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A problem (Momentum)

A problem from my Physics book, written by stupid Tsokos.

A helicopter rotor whose length is R = 5.0 m. pushes air downwards with a speed v.

Assuming that the density of air is constant at p = 1.20 kg m ^ -3 and the mass of the helicopter is 1200kg, find v.

You may assume that the rotor forces the air in a circle of radius R (spanned by the rotor) to move with the downward v.


The momentum of the air under the rotor is mv, where m is the mass of air in a circle of radius 5.0m

In time (change in time), the mass is enclosed in a cylinder of radius R and height v * (change in time).

Thus, the momentum of this is p * 3.14 * R^2 * v^2 * (change in time), and its rate of change is p * 3.14 * R^2 * v^2.

This is the upward force on the helicopter, which must equal the helicopter's weight of 12 000N. So.

p * 3.14 * R^2 * v^2 = Mg

v = ( (Mg)/(p*3.14*R^2) ) ^ 0.5

Thus , v = 11m s^-1

My question : Could somebody explain how he put the velocity into the formula and why it became powered to 2?

Momentum = m * v, so if I have understood it correctly p * 3.14 * R^2 is in this case the mass.

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the height of the column is v*(change in time)

therefore volume of column is 3.14*R2*v*(change in time)

mass is p*V = p*3.14*R2*v*(change in time)

momentum is m*v = p*3.14*R2*v*(change in time)*v = p*3.14*R2*v2*(change in time)

v2 comes from v used once for calculation of height of the column and used again for momentum.

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It makes sense and I agree with you :(

But what happens to (change in time)?

Because the book never used it in the calculation to find velocity --> v = ( (Mg)/(p*3.14*R^2) ) ^ 0.5

Is it because we don't know the time?

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momentum = p*3.14*R2*v2*(change in time)

momentum/(change in time) = p*3.14*R2*v2

but momentum/(change in time) can also be called the "rate of change of momentum" which is directly proportional (and equal when we use SI units) to the force (mg i.e. helicopter's weight) by Newton's Second Law of Motion

therefore mg = p*3.14*R2*v2 (since both are equal to the rate of change of momentum)

make v subject of the formula and we get the final formula given.

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