Centripetal Force: Solving with Tension in a Cylinder

[raman911]

Homework Statement

http://img147.imageshack.us/img147/7830/scan0001ym3.png

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]

please 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]

please hep