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MAVERICK [17]
3 years ago
11

A rubber wheel on a steel rim spins freely on a horizontal axle that is suspended by a fixed pivot at point P. When the wheel sp

ins at a rate of 4.00 rev/s, it precesses smoothly about point P in a horizontal plane with a period of 3.50 s. The wheel's outer radius is 15.0 cm, and its total mass is measured to be 1.12 kg, 60% of this being the spinning wheel, and 40% its axle assembly. The wheel and its axle are both symmetric about the middle of the wheel, and the pivot point P is 4.70 cm from the wheel's middle.
a. its axle considered as a body.
b. What is the direction of the net torque on the wheel about the pivot point P?
c. If the wheel is spinning counterclockwise (0) as seen from point B on the wheel's axle, what is the direction of its spin angular momentum referenced to the diagram?
d. As seen from point A above the pivot, which way is the wheel precessing (C or O)? Please give your reasoning.
e. Using this information, determine the moment of inertia I of the wheel about its axle. Please be sure your work is clear.
Engineering
1 answer:
Nimfa-mama [501]3 years ago
3 0

Answer:

stan lee

Explanation:

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Answer:

the percent increase in the velocity of air is 25.65%

Explanation:

Hello!

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

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α=density

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Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}

Now we can use the equation that defines the percentage of increase, in this case for speed

i=(\frac{V2}{V1} -1) 100

Now we use the equation obtained in the previous step, and replace values

i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65

the percent increase in the velocity of air is 25.65%

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