What must be its frictional force?

[crosbykins]

1. A 0.20g flea sits at a distance of 5.0cm from the centre of a rotating record.
a) If the record rotates at 77rpm, what is the centripetal force.
b) For the flea to remain at this spot, what must be its frictional force?

2. A pilot of mass 60.0kg is flying her plane in a vertically oriented circular loop. Just at the bottom of the loop, the plane's speed is 1.8*10^2km/h and the pilot feels exactly four times as heavy as she normally does.
a) What is the radius of the loop.
b)At what speed must she be flying at the top of the loop in order to feel weightless?

2. Homework Equations

Fnet = 4pi^2mrf^2
Fnet = mv^2/r

3. The Attempt at a Solution
a)
77rpm * (1/60s)
=1.3s^-1

Fnet=4pi^2(.0002kg)(.005m)(1.3s^-1)^3
=6.7*10^-5N

Therefore, the centripetal force is 6.7*10^-5N

b)
For the flea to stay in the same spot the firction must be 6.7*10^-5N also.

2.
a)
4mg = mv^2/r
r = v^2/4mg
=64m

Therefore, the radius is 64m.

b)
mg=mv^2/r
[(60.0kg)(9.8m/s^2)(64m)]/60.0kg =v^2
25m/s=v

Therefore, she must be flying at a speed of 25m/s.

DID I DO THESE QS RIGHT! ANY HELP APPRECIATED!

 

[collinsmark]

a)
77rpm * (1/60s)
=1.3s^-1

I suggest that you keep more significant figures for now. You can round later when you get to your final answers. But the rotation rate is not your final answer, so you should keep at least one more significant figure.

Fnet=4pi^2(.0002kg)(.005m)(1.3s^-1)^3
=6.7*10^-5N

Two things:

• Check your conversion. The problem statement said the radius is 5.0 cm, not 5.0 mm.
• Why are you cubing the rotational frequency?

2.
a)
4mg = mv^2/r
r = v^2/4mg
=64m

You're forgetting about the force/acceleration due to gravity. Even if she wasn't traveling in a circle at all, but rather traveling in a straight line, she would still feel 1g of acceleration (i.e. her weight would be what it is normally). The rest of the 4g's originates from the centripetal acceleration.

 

[crosbykins]

Thank you so much for your help!