The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C ,
which delivers 3 kW of power to the shaft while it is turning at 50rev/s. Gears A and B remove 1 kW and 2 kW, respectively. The shaft is free to turn in its support bearings D and E.
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Answer:
The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s
Explanation:
The rotational inertia of the merry-go-round = 600 kg·m²
The radius of the merry-go-round = 3.0 m
The mass of the boy = 20 kg
The speed with which the boy approaches the merry-go-round = 5.0 m/s
Where;
= The tangential force
I = The rotational inertia
m = The mass
α = The angular acceleration
r = The radius of the merry-go-round
For the merry go round, we have;
= The rotational inertia of the merry-go-round
= The angular acceleration of the merry-go-round
= The linear velocity of the merry-go-round
t = The time of motion
For the boy, we have;
Where;
= The rotational inertia of the boy
= The angular acceleration of the boy
= The linear velocity of the boy
t = The time of motion
When the boy jumps on the merry-go-round, we have;
Which gives;
From which we have;
The velocity of the merry-go-round, , after the boy hops on the merry-go-round = 1.5 m/s.