Components of torque in a solid sphere

[KiNGGeexD]

A uniform sphere of mass M and radius R has a point on its surface fixed at the origin. Its centre lies along a line in the direction of the position vector r = i + 2k + 3k at length R. Find the components of the torque acting on it due to gravity if the z-direction is upwards and gravity acts downwards.

Solution:

First thing I noticed was that there was no y component in the vector r so I figured perhaps there is no torque in that component!

I tried τ= r x sinθ

And simply used r as 5k for the z component and i for the x component!

I used mg as the force and used θ as 90 because all axis are mutually perpendicular, but I think this is the wrong approach! I think the vector r would play more of role!

Also would I use τ= Iαθ

Because it's a sphere? We only touched on rotational torque very briefly in lectures and for this question there is no model answer!:(Thanks for any help:)

 

[Tanya Sharma]

A uniform sphere of mass M and radius R has a point on its surface fixed at the origin. Its centre lies along a line in the direction of the position vector r = i + 2k + 3k at length R. Find the components of the torque acting on it due to gravity if the z-direction is upwards and gravity acts downwards.

Solution:

First thing I noticed was that there was no y component in the vector r so I figured perhaps there is no torque in that component!

I tried τ= r x sinθ

And simply used r as 5k for the z component and i for the x component!

I used mg as the force and used θ as 90 because all axis are mutually perpendicular, but I think this is the wrong approach! I think the vector r would play more of role!

Also would I use τ= Iαθ

Because it's a sphere? We only touched on rotational torque very briefly in lectures and for this question there is no model answer!:(


Thanks for any help:)

It seems there is a typo in the position vector 'r' given in the question.It should be r = i + 2j + 3k .

What is the position vector of the point of application of gravity ?
Write the force of gravity in vector form .
Calculate the cross product of 'r' and 'F' .This would give you the torque .

 

[KiNGGeexD]

The vector for gravity would be

mg in the downward k direction

So -2k?

 

[Tanya Sharma]

The vector for gravity would be

mg in the downward k direction

So -2k?

Why -2k ?

 

[KiNGGeexD]

If the torque due to gravity is z downwards! Is the torque in the positive z direction 0 as they are parallel or antiparallel rather

 

[Tanya Sharma]

If the torque due to gravity is z downwards! Is the torque in the positive z direction 0 as they are parallel or antiparallel rather

You are misinterpreting the question.

The question asks you to consider positive direction of z-axis upwards and gravity downwards . If the positive z-direction is ## \hat {k} ## ,then gravity points in ## -\hat {k} ## .

What is the magnitude of force of gravity on the sphere ?
How do you represent force of gravity in vector form ?

 

[KiNGGeexD]

The force =mg -z?

 

[Tanya Sharma]

The force =mg -z?

No...

## F = -Mg\hat{k} ## .

 

[KiNGGeexD]

Sorry I didn't mean z!
Do I use τ = r x Fsinθ

 

[nasu]

You could but will be a roundabout way. You don't know the angle so you will need to find the angle first.
It would be easier to just cross multiply the two vectors in their component form.

 

[KiNGGeexD]

Do the cross product of the two vectors?

 

[KiNGGeexD]

And my component vector for force would just be -mg k?

So there would be no torque in the x and y

 

[nasu]

It will be. It's a cross product.
What is i x j, for example? Is not the same as i.j (dot product).

 

[KiNGGeexD]

It will be. It's a cross product.

What is i x j, for example? Is not the same as i.j (dot product).

The way I was taught to do the cross product was in the form of a matrix, not sure!

i x j is equal to z isn't it?

 

[nasu]

i x j = k
j x k =i
k x i = j

And if the order of the terms is changed, you have a minus sign.
For example, j x i = -k.

 

[KiNGGeexD]

So for the x component I would do

τ= r x F

And use just the i component of the vector r?

 

[nasu]

Why just the i component?
Multiply term by term.
Like you do in algebra when you have something like (a + b +c)x(p+q+r).
Then you will see what components are present in the result.

 

[KiNGGeexD]

But then the result will be the same for every component will it not? What is changing?

 

[nasu]

I don't understand your question.
Just do the multiplication.

( i + 2j + 3k)x(-mgk)

and see what you get.
You are worrying too much for a trivial question.

 

[KiNGGeexD]

I would get

mg j - 2mg i + 0

 

[nasu]

OK. So now you have all the components of the torque. The k component is zero. The other two are not.

And you should take into account some units in the expression of r.
It should be in meters or some other unit of length.
I mean, the i component should be 1m and not just 1. (for example)

 

[KiNGGeexD]

So is that the question finished??

All would be in Nm^-1 wouldn't they

 

[nasu]

You mean N/m?
How would r multiplied by F be in N/m?

The part about "calculating the components of the torque" is done, yes.

 

[KiNGGeexD]

Ah ok! Torque would be in N/m^2?

 

[nasu]

What is the unit for r?
What is the unit for F?

 

[KiNGGeexD]

Units for r are metres?

Units for force are Newtons?

So what I said before

N/m ?

 

[nasu]

What do you get if you multiply Newtons by meters?

Or what do you mean by the "/"? Maybe I misunderstood you. Usually the "/" is used for division.

 

[KiNGGeexD]

Yea sorry haha

Newton metres would be the units! To avoid confusion!

This is an extremely trivial question!