The aircraft link is made from an a992 steel rod. determine the smallest diameter of the rod, to the nearest 1 /16 in., that wil
1 answer:
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Answer:
(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Explanation:
Given that,
Initial spinning = 50.0 rad/s
Time = 20.0
Distance = 2.5 m
Mass of pole = 4 kg
Angle = 60°
We need to calculate the angular acceleration
Using formula of angular velocity
The angular acceleration is -2.5 rad/s²
We need to calculate the number of revolution
Using angular equation of motion
Put the value into the formula
The number of revolution is 500 rad.
(II). We need to calculate the torque
Using formula of torque
Put the value into the formula
Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Using conservation of energy law:-
∑ work in = ∑ work out
and work= force* displacement
so when we wanted to move a 100kg a distance of 1m
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m
by the same way:-
------------------------
work in = 100kg * 2m * g (m/
)= work out
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m
∑ work in = ∑ work out
and work= force* displacement
so when we wanted to move a 100kg a distance of 1m
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m
by the same way:-
------------------------
work in = 100kg * 2m * g (m/
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m