Definition of work done by torque

[Cedric Chia]

I' m trying to derive the work done by a torque from W = ∫ F ⋅ ds and I' ve looked up the internet, it said:

W = ∫ F ⋅ ds ( since ds = dθ × r ) ---------------------------------------- ( Line 1 )

it can be written as

W = ∫ F ⋅ dθ x r

this is a vector triple product , thus can also be written as

W = ∫ r × F ⋅ dθ

W = ∫ Torque ⋅ dθ ----------------------------------------------------- ( Line 2 )

My question is :
In what direction is dθ pointing so that when I cross-product dθ and r ( Line 1 ), it become ds ? And also, when I dot-product Torque and dθ ( Line 2 ) , what is the angle between them ?

 

[haruspex]

I' m trying to derive the work done by a torque from W = ∫ F ⋅ ds and I' ve looked up the internet, it said:

W = ∫ F ⋅ ds ( since ds = dθ × r ) ---------------------------------------- ( Line 1 )

it can be written as

W = ∫ F ⋅ dθ x r

this is a vector triple product , thus can also be written as

W = ∫ r × F ⋅ dθ

W = ∫ Torque ⋅ dθ ----------------------------------------------------- ( Line 2 )

My question is :
In what direction is dθ pointing so that when I cross-product dθ and r ( Line 1 ), it become ds ? And also, when I dot-product Torque and dθ ( Line 2 ) , what is the angle between them ?

A rotation can be represented as a vector along the axis of rotation, so it is normal to the plane of rotation. If the force lies in that plane then it will be parallel to the rotation vector. But in general it need not be. E.g. consider a car skidding at an angle, brakes off. The frictional force wIll be at any angle to the rotation axis.

 

[Cedric Chia]

A rotation can be represented as a vector along the axis of rotation, so it is normal to the plane of rotation. If the force lies in that plane then it will be parallel to the rotation vector. But in general it need not be. E.g. consider a car skidding at an angle, brakes off. The frictional force wIll be at any angle to the rotation axis.

Thanks for the reply, so the dθ is not in the same direction with the unit vector θ hat ? ( which we introduced in the polar coordinates, θ hat is tangent to the circular motion ) . Instead, dθ is perpendicular to the plane of circular motion ( same direction as angular velocity ) ?

 

[Orodruin]

Instead, dθ is perpendicular to the plane of circular motion ( same direction as angular velocity ) ?

Yes. In fact, $d\theta/dt$ is the angular velocity.

 

[Cedric Chia]

Yes. In fact, $d\theta/dt$ is the angular velocity.

Thank you