Answer:
201.6 N
Explanation:
m = mass of disk shaped merry-go-round = 125 kg
r = radius of the disk = 1.50 m
w₀ = Initial angular speed = 0 rad/s
w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s
t = time interval = 2 s
α = Angular acceleration
Using the equation
w = w₀ + α t
4.296 = 0 + 2α
α = 2.15 rad/s²
I = moment of inertia of merry-go-round
Moment of inertia of merry-go-round is given as
I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²
F = constant force applied
Torque equation for the merry-go-round is given as
r F = I α
(1.50) F = (140.625) (2.15)
F = 201.6 N
Answer:
Initial velocity = 10 m/s
θ = 60°
This is the case of projectile motion
So the horizontal component of velocity 10 m/s = 10 cosθ
u = 10 cosθ
u = 10 cos 60°
u=5 m/s
x= 5 m
So in the horizontal direction
x = u .t
5 = 5 .t
t = 1 sec The vertical component of velocity 10 m/s = 10 sinθ
Vo= 10 sinθ
Vo= 10 sin 60°
Vo = 8.66 m/s
h=3.75 m
So height of robot = 3.75 - 0.75 m
height of robot =3 m
Answer:
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Answer:

Explanation:
From the question we are told that:
Crane Length 
Crane Mass 
Arm extension at lifting end 
Arm extension at counter weight end 
Load 
Generally the equation for Torque Balance is mathematically given by



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