Help with apparent weight problem

In summary, the conversation is about a student needing help with a physics problem involving centripetal force and an airplane. The problem is asking for the ratio between the apparent weight and the true weight of the pilot. The solution involves using the equation F=ma_c and the ratio a_c=v^2/r to find the answer of 19800.
  • #1
vee123
13
0
I need help with a physics problem ASAP please :)

I am currently in AP Physics and we are studying centripetal force. My teacher gave me a problem that is driving me crazy because I don't know where to begin with it, and I have a test tomorrow...

Here's the problem:

An airplane pulls out of a dive in a vertical circle of radius 1.0 km traveling with a speed of 550km/h. How many times greater is the apparent weight of the pilot than his true weight?

Can someone please help me?
 
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  • #2
The true weight would be your weight if you were standing on the surface of the earth. This is because your weight is the normal force acting on you. If you were just standing on the surface, the normal force would oppose gravity which is the only other force acting on you. Thus, your weight would be the force of gravity, mg. In this situation, why would that be any different?
 
  • #3
vee123 said:
I am currently in AP Physics and we are studying centripetal force. My teacher gave me a problem that is driving me crazy because I don't know where to begin with it.

Here's the problem:

An airplane pulls out of a dive in a vertical circle of radius 1.0 km traveling with a speed of 550km/h. How many times greater is the apparent weight of the pilot than his true weight?

Any suggestions?

Here's another hint: It's important where in the loop the plane is (does the question specify that?)
 
  • #4
It doesn't specify where in the loop the plane is. I just want to know how to get through this problem. I'm so confused! :(
 
  • #5
[tex]
F = ma_c
[/tex]

and

[tex] a_c = v^2/r [/tex]

so the ratio between the centripetal Force and the normal weight 9.8m is

[tex]\frac{m(v^2/r)}{9.8m} = \frac{1980000000}{1000*9.8} = 19800[/tex]
 

1. What is apparent weight?

Apparent weight is the perceived weight of an object when it is in motion or in a non-inertial reference frame. It is affected by factors such as acceleration and gravitational pull.

2. Why is apparent weight important?

Understanding apparent weight is important in various fields such as physics, engineering, and sports. It helps us to accurately calculate forces and design structures that can handle different forces.

3. How is apparent weight different from actual weight?

Actual weight is the force of gravity acting on an object, while apparent weight is the force perceived by an object in motion. This means that an object's apparent weight can be different from its actual weight in certain situations.

4. How can I calculate apparent weight?

To calculate apparent weight, you need to know the object's mass, its acceleration, and the gravitational pull. The formula is Apparent Weight = Mass x (Acceleration due to gravity - Acceleration of reference frame).

5. Can apparent weight be negative?

Yes, apparent weight can be negative. This happens when the acceleration of the reference frame is greater than the acceleration due to gravity, causing the object to feel like it has less weight. This is common in situations like free-fall or in elevators that are accelerating downwards.

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