Central force

[raman911]

1. The problem statement, all variables and given/known data
http://img147.imageshack.us/img147/7830/scan0001ym3.png (http://imageshack.us)





3. The attempt at a solution

A)
i think force of Tension give them necessary centripetal force directed toward the center of the cylinder.

C)
Centripetal Force

Plz help me

[raman911]

Can Any One Help Me?

[raman911]

I Am Wating From Last Hour

[raman911]

Plz Help Me

[raman911]

Plz Help Me

[raman911]

Can Any One Help Me?

[PiratePhysicist]

First, please, please, do not self bump like this.
Second, for part A it is asking what physical object is exerting the force.
Part B: If the riders vertical velocity is zero (they're not falling or rising) and that doesn't change, what must be true of their acceleration and thus the vertical force?
Part C: Centripetal force is perpendicular to gravity here. Think what is happening between the person and the ride. (Read Part E for a hint)

[raman911]

what to do in d ?

[raman911]

Can Any One Help Me?

[raman911]

what to do in d ?

[raman911]

plz help hurry its due after 3 hours

[PiratePhysicist]

For D, draw a Free body diagram. Just draw the force vectors that act on a person. Please don't keep bumping this.

[raman911]

For D, draw a Free body diagram. Just draw the force vectors that act on a person. Please don't keep bumping this.
than u . can u help me in e. what formula i need to use in e.

[PiratePhysicist]

Well, in this case we need a frictional force equal to the gravitational force (atleast). And a frictional force is dependent on the normal force, which in this case will be the centripetal force. So:
F_f=F_g

\mu F_C=F_g

\mu m \omega r = mg

\mu \omega r = g

\omega = \frac{g}{\mu r}

Then you just need to convert the angular frequency to a frequency using
\omega = 2 \pi f

[raman911]

what is w is that v^2

[PiratePhysicist]

\omega is the angular velocity:
\omega = \frac{v^2}{r}

[raman911]

\omega is the angular velocity:
\omega = \frac{v^2}{r}
ok

\mu m \omega r = mg

why u wrote one more r

[PiratePhysicist]

Whoops, I typo'ed twice
It's:
\omega = \frac{v}{r}
and
F_c=m\frac{v^2}{r}=m\omega^2 r
So
\mu m\omega^2r=mg
\omega=\sqrt{\frac{g}{\mu r}}

[raman911]

plz hep