# Homework Help: Central force

1. Oct 14, 2007

### raman911

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

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

Last edited by a moderator: May 3, 2017
2. Oct 14, 2007

### raman911

Can Any One Help Me?

3. Oct 14, 2007

### raman911

I Am Wating From Last Hour

4. Oct 14, 2007

### raman911

Plz Help Me

5. Oct 14, 2007

### raman911

Plz Help Me

6. Oct 14, 2007

### raman911

Can Any One Help Me?

7. Oct 14, 2007

### PiratePhysicist

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)

8. Oct 14, 2007

### raman911

what to do in d ?

9. Oct 14, 2007

### raman911

Can Any One Help Me?

10. Oct 14, 2007

### raman911

what to do in d ?

11. Oct 14, 2007

### raman911

plz help hurry its due after 3 hours

12. Oct 14, 2007

### PiratePhysicist

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

13. Oct 14, 2007

### raman911

than u . can u help me in e. what formula i need to use in e.

14. Oct 14, 2007

### 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
$$\omega = 2 \pi f$$

15. Oct 14, 2007

### raman911

what is w is that v^2

16. Oct 14, 2007

### PiratePhysicist

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

17. Oct 14, 2007

### raman911

ok

$$\mu m \omega r = mg$$

why u wrote one more r

18. Oct 14, 2007

### 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
$$\mu m\omega^2r=mg$$
$$\omega=\sqrt{\frac{g}{\mu r}}$$

19. Oct 22, 2007

plz hep