# Rotational Motion Problem - 14

1. Dec 16, 2013

### coldblood

Hi friends,

The problem is as:

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1470127_1461728164054289_845411707_n.jpg

Attempt -

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1468786_1461728244054281_988835356_n.jpg

2. Dec 17, 2013

### tiny-tim

hi coldblood!

you're only considering the torque of F (about the centre) … what about the torque of the friction?

hint: the clue is in the diagram

what do you think that dotted line is there for (the extension of F)?

what would happen if the angle was larger, and that dotted line went exactly through the bottom of the spool?

3. Dec 17, 2013

### coldblood

thread unwinds, spool rotates counter-clock and friction would act right wards. Is that correct?

4. Dec 18, 2013

### tiny-tim

(just got up :zzz:)
(ie if the force goes exactly through the point of contact with the horizontal surface)
nooo (and what was your reason?)

hint: where is the centre of rotation?

what is the total torque about the centre of rotation?

5. Dec 18, 2013

### coldblood

If the external force acts at the bottom point horizontally,

It'll provide spool an anti clockwise torque, friction will act leftwards. And I think thread should unwind.

Because I found some equations: if the force acts at the top point horizontally in + x direction, It generates a torque in clockwise manner and direction of friction is rightwards.

6. Dec 18, 2013

### tiny-tim

no, you're not understanding what i'm saying …

i'm saying, if the angle α is increased so that the pulling force F is further round, and steeper, and so the line of action of F (as well as f) goes through that bottom point

7. Dec 18, 2013

### coldblood

Then the spool will be rotating clockwise, friction will act right right and thread unwinds.

8. Dec 18, 2013

### tiny-tim

??

how many forces are there on the spool?

what is the torque of each of them about the bottom point?

so what is the total torque?

9. Dec 18, 2013

### coldblood

About bottom point only the torque of F will act because weight, frictional force and the reaction from the ground will pass through that point only.

10. Dec 18, 2013

### tiny-tim

exactly!

so if the line of action of F passes
(a) to the left of
(b) to the right of
(c) directly through​
the bottom point, which way will the spool turn?

11. Dec 18, 2013

### coldblood

12. Dec 18, 2013

### tiny-tim

(a) is the diagram in the top-left corner of your printed question

for (b) and (c), if you move the string further round, α will increase, and the dotted line will slide over to the right

(eg if α = 90°, it's obviously to the right!)

13. Dec 19, 2013

### coldblood

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-prn2/1476572_1462877957272643_1028005643_n.jpg
https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/936643_1462877960605976_1351577648_n.jpg

14. Dec 19, 2013

### tiny-tim

hi coldblood!

your first equation, α = F(b - a)/IP is correct only if the angle of F is 0

you have drawn F in the wrong place, F does not come out of the bottom of that small circle, it has to be tangential, so it comes from a point at the same angle from the vertical as F is from the horizontal

and you do not really need to calculate it exactly, or to use k

it's enough to say "the only torque about P is from F, which is clearly clockwise, and so the spool rotates clockwise"

similarly, the only torques about C are F (anticlockwise) and f … since we know the spool rotates clockwise, that means that the torques of f must be … ?

now do the other two cases, (b) and (c) (with a corrected diagram)​