Rotational motion question -- Wire wound around a rolling spool

[anonymous99]

Homework Statement

A spool with inside radius R1 =80mm and outer radius R2= 160mm rolls without slipping on a rough horizontal surface. Mass of spool =5kg and moment of inertia = 0.072kg m^3. A wire is wound round the inner radius and a force, F, of 20N is applied horizontally to the right. Calculate the linear/angular accelerations of the spool.

Relevant Equations

Sum of forces in x direction= F - Fr where Fr is friction force on the rim.

Sum of torques = F*R1 - Fr*R2 = I*a/R2
I did some substitution and rearranging to get rid of Fr and got F*R1-F*R2 = a(I/R2 - M*R2) which gives a= 4.57 ms^-2 which is wrong. If i take clockwise torque as positive instead, I get -F*R1+Fr*R2= I*a/R2 which rearranges to F*R2-F*R1 = a(I/R2 + M*R2) which gives me the right value of 1.28.

Shouldn't the answer have the same magnitude regardless of sign convention?

 

[BvU]

Hello John, !

Sum of torques = F*R1 - Fr*R2 = I*a/R2

Is that so ? Can you show a diagram (that is consistent with 'Sum of forces in x direction= F - Fr ') ?

Shouldn't the answer have the same magnitude regardless of sign convention?

Definitely, yes.

PS: 'I did some substitution and rearranging ' doesn't help me to help you !

 

[anonymous99]

Hello John, !

Is that so ? Can you show a diagram (that is consistent with 'Sum of forces in x direction= F - Fr ') ?

Definitely, yes.

PS: 'I did some substitution and rearranging ' doesn't help me to help you !

Yes, I've attached a diagram of the spool. From the diagram, if CCW rotation is positive then F should have a positive torque and Fr negative. And sorry, I'll expand on my working.

F*R1 -(F- ma)R2 = I*a/r
F*R1 -F*R2+ma*R2= I*a/r
F(R1-R2)= I*a/r - ma*R2 = a(I/R2- mR2)
a= F(R1-R2)/(I/R2-mR2)

This gives me the wrong answer. What's wrong with my working here?

 

[haruspex]

F*R1 -(F- ma)R2 = I*a/r

You problem is consistency of signs relating to the rotation.
On the LHS you are taking anticlockwise as positive, so on the right, if you are taking acceleration to the right as positive, it should be -a/r.

For problems like these, where you do not need to find the value of the frictional force, it is generally simpler to take the point of rolling contact as the axis.

 

[BvU]

Interesting: you let F act at the lower side of the core. How do you know it doesn't act at the side above ?

 

[anonymous99]

Why would acceleration be negative, is it not accelerating in the positive x direction since that's where its direction of motion is?

 

[anonymous99]

And I let it act at the bottom because that's where the frictional force acts on the wheel to stop it from sliding.

 

[BvU]

So you want to roll up a wire on a spool by pulling it towards you. Should be difficult

Re

Why would acceleration be negative, is it not accelerating in the positive x direction since that's where its direction of motion is?

That's circular reasoning. If the thing is accelerating in the positive direction, that means that angular acceleration is anti-clockwise.

Also, the way you are treating the forces, F_r must come out positive. That doesn't happen when a > F/m (=4 m/s²).

 

[DEvens]

So you want to roll up a wire on a spool by pulling it towards you. Should be difficult

Are you sure it's difficult? Maybe this spool looks a lot like a popular toy.



It's actually a very old question in this genre. One variation on the question is, what ratios of inner and outer radii work so that this does wind up the wire?

 

[DEvens]

And I let it act at the bottom because that's where the frictional force acts on the wheel to stop it from sliding.

The question is about where the wire attaches to the central core, not the place the friction acts on the outside of the spool.

You put it at the bottom of the core. The question is, how do you know it should be at the bottom and not the top. If I read the question correctly, it does not say whether it's at the top or the bottom. So you need to make it explicit you assumed it. Or work the question twice, once with top and once with bottom.

At the bottom you get "wind the dog." At the top you get "unwind the dog."

 

[BvU]

Since we are given the answer: this is indeed a 'wind the dog' case ... it may even be realistic if the friction coefficient is > 0.277

 

[haruspex]

Why would acceleration be negative, is it not accelerating in the positive x direction since that's where its direction of motion is?

The acceleration, a, is positive, but you are taking anticlockwise as a positive rotation. If the disc rotates positively the disc goes to the left, so the relationship is a=-αr.

 

[haruspex]

Interesting: you let F act at the lower side of the core. How do you know it doesn't act at the side above ?

View attachment 252871

I had been going to ask the same, but since the image in post #3 has the frictional force added by hand I presumed the rest of it was provided with the question.

 

[anonymous99]

Doesn't the frictional force have to act there though as that's where the wheel comes into the road and the static friction force pushes it to the left to stop it from sliding?

 

[anonymous99]

The acceleration, a, is positive, but you are taking anticlockwise as a positive rotation. If the disc rotates positively the disc goes to the left, so the relationship is a=-αr.

Ah ok, that gives me the right answer now. Thanks :)