# Sth about centripetal force. pls help ><

1. Oct 25, 2004

sth about centripetal force. pls help!! ><

i got some problem about centripetal force. in this experiment, " http://www.smcc.edu.hk/phy/Experiments.files/Centripetal Force.doc "
my hand apply a horizontal force, to give the velocity 'ω' to the stopper thing. when i decompose ω into vectors, i got a centripetal force to the centre. but then the problem comes. i don't undrstand y Mg doesn't fall down. as the centripetal force (Tension) has a direction to centre and Mg is pointing downwards, y don't it fall?

is that becoz when there is a 'ω', there will be a centripetal force(T) with a magnitude, which hv a length 'r' (radius)? so when the bung is whirled, there will be a length 'r', which make Mg remain at the same level?

but then there'll be another problem... if Mg doesn't fall, according to Newton's 3rd law, there is another upward force. is that real?

i'm just thinking when the force applied is small, the deflected angle of the bung is large. is it becoz the centripetal force is small? if it is, the vertical component of the bung is then smaller! how can it keep 'mg' at the same level then?

pls help!! ><

2. Oct 25, 2004

### arildno

Now, would you agree that the tension force from the string acts parallell to the string?

3. Oct 25, 2004

isn't tension acts along the string??

4. Oct 25, 2004

### arildno

True, that's what I meant!
The string was never EXACTLY horizontally aligned, was it?

5. Oct 25, 2004

yes. but is there a chance for the string to move up instead of slightly deflected?

6. Oct 25, 2004

### arildno

If you start accelerating the system from an equilibrium state, so that the angular velocity increases, then you'll see that the string (and the stopper) moves upwards in the sense of becoming "more" horizontal.

7. Oct 25, 2004

why can't i make it more horizontal if i tried to keep L constant throughout the experiment ?? when i'm doing this 'experiment', the deflection is too large that the bung hit my head!

8. Oct 25, 2004

### arildno

Hmm..I'm a bit unsure about the last comment, but I'll answer your question "why doesn't it fall down?"

Since we agree that the string is NOT exactly horizontal, the tension force has an UPWARDS component (opposite the direction of gravity!).
Now, if the angle between the rod and the taut string is $$\theta$$, we let "T" be the magnitude of the tension force, (Or simply the tension in the string).
Can you agree that the requirement that the rubber bung does NOT fall down must be:
$$T\cos\theta-mg=0$$
(m is the mass of the rubber bung)
The equation is simply Newton's 2.law in the vertical with the requirement of no vertical acceleration.

9. Oct 25, 2004

yes i understand the equation. how can we make the forces be balanced??

10. Oct 25, 2004

### arildno

You see that the given equation implies that balance in the VERTICAL, requires that the magnitude of the tension force must satisfy:
$$T=\frac{mg}{\cos\theta}$$
Agreed?

11. Oct 25, 2004

what is the meaning of this equation??

12. Oct 25, 2004

### arildno

The meaning of this equation is that the MAGNITUDE of the tension force nust satisfy this relation, in order for us to have balance of forces in the VERTICAL.

13. Oct 25, 2004

oh i see. : )
just now i realized the centripetal force is from Mg and Mg=T...
so the relation between M and m is M=m/cosθ ??

14. Oct 25, 2004

### arildno

With "m" I meant the mass, which you denoted by "M"

Now let us talk about the CENTRIPETAL force!
Since you move the bung in a HORIZONTAL circle, the CENTRIPETAL ACCELERATION must lie in the HORIZONTAL plane, and the CENTRIPETAL FORCE must be in the horizontal as well.

Hence, the centripetal force is nothing but THE HORIZONTAL COMPONENT OF THE TENSION FORCE.
The tension force itself has TWO components:
1)The vertical component: This balances gravity so the bung doesn't fall down.
2)The horizontal component: This is simply the centripetal force which makes the bung travel in a CIRCLE

Did you get that?

15. Oct 25, 2004

oh...
one more question: where is the tension force comes from? if it comes from Mg, why can it balance the gravity?

16. Oct 25, 2004

### arildno

It does NOT "come from" Mg (that's gravity), it comes from that your string stretches out a bit; the constituent molecules in the string opposes that stretch, and tries as best as they can to keep the length as close to the equilibrium length as possible.

Think of the string as a minuscule chain if you like; you can coil and bend a chain quite a lot, but what you won't be able to do is to lengthen it..

17. Oct 25, 2004

then i see. i should mind the wordings, right? : )
i heard sb sayin' if the angle of deflection is high, the friction between the glass rod and string will increase. but the contact surface area is the same!! how can we explain that??

18. Oct 25, 2004

### arildno

1)Now consider a box of mass M standing on the floor.
A NORMAL FORCE of value Mg works upwards to keep the books from falling through, right?
2)Now, put another box of mass M on top the other; clearly the normal force is now 2Mg; but the contact area between the floor and the box hasn't appreciably increased, has it?
(On a tiny, microscopic scale, that which corresponds to surface area will probably have increased somewhat, but macroscopically, we won't notice that)
3) To the question:
When the system whirls extremely fast, the string is pushed more strongly against the glass surface.
Hence, the NORMAL FORCE from the glass on the string increases.
Since it often is true that the FRICTION is proportional to the normal force, the frictional force increases as well..

19. Oct 25, 2004

### arildno

I have to retract 3) a bit:
Assume that the lower mass is at rest.
Then the tension in the rope segment attached to it must be:
$$T_{low}=m_{low}g$$
Where $$T_{low}$$ is the tension is the lower, free-hanging end of the string, whereas $$m_{low}$$ is the mass of the object dangling from it.
Consider now the rope segment inside the glass rod.
At the top, we know that the tension is what is needed to whirl the system around, that is:
$$T_{up}=\frac{Mg}{\cos\theta}$$
A STATIC friction F (from the glass rod) must keep this rope segment at rest in the vertical direction; hence:
$$T_{up}-T_{low}-F=0$$
Or:
$$F=\frac{Mg}{\cos\theta}-m_{low}g$$
From this expression, we see that F increases with the angle.