How Do You Solve This Rotational Motion Problem?

[coldblood]

Hi friends,
Please help me in solving this problem, I'll appreciate the help.

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


Thank you all in advance.

 

[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?

 

[coldblood]

hi coldblood!

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

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

 

[tiny-tim]

(just got up :zzz:)

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

(ie if the force goes exactly through the point of contact with the horizontal surface)

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

nooo (and what was your reason?)

hint: where is the centre of rotation?

what is the total torque about the centre of rotation? ​

 

[coldblood]

](ie if the force goes exactly through the point of contact with the horizontal surface)

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.

 

[tiny-tim]

If the external force acts at the bottom point horizontally,

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

 

[coldblood]

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

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

 

[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?

 

[coldblood]

??

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?

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.

 

[tiny-tim]

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.

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?

 

[coldblood]

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?

Can you please make a figure for the three? Please

 

[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!)

 

[coldblood]

(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!)

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

 

[tiny-tim]

hi coldblood!

your first equation, α = F(b - a)/I_P 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)​