-1

A 3400 kg plane flying at a constant speed of 170 m/s is to do a vertical loop. What is the radius of the loop if the pilot feels three times his normal weight when he is at the top of the loop?

### 1 Answer by Expert Tutors

John R. | Physics and Math Tutor – Kind, Easy-going, PatientPhysics and Math Tutor – Kind, Easy-goin...
4.9 4.9 (140 lesson ratings) (140)
0
Hi Chris,

1. First draw a free-body diagram of the airplane at the top of the loop.
2. The only forces acting on the plane is gravity (mg) and the force on the wing. Neglect the thrust and friction in this case. Or if you include them they will cancel.
3. The acceleration of the plane is due only to circular motion, v^2/r, and acts in the nadir (earthward or down) direction.
4. Apply Newtons second law, F=ma. The left side is -mg-3g, the 3g comes from the requirement of the sensation of 3 times his normal weight.
5. Te right side of the equation is -m*v^2/r.
6. Solve the equation for r

If he feels 3 times his normal weight at the top of the loop, I think the left side should be -3mg, and then

3mg = mv2/r,

or

3g = v2/r.

If the left side is -mg - 3mg, should that not mean he feels 4 times his normal weight?
I guess it depends on the interpretation of the question.  Should it be interpreted as 3mg total, or 3mg in addition to the normal weight?
It may be helpful to think of the person sitting on a scale that reads 3 times his weight.
A good question to ask would be: How fast would the plane be going if the pilot felt weightless? In this case the left side would be -mg.

Correction on my original left side; -mg-3mg. (I forgot the m in the second term). But I think you figured that out.
From the wording of the problem, the pilot feels a total downward force of 3mg at the top, hence the total force pointing toward the center of the loop would be 3mg.  That includes his normal weight of mg, unless the problem intended to mean that the downward "push" felt like it added 3mg to his weight.
I think you're on the right track. But the force due to the plane is 3mg and then you must add the gravitational force.
OK, I see it now. Your example of feeling weightless helped. If he feels weightless, there is no additional force from the plane, but the mg is always there, whether he feels it or not. Then the total centripetal force at the top is mg. But if he feels 3mg at the top, there is still the mg that is already there, so the total centripetal force magnitude at the top is

(his true weight) + (the weight he feels) = mg + 3mg