Rod's Physics 3/4 Questions Thread

[PB]

Yes, it is only for finding the angle of banked, circular roads. It makes solving this particular type of question so much easier!

[Rishi97]

A stunt pilot appearing at an air show decides to perform a vertical loop so that she is upside down at the top of the loop. She maintains a constant speed of 35m/s while completing the 100m radius loop
Calculate the apparent weight of the 80kg pilot when she is at the top of the loop
Thanks

[katie101]

If a dead battery and a new battery were used in series, would the well used battery have any effect on the output voltage of the new battery? Why/why not?

[Homer]

A stunt pilot appearing at an air show decides to perform a vertical loop so that she is upside down at the top of the loop. She maintains a constant speed of 35m/s while completing the 100m radius loop
Calculate the apparent weight of the 80kg pilot when she is at the top of the loop
Thanks

180N ?

[lzxnl]

If a dead battery and a new battery were used in series, would the well used battery have any effect on the output voltage of the new battery? Why/why not?

Not directly, but the dead battery would then just be an extra resistor in the circuit. Although the new battery's output voltage itself wouldn't be affected, the current in the circuit would be.

Hey Rishi,
Good work on part a.
As for Part B, try using this handy formula for finding the angle of banked roads -> tan(theta) =v^2/gr

v=design speed only

hopefully that gets you the right answer.

I would suggest you learn where that formula comes from, or else you risk becoming the robotic drone that VCE physics turns its students into.

A stunt pilot appearing at an air show decides to perform a vertical loop so that she is upside down at the top of the loop. She maintains a constant speed of 35m/s while completing the 100m radius loop
Calculate the apparent weight of the 80kg pilot when she is at the top of the loop
Thanks

So, the net force is directed to the centre of the circle, which is downwards. The normal reaction is pointing downwards, as is the weight force. Hence, mv^2/r = mg + N. N= mv^2/r - mg = 80 kg (v^2/r - g) = 80(1225/100 - 10) = 80*2.25 = 180 N, as stated above

[Rishi97]

Thanks Homer and lzxnl for trying But the answer is 200 N. Could you try again?

[Homer]

pretty sure its 180N, unless the book used 9.8 as g and rounded up? :s

[lzxnl]

This is why I think VCE physics makes no sense. Why don't VCAA and textbooks just agree on a value for g to use? In the exam, you have to use g = 10 m/s^2, although that's not quite accurate. Here, if you replace g with 9.8, you see that we have 80(11.25 - 9. which is 80*0.2 = 16 more than the answer we got before; after rounding up, you get 196 N => 200 N.

[Rishi97]

This is why I think VCE physics makes no sense. Why don't VCAA and textbooks just agree on a value for g to use? In the exam, you have to use g = 10 m/s^2, although that's not quite accurate. Here, if you replace g with 9.8, you see that we have 80(11.25 - 9. which is 80*0.2 = 16 more than the answer we got before; after rounding up, you get 196 N => 200 N.

I totally agree. This is why I'm confused in physics 90% of the time. lol
Anyway, thanks for helping me out
I'll post many other questions throughout the year

[Rod]

help here;

a train goes around a bend of radius 350m that is banked at an angle of 8 degrees. At what maximum spped can the train safely travel around the bend?

[Rishi97]

help here;

a train goes around a bend of radius 350m that is banked at an angle of 8 degrees. At what maximum spped can the train safely travel around the bend?

try using this formula
tan(angle) = v^2/gr

[lzxnl]

Please learn where this comes from though

[Rishi97]

Please learn where this comes from though

I've shown this formula to my teacher and he doesn't know where it came from.
Could u kindly explain it?

thanks

[lzxnl]

I've shown this formula to my teacher and he doesn't know where it came from.
Could u kindly explain it?

thanks

You kidding!? Your 3/4 physics teacher can't explain a formula!? Dear me.
Look at the attached diagram. Now, the vertical component of the normal reaction force must balance the weight force as the object isn't moving up or down (horizontal circle?). The horizontal component of the normal reaction force is the only horizontal force component, so it is equal to mv^2/r. Using trig, we find that tan theta = mv^2/r divided by mg, or mv^2/mgr = v^2/gr as requested

[Rod]

I've shown this formula to my teacher and he doesn't know where it came from.
Could u kindly explain it?

thanks

Rishita, pleasssssseeeeeeee make your own thread