Answer:
the child have to exert this much amount of force radially to stay on the wheel.
Explanation:
Given:
mass of the child,
radius of the merry-go round wheel,
moment of inertia of the wheel,
angular velocity of the wheel,
Since the child is sitting on the edge of the rotating wheel the child will feel and outward throwing force called centrifugal force.
Centrifugal force is mathematically given as:
the child have to exert this much amount of force radially to stay on the wheel.
Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
