Torque , Pseudo Torque

Torque :

Torque is the turning or twisting action of a force, acting on a body, about its axis of rotation.

If a force $\vec{F}$  is acting at a point P on a rigid body, then the torque due to the force $\vec{F}$ acting on the body, measured about the axis of rotation (n), is given by

$ \displaystyle \vec{\tau} = \vec{r}\times \vec{F} $

Where $\vec{r}$ is position vector of point P w.r.t. origin O

Magnitude of the torque is equal to the product of the force and the shortest distance between the force and the axis of rotation.

$ |\vec{r} | = r F sin\theta $  ; where θ = angle between $\vec{r}$ and $\vec{F}$

Illustration : A force $ \displaystyle \vec{F}= (2\hat{i} + 3\hat{j} + 4\hat{k}) $ is applied to a point having position vector $ \displaystyle \vec{r}= (3\hat{i} + 2\hat{j} + \hat{k}) $
Find the torque due to the force about the origin.

Solution: $ \displaystyle \vec{\tau} = \vec{r}\times \vec{F} $

$ \displaystyle = (3\hat{i} + 2\hat{j} + \hat{k}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) $

$ \displaystyle \vec{\tau}= (5\hat{i} – 10\hat{j} + 5\hat{k}) $

$ \displaystyle \tau= \sqrt{5^2 + (-10)^2 + 5^2} = 5 \sqrt{6}  N-m$

Exercise : A particle describes uniform circular motion. Find the torque acting on it about the centre of the circle.

Work done by a Torque

Work done by torque τ is W = τ θ ;  θ = angular displacement

For variable torque

$ \displaystyle dW = \tau d\theta $

$ \displaystyle W = \int\tau d\theta $

Illustration : A tangential force of F = 1.5 N acts on a particle of mass m = 2 kg revolving in a circular path of radius r = 3m. What is the work done by the torque for a complete revolution of the particle?

Solution: Work done by the torque is given by

$ \displaystyle W = \int\tau d\theta $ ; where τ = rF

$ \displaystyle W = r F \int d\theta $

Putting θ = 2π for a complete revolution we obtain

W = 2πrF

⇒ W = 2π × 3 × 1.5 = 9π

Pseudo Torque :

If motion of the body is observed from a non-inertial frame of reference having an acceleration a in a fixed direction with respect to an inertial frame, one has to apply a pseudo force (-ma) to each particle. These pseudo forces produce a pseudo torque about the axis.

In such a case
τext = Iα does not hold.

But there exists a very special and very useful case when τext = Iα does hold even if the angular acceleration α is measured from a non-inertial frame A.

That special case occurs when the axis of rotation passes through the centre of mass.

Also Read :

→ Motion of Rigid body & Rotational Kinematics
→ K.E of Rotation & Moment of Inertia
→ Moment of inertia of a thin ring & Radius of Gyration
→ Parallel & Perpendicular axes Theorem
→ Angular momentum & its Conservation
→ Relation b/w torque & angular acceleration
→ Combined Rotation & Translation

Next Page → 

←Back Page

Leave a Reply