Pulling out of a dive, a 51.0kg pilot on an airplane guides into a vertical circle with a radius of 4.91 m. At the bottom of the dive, the speed of the airplane is 129m/s. What is the apparent weight of the pilot at that point?

Question

xishem · Answer

You're going to need to calculate the centripetal acceleration, using this formula:
$$a_c=\frac{v^2}{r}$$So try that, and we can work from there.

anonymous · Answer

Ac = (129m/s)^2 /  4.91m
= 3389.205
or 3.39 x 10^3?

xishem · Answer

That's right. Now try drawing a free-body diagram of what's happening to the diver at the bottom of the circle of motion. There should be two forces acting upon him.

anonymous · Answer

|dw:1352514275144:dw|

Im not really sure how to draw this one because the way its worded, I have no clue whos going where and etc...

xishem · Answer

Well, the diver is at the BOTTOM of the circle, right? And which direction is a centripetal acceleration always in?

anonymous · Answer

inwards

anonymous · Answer

But I wasnt sure if I shouldve drawn that along the radius on the bottom?

xishem · Answer

Alright, so there are two forces acting upon the diver. One upward (centripetal acceleration) and one downward (gravitational acceleration). The free-body diagram will look like this:|dw:1352514696931:dw|

anonymous · Answer

Ok I sort of get it.... still unsure how to do the entire problem though...

xishem · Answer

Well, the sum of the accelerations is the net acceleration. So:

$$\sum a = a_c-a_g=3.39 \times 10^3 \frac{m}{s^2}-9.81 \frac{m}{s^2}=+3.38 \times 10^3 \frac{m}{s^2}$$This means that the net acceleration is upwards by that amount, and this can be converted to an apparent weight (although the apparent weight will be negative!)

xishem · Answer

Also, take note that that is a HUGE acceleration, but it makes sense. Trying to fly in a circle of radius of <5 meters at a tangential speed of 129 m/s is absolutely insane. In a real situation, the pilot would certainly die.

anonymous · Answer

Well I tried finding the weight of the pilot and it said I ws incorrect.

xishem · Answer

I get$$-1.72 \times 10^5 N$$

anonymous · Answer

Thats the answer I put in and it said I was incorrect.

xishem · Answer

Are you positive that the question is copied correctly. Because this comes out around to a really, really unreasonable 345 g-force.

anonymous · Answer

491 for radius... sorry
:S

anonymous · Answer

I had a scribble in between the 4 and 9 and read it as 4.91

xishem · Answer

That makes a lot more sense :P. Just redo the calculations in the way they've already been done, and you should be good to go.

anonymous · Answer

It ended up marking me wrong for the question, oh well. thanks anyways!