A bar of mass $m$ resting on a smooth horizontal plane starts moving due to a constant force $F$. In the process of its rectilinear motion the angle $ heta$ between the direction of this force and the horizontal varies as $ heta=k x$, where $k$ is a constant and $x$ is the distance traversed by…

Question

A bar of mass $m$ resting on a smooth horizontal plane starts moving due to a constant force $F$. In the process of its rectilinear motion the angle $	heta$ between the direction of this force and the horizontal varies as $	heta=k x$, where $k$ is a constant and $x$ is the distance traversed by the bar from its initial position. The velocity $(v)$ of the bar as a function of the angle $	heta$ is

Aicharya · Accepted Answer

Component of force responsible for horizontal motion is $F \cos \theta$.

So, acceleration of bar, $a=\frac{F \cos \theta}{m}$ As acceleration, $a=\frac{d v}{d t}$, we have $\Rightarrow \quad \frac{d v}{d t}=\frac{F \cos \theta}{m}$ we can also express this as, $\begin{array}{ll}\Rightarrow & \frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{F \cos \theta}{m} \\ \Rightarrow & v d v=\frac{F \cos \theta}{m} d x\end{array}$ Integrating both sides and putting $\theta=k x$, we get $\int v d v=\int \frac{F \cos k x}{m} d x$ $\begin{array}{ll}\Rightarrow & \frac{v^2}{2}=\frac{F}{m} \int \cos k x \cdot d x \\ \Rightarrow & \frac{v^2}{2}=\frac{F}{m} \cdot\left(\frac{\sin k x}{k}\right) \\ \Rightarrow & v^2=\frac{2 F \sin k x}{m k} \\ \Rightarrow & v=\sqrt{\frac{2 F \sin \theta}{m k}}\end{array}$

A bar of mass $m$ resting on a smooth horizontal plane starts moving due to a constant force $F$. In the…

Solution