physics centripetal force question

[tshank]

A 70 kg pilot is cruising at 200 m/s. He then performs a hook maneuver by flying a lower semicircle at 100 m/s and an upper semicircle at 200 m/s. The radius of the circle is 500 m and the pilot ends up at his original position. What is the total amount of work done on the pilot?

1. 0 kJ
2. 700 kJ
3. 1050 kJ
4. 1750 kJ

Highlight for answer: A



Kaplan explanation:

The total change in potential energy for the pilot is 0, since his total displacement is 0 m. The change in kinetic energy is negative for the lower portion of the hook. By the same token, the change over the upper portion of the hook is positive, and the total change in kinetic energy is 0. This is always the case when the initial speed (before the hook) and the final speed are the same. Note that the radius of the circle allows you to calculate the forces acting on the pilot, but since the movement of the plane is always perpendicular to the centripetal forces acting on it, centripetal work is always equal to 0. The definition of work is F * d * cos (angle between F and d). If the pilot’s final speed had been different from his initial speed of 200 m/s, then the change in kinetic energy would be nonzero, and if his final altitude had been different from his initial altitude, then there would also be a nonzero change in potential energy. The total amount of work done on the pilot is zero, making choice (A) the correct answer.

Looking at the equations, there explanation makes sense. But, it doesn't make sense intuitively for me. Can someone explain how this works intuitively, beyond the equations? Doesn't the plane do work and therefore there is work done on the pilot? Or, is the work done by the plane separate from the work done on the pilot?

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[Cawolf]

Work is independent of path.

You could restate the question to ask how the pilot's energy is different at the start and end of the maneuver.

If he is in the same position then his PE is the same and if his speed is the same, then so is his KE.

There is no change in energy, so there is no Work done.

That wordy explanation simplifies down to W = F dot D. There is no displacement, so the work is zero.

 

[tshank]

Oooohhhh. I thought in W = F dot D that the D was distance, not displacement. Thanks!!

 

[Cawolf]

Oooohhhh. I thought in W = F dot D that the D was distance, not displacement. Thanks!!

Np.

Just remember that any formula involving dot products or cross products use vectors.

 

[tshank]

I've never heard of a "dot" product. What is that? So, displacement is a vector, speed is a scaler?

 

[Cawolf]

Did you take physics?

If not or if it's old here is a good resource.
http://hyperphysics.phy-astr.gsu.edu/hbase/vsca.html

They are ways of multiplying vectors.

The formula for Work is a dot product.

So, displacement is a vector, speed is a scaler?

Yes

 

[tshank]

I did take physics, but we never used the "dot" product terminology. It wasn't too in depth regarding these basics.

So, when it comes to work, the vector is changed. But when it come to force, the speed is changed.
For example, a charged particle in a magnetic field undergoes no change to its work from the magnetic field (its vector quantity won't change), but it can have force acted on it because its speed (a scaler quantity) can be changed by the magnetic field. Is this right? The magnet is acting as a dot or scaler, not a vector, and therefore can't increase work to the vector (speed x distance) of the charged particle?

Thanks for the response. That website is good and makes sense.

 

[Cawolf]

Whoa, I am pretty confused by the middle paragraph you typed.

So, when it comes to work, the vector is changed. But when it come to force, the speed is changed.

I am not sure what you mean by this. Work is a scalar quantity, it has no direction. Force is a vector, it does not relate to speed really....

For example, a charged particle in a magnetic field undergoes no change to its work from the magnetic field (its vector quantity won't change), but it can have force acted on it because its speed (a scaler quantity) can be changed by the magnetic field. Is this right? The magnet is acting as a dot or scaler, not a vector, and therefore can't increase work to the vector (speed x distance) of the charged particle?

A magnetic field changes the direction of the particle (velocity) but does not change the magnitude of the velocity (speed). It does no work.

Something can't act as a dot or scalar or vector. These are terms applied to quantities, not things. A dot product is a mathematical operation. A scalar is something that has no direction. A vector has magnitude and direction.

 

[tshank]

Okay, sorry for the confusion.

I watched some kahn academy videos on dot vs. cross - its much more clear now.

So, the magnetic field affects the force, because it affects the direction of velocity which is a scalar (cross product formula F = qvBsin(theta) , but it doesn't change the work because the speed (magnitude of velocity) is not affected in a cross product situation. The speed doesn't increase, but the direction does. Therefore acceleration increases, affecting the force. But the displacement of the particle doesn't change, therefore the work doesn't change.

Work is a dot product, magnetic force is a cross product. Force of gravity is neither. How does this sense make to you?

 

[Cawolf]

Yah they are just vector multiplication operations.

The magnetic field exerts a force by the formula you wrote....ie F = qv x B but it doesn't DO work.

The speed doesn't increase but the direction can be changed - yes.

I don't know what "Therefore acceleration increases, affecting the force" means.

I think you are sort of using a lot of terms you aren't really familiar with so it makes it confusing to try and follow. I think you have the general idea, but work isn't something that objects can "have". It is the force being exerted on the object over some displacement.

 

[tshank]

Sorry, I was just trying to explain the same reality with two different connecting theories.

1st way to think about it:
Acceleration can be seen as a change in speed, or a change in direction. In this case, we are talking about angular velocity and angular acceleration, in which the direction of the velocity changes, relating to angular acceleration. Since the particle is accelerating, it has force being acted on it. F = ma. This leads you to the second idea.

Second thought relating force to the work done/not done:
Right, the magnetic field exerts a force, but it doesn't exert a force that is in the same direction of the displacement displacement, therefore it doesn't do work. Work requires force to be acted in the same direction of the displacement.