Using the formula of torque `tau =xF_(y)-yF_(x),` derive the polar formula of torque.

Explanation

Let an external force `vecF` acting on particle P . Suppose the line of action of `vecF` makes an angle `alpha` with the x-axis.
From Fig.
`F_(x)= Fcos alpha" "cdots (1)`
`F_(y) = F sin alpha" "cdots(2) `
If the coordinates of P are denoted by (x,y) where `vec(OP)=vecr" and"" "anglePOX=theta `, then
x `=rcostheta " "cdots(3)`
and y= `rsintheta" "cdots(4)`
Putting the values of `F_(x), F_(y),` x and y in expression of torque `tau= xF_(y)-yF_(x)` we get,
`tau = rF [ sin alphacos theta -cosalpha sin theta] = rFsin (alpha -theta) " " cdots(5)`
Let the line of action of `vec F` makes an angle `phi` with the position vector `vecr`.
From figure `theta +phi=alpha`
or, `phi= alpha-theta " "cdots (6)`
Putting the value of `phi` in equation (5),
`tau = rFsin phi`
which is the expression of torque in polar coordinates.