I know that the basic definition of the work done on a particle is: $$W_{12}=\int_1^2\mathbf{F}\cdot\mathrm{d}\mathbf{r}$$ but what if I have want to calculate the work done on a rigid body? is the formula the same?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
The rigid body, in addition to translation, can also rotate. However, rotation isn't possible in the case of a point particle. So when we calculate the work done on a rigid body, we also need to account for the work done by the torque acting on it, in addition to the work done by the force. Thus the general expression for a work done on a rigid body by a force $\mathbf F$ and a torque $\boldsymbol{\tau}$ is
$$W=\underbrace{\int \mathbf F\cdot \mathrm d \mathbf r}_{\text{work done by the force}}+\underbrace{\int \boldsymbol{\tau}\cdot \mathrm d \boldsymbol{\theta}}_{\text{work done by the torque}}$$
where $\mathbf r$ is the position vector of the rigid body's center of mass and $\boldsymbol{\theta}$ is the angle rotated by the body.
-
$\begingroup$ Are $\mathrm{F}$ and $\mathrm{\tau}$ the net force and net torque? $\endgroup$– herosaiCommented Jun 7, 2020 at 18:04
-
$\begingroup$ @herosai Yes, $F$ and $\tau$ are the net force and net torque respectively. Also, see my edit. $\endgroup$– user258881Commented Jun 7, 2020 at 18:07
-
$\begingroup$ I think $\mathbf{r}$ should be the position vector of the center of mass. $\endgroup$– FelipeCommented Jun 7, 2020 at 19:11
-
$\begingroup$ But in case of rigid body how the energy that is transferred through work done on a particle of the body (the force acts at a point of the rigid body) translates into energy of the whole system? $\endgroup$ Commented Oct 18, 2020 at 18:53