M2 Power Question

This is from the OCR Mechanics 2 specification, Q15 on Miscellaneous Exercise 2 of the textbook.

15. If the air resistance to the motion of an airliner at speed v ms-1 is given by kv² newtons at ground level, then at 6000 metres the corresponding formula is 0.55 kv², and at 12000 metres it is 0.3 kv². If an airliner can cruise at 220 ms-1 at 12000 metres, at what speed will it travel at 6000 metres with the same power output from the engines?

Suppose that k=2.5 and that the mass of the airliner is 250 tonnes. As the airliner takes off its speed is 80 ms-1 and it immediately starts to climb with the engines developing three times the cruising power. At what angle to the horizontal does it climb?

I can answer the first part to get an answer of approximately 180 ms^-1. This answer is confirmed as right. But how do I go about solving the second part?

Reply 1

ghostwalker

Original post by Big-Daddy

But how do I go about solving the second part?

Consider.

What's the power developed when the airliner is crusing?

Hence what's the power developed at takeoff?

This power is consumed in, work done against air resistance plus work done in lifting the plane. This will allow you to work out the vertical velocity.

~~Then (vertical velocity)/(horizontal velocity) = tan(desired angle)~~

Edit: See below.

(edited 11 years ago)

Reply 2

Big-Daddy

Original post by ghostwalker

This power is consumed in, work done against air resistance plus work done in lifting the plane. This will allow you to work out the vertical velocity.

But doesn't the power also have to be used in maintaining the cruising? I don't get how it's split into the vertical force and horizontal force (nor do I understand how air resistance is split into these two components).

Reply 3

ghostwalker

Original post by Big-Daddy

Part of the power maintains the cruising when it's taking off, and that's your horizontal force.

Any unused power goes into the vertical force.

Air resistance is not split into two components.

Reply 4

Big-Daddy

Original post by ghostwalker

Part of the power maintains the cruising when it's taking off, and that's your horizontal force.

Any unused power goes into the vertical force.

Air resistance is not split into two components.

So if we call horizontal velocity vx and vertical vy, vertical air resistance = k*v_y and vertical = k*v_x? I can't think of how else you might model it.

Reply 5

ghostwalker

Original post by Big-Daddy

So if we call horizontal velocity vx and vertical vy, vertical air resistance = k*v_y and vertical = k*v_x? I can't think of how else you might model it.

I think you can safely ignore vertical air resistance.

On reflection my previous post wasn't entirely correct.

The velocity of 80ms^-1 against which we have air resistance would be in the direction the air craft is travelling.

You don't need to deal with vertical/horizontal air resistance, just the resistance against the motion in the direction of travel.

Also angle will be sin^-1 "vertical air speed"/80.

(edited 11 years ago)

Reply 6

Big-Daddy

Original post by ghostwalker

If it's travelling at 80 ms^-1 in v_x then shouldn't the angle be arctan(v_y/v_x)?

The problem with "resistance against the motion in the direction of travel" is that the direction is at an angle, so there will be two components of motion (v_x and v_y) so there will be two components of resistance, surely? Unless you're saying I should just take resistance to x motion?

Reply 7

ghostwalker

Original post by Big-Daddy

I think that whatever the angle the resistance and the power required to overcome it remains the same.

IMHO, it's highly debatable what is going on as the plane takes off. Is it 80 m/s horizontally, or at the angle the plane takes off at? I think, but could be wrong, that it's 80 horizontal before it takes off, and becomes 80 at an angle as it takes off.

Either way sin of a small angle is approximately equal to the tan of the same small angle, so hopefully it makes no difference.

Sorry I can't give a definitive answer.