.20 * 0.0784

I = 0.01568

I = 0.016 kgm²

a and c

c is 100% but i am confused in a

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west