IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# Torque

## Problems from IIT JEE

Problem (IIT JEE 1998): The torque $\vec{\tau}$ on a body about a given point is found to be equal to $\vec{A}\times\vec{L}$, where $\vec{A}$ is a constant vector and $\vec{L}$ is the angular momentum of the body about that point. From this its follows that

1. $\frac{\text{d}\vec{L}}{\text{d}t}$ is perpendicular to $\vec{L}$ at all instants of time.
2. the component of $\vec{L}$ in the direction of $\vec{A}$ does not change with time.
3. the magnitude of $\vec{L}$ does not change with time.
4. $\vec{L}$ does not change with time.

Solution: The torque is equal to the rate of change of angular momentum, \begin{align} \label{job:eqn:1} {\mathrm{d}\vec{L}}/{\mathrm{d}t}=\vec{\tau}=\vec{A}\times\vec{L}. \end{align} The cross product $\vec{A}\times\vec{L}$ is perpendicular to both $\vec{A}$ and $\vec{L}$ and hence ${\mathrm{d}\vec{L}}/{\mathrm{d}t}$ is perpendicular to $\vec{L}$. Let $\vec{L}=L_x\hat\imath+L_y\hat\jmath+L_z\hat{k}$ and $\vec{A}=A\,\hat\imath$ ($\vec{A}$ being a constant vector allows us to define $x$ axis along $\vec{A}$). Substitute in above equation to get, \begin{align} \label{job:eqn:2} {\mathrm{d}L_x}/{\mathrm{d}t}\;\hat\imath+{\mathrm{d}L_y}/{\mathrm{d}t}\;\hat\jmath+{\mathrm{d}L_z}/{\mathrm{d}t}\;\hat{k}=\vec{A}\times\vec{L}=-AL_z\hat\jmath+AL_y\hat{k}, \end{align} which gives, \begin{align} \label{job:eqn:3} &{\mathrm{d}L_x}/{\mathrm{d}t}=0,\\ \label{job:eqn:4} &{\mathrm{d}L_y}/{\mathrm{d}t}=-AL_z,\\ \label{job:eqn:5} &{\mathrm{d}L_z}{\mathrm{d}t}=AL_y. \end{align} The equation gives $L_x=\text{constant}$. Differentiate $\vec{L}\cdot\vec{L}=L^2$ and use above equations to get, \begin{align} \label{job:eqn:6} \frac{\mathrm{d}L^2}{\mathrm{d}t}&=\frac{\mathrm{d}}{\mathrm{d}t}\left(L_x^2+L_y^2+L_z^2\right)=2L_x\frac{\mathrm{d}{L_x}}{\mathrm{d}t}+2L_y\frac{\mathrm{d}{L_y}}{\mathrm{d}t}+2L_z\frac{\mathrm{d}{L_z}}{\mathrm{d}t}\nonumber\\ &=2L_x (0)+2L_y(-AL_z)+2L_z(AL_y)=0. \end{align} From above equation $\frac{\mathrm{d}L^2}{\mathrm{d}t}=2L\frac{\mathrm{d}L}{\mathrm{d}t}=0$, which gives $L=\text{constant}$.