You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). your mass is 50 kg. your friend pushes the edge of the merry go round with 100 n of force, applied at a 90 deg angle. (hint: the merry-go-round is a disk).
a. how much torque did your friend apply?
b. what is your angular acceleration if you are sitting in the middle of the merry-go-round?
c. what is your angular acceleration if you are sitting at the edge of the merry-go-round?
d. which angular acceleration is larger and why?

a. b. c. d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by , being r the radius, F the force aplied and the angle between the vector force and the vector radius. Since and so .

b. Since the relation hols, being I the moment of inertia, the angular acceleration can be calculated by . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by , being the mass of the person and the distance from the person to the center. Given all of this, we have

.

c. Similar equation to b, but changing , so

.

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14