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 \(\theta\) between the direction of this force and the horizontal varies as \(\theta = kx\), 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 \(\theta\) is

• A. \(v=\sqrt{\dfrac{2F\sin\theta}{mk}}\)
• B. \(v=\sqrt{\dfrac{2F}{mk\sin\theta}}\)
• C. \(v=\dfrac{2F\sin\theta}{mk}\)
• D. \(v=\dfrac{2F}{mk\sin\theta}\)

Hint:

When force acts at a variable angle with displacement, use the work-energy theorem: \[ dW=F\cos\theta\,dx. \] Then use the given relation between angle and displacement to integrate properly.

Answer 1

The Correct Option is A

Step 1: Understand the motion of the bar.
The bar is moving on a smooth horizontal plane.
Since the surface is smooth, there is no friction.
The displacement of the bar is along the horizontal direction.
The force \(F\) makes an angle \(\theta\) with the horizontal direction.
Therefore, only the horizontal component of force does work on the bar.
The horizontal component of force is \[ F\cos\theta \]

Step 2: Use the work-energy theorem.
According to the work-energy theorem, \[ dW=dK \] For a small displacement \(dx\), the work done is \[ dW=F\cos\theta \, dx \] The change in kinetic energy is \[ dK=d\left(\frac{1}{2}mv^2\right) \] Hence, \[ d\left(\frac{1}{2}mv^2\right)=F\cos\theta \, dx \]

Step 3: Use the given relation between \(\theta\) and \(x\).
Given that \[ \theta=kx \] Therefore, \[ x=\frac{\theta}{k} \] Differentiating both sides, \[ dx=\frac{d\theta}{k} \]

Step 4: Substitute \(dx\) in the work-energy equation.
Now, \[ d\left(\frac{1}{2}mv^2\right)=F\cos\theta \cdot \frac{d\theta}{k} \] So, \[ d\left(\frac{1}{2}mv^2\right)=\frac{F}{k}\cos\theta \, d\theta \] Integrating from initial position to any position,
At the initial position, \[ \theta=0 \] and since the bar starts from rest, \[ v=0 \] Thus, \[ \int_0^v d\left(\frac{1}{2}mv^2\right) = \frac{F}{k}\int_0^\theta \cos\theta \, d\theta \] This gives \[ \frac{1}{2}mv^2 = \frac{F}{k}\sin\theta \]

Step 5: Solve for velocity.
Multiplying both sides by \(2\), \[ mv^2=\frac{2F\sin\theta}{k} \] So, \[ v^2=\frac{2F\sin\theta}{mk} \] Taking square root, \[ v=\sqrt{\frac{2F\sin\theta}{mk}} \]

Step 6: Final conclusion.
Therefore, the velocity of the bar as a function of \(\theta\) is \[ \boxed{v=\sqrt{\frac{2F\sin\theta}{mk}}} \]