Find the average acceleration, in meters per second squared, in the x-direction between t1 = 0.15 s and t2 = 1.9 s.
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Answer:
Part 1
The angular speed is approximately 1.31947 rad/s
Part 2
The change in kinetic energy due to the movement is approximately 675.65 J
Explanation:
The given parameters are;
The rotation rate of the merry-go-round, n = 0.21 rev/s
The mass of the man on the merry-go-round = 99 kg
The distance of the point the man stands from the axis of rotation = 2.8 m
Part 1
The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s
The angular speed is constant through out the axis of rotation
Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s
Part 2
Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;
I = m·r²
Where;
m = The mass of the man alone = 99 kg
r = The distance of the point the man stands from the axis, r = 2.8 m
v = The tangential velocity = ω/r
ω ≈ 1.31947 rad/s
Therefore, we have;
I = 99 × 2.8² = 776.16 kg·m²
= 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J
