PHYSICS! PLEASE SHOW WORK. VERY URGENT. An airplane propeller is 1.78m in length

Question

PHYSICS! PLEASE SHOW WORK. VERY URGENT.

An airplane propeller is 1.78m in length (from tip to tip) with mass 124kg and is rotating at 2900rpm (rev/min ) about an axis through its center. You can model the propeller as a slender rod.

Part A

What is its rotational kinetic energy?

Part B

A particle of mass 3 m is located 1.30m from a particle of mass m

Part A

Where should you put a third mass M so that the net gravitational force on M due to the two masses is exactly zero?

Part B

Is the equilibrium of M at this point stable or unstable for points along the line connecting m and 3m ?

Part C

Is the equilibrium of M at this point stable or unstable for points along the line passing through M and perpendicular to the line connecting m and 3 m ?

PHYSICS! PLEASE SHOW WORK. VERY URGENT. An airplane propeller is 1.78m in length (from tip to tip) with mass 124kg and is rotating at 2900rpm (rev/min ) about an axis through its center. You can model the propeller as a slender rod. Part A What is its rotational kinetic energy? Part B Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm rpm ? A particle of mass 3 m is located 1.30m from a particle of mass m Part A Where should you put a third mass M so that the net gravitational force on M due to the two masses is exactly zero? Part B Is the equilibrium of M at this point stable or unstable for points along the line connecting m and 3m ? Part C Is the equilibrium of M at this point stable or unstable for points along the line passing through M and perpendicular to the line connecting m and 3m ?

Explanation / Answer

1

a) Rotational kinetic energy = 1/2 x I x angular velocity^2

I = moment of inertia

sooooo....

diameter = 1.78 m, mass = 124 kg

moment of inertia of a rod = m x L^2/12

124 x 1.78^2/12 = 32.72 = moment of inertia

angular velocity = 2900 x 2pi / 60 = 303.5 rads-1

0.5 x 32.72 x 303.5^2 = 1507287 J

b) Reduce the mass by 75% = 31 Kg

substitute in the formula for rotational kinetic energy and m= 31 Kg remaining values are same

we will get the final answer is 1151 rpm

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2

A)M must be in between m and 3m.
Let x be the distance of M right from 3m.
Then force of attraction due to 3m on M ie F1 = G M 3m / x^2
Now the same M is at a distance 1.3 - x right from m.
So force of attraction on M due to m is given as F2 = G M m / (1.3 - x)^2
As M has to feel zero force then G M 3m / x^2 , G M m / (1.3 - x)^2 have to be equal to each other
ie F1 = F2
Equating, 3/x^2 = 1 / (1.3-x)^2
x = ./3 *1.3 / (1+./3) = 0.824 m
Equilibrium at this point is stable but for other points both on the line joining m and 3m as well as perpendicular line there will be equilibrium.

B) the mass M will be in stable equilibrium along the line passing thpough m and 3m

C) the mass M will be in unstable equilibrium along the line perpendicular to the ine jining m and 3m

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