DISPLACEMENT WITH CONSTANT ACCELEration. Given information: displacement=64m, acceleration=9.81m\s and time interval=3s. calcula
1 answer:
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Answer: 42.49
Explanation:
To solve this, we need to keep in mind the following:
While the sphere hangs it is under the effect of gravity. It is creating a Angle of 90° taking the roof as a reference.
Gravity can be noted as a Acceleration Vector. The magnitud for Earth's Gravity is a constant: 9.81
The acceleration of the Van will affect the sphere also, but this accelaration will be on the X-axis and perpendicular to the gravity. Because this two vectors are taking action under the sphere they will create a angle. This angle can be measured as a relation of the two magnitudes.
Tangent (∅) = Opossite Side / Adyacent Side
By trigonometry, we know the previous formula. This formula allows us to find the Tangent of a angle as a relation between the two perpendiculars magnitudes. In this case the Opossite Side will be the Gravity Accelaration, while the Adyancent Side is the Van's Acceleration.
(1) Tangent (∅) = Gravity's Acceleration (G) / Van's Acceleration (Va)
Searching for the Va in (1)
Va = G/Tan(∅)
Where ∅ in this case is equal to 13.0°
Va = 9.81 / Tan(13.0°)
Va = 42.49
The vans acceleration need to be 42.49 to create an angle of 13° with the Van's Roof
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s
By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s
Position #2:
Radius, r₂ = 2 ft
By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²
Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf
The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer: 44.25 ft-lb
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s
By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s
Position #2:
Radius, r₂ = 2 ft
By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²
Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf
The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer: 44.25 ft-lb