Question

A simple bob of mass m and charge q is suspended in a uniform horizontal elec-tric field E. The bob is held in its least Potential point from which it is projected horizontally with a speed such that it continue moving in uniform circular motion. Find the projection speed of the bob

• A. sqrt((L/m)*sqrt((mg)^2 + (qE)^2))

Answer 1

Answer: (A)

sqrt((L/m)*sqrt((mg)^2 + (qE)^2))

Answer 2

The bob of mass $m$ and charge $q$ is suspended in a uniform horizontal electric field $E$. The forces acting on the bob are gravity $m\vec{g}$ downwards and electric force $q\vec{E}$ horizontally. The effective gravitational acceleration is $\vec{g}_{eff} = \vec{g} + \frac{q\vec{E}}{m}$. The magnitude of the effective acceleration is $g_{eff} = \sqrt{g^2 + (qE/m)^2}$. The direction of the effective force $m\vec{g}_{eff} = m\vec{g} + q\vec{E}$ makes an angle $\theta_0$ with the downward vertical, where $\tan \theta_0 = \frac{|qE|}{mg}$. The magnitude of the effective force is $F_{eff} = m g_{eff} = \sqrt{(mg)^2 + (qE)^2}$. The least potential energy point is the equilibrium position, where the string aligns with the direction of the effective force, making an angle $\theta_0$ with the downward vertical.

The problem states that the bob is projected horizontally from this least potential point (equilibrium position) with a speed $v_0$ such that it moves in uniform circular motion centered at the pivot. However, uniform circular motion centered at the pivot in the presence of constant gravity and a constant horizontal electric field is generally not possible unless both gravity and the electric field are zero. The tangential force $F_{tan} = mg\sin\theta + qE\cos\theta$ is not zero for all angles $\theta$, which is required for uniform circular motion.

Given the likely flawed nature of the question, we interpret it as asking for the speed $v_0$ such that the required centripetal force at the point of projection (the equilibrium position) is equal to the magnitude of the effective force $F_{eff} = \sqrt{(mg)^2 + (qE)^2}$. This is a plausible interpretation that leads to a unique answer involving the given parameters, although the premise of uniform circular motion is inconsistent with the forces.

At the equilibrium position, the string makes an angle $\theta_0$ with the downward vertical, where $\tan \theta_0 = qE/mg$. The forces are gravity $m\vec{g}$, electric force $q\vec{E}$, and tension $\vec{T}$. The net force $m\vec{g} + q\vec{E}$ has magnitude $F_{eff} = \sqrt{(mg)^2 + (qE)^2}$ and is directed along the string away from the pivot at equilibrium.

For circular motion centered at the pivot with speed $v_0$ at this point, the net force towards the center (along the string towards the pivot) must provide the centripetal force $mv_0^2/L$. The forces in the radial direction are tension $T$ (towards the pivot) and the component of $m\vec{g} + q\vec{E}$ along the string (away from the pivot), which has magnitude $F_{eff}$. The radial equation of motion at the equilibrium position is $T - F_{eff} = \frac{mv_0^2}{L}$.

If we assume the intended meaning is that the required centripetal force at the equilibrium point is equal to the magnitude of the effective force, then $\frac{mv_0^2}{L} = F_{eff}$. $\frac{mv_0^2}{L} = \sqrt{(mg)^2 + (qE)^2}$. $v_0^2 = \frac{L}{m} \sqrt{(mg)^2 + (qE)^2}$. $v_0 = \sqrt{\frac{L}{m} \sqrt{(mg)^2 + (qE)^2}}$.

This speed, when projected horizontally from the equilibrium position, would result in a centripetal force requirement equal to the magnitude of the effective force at that instant. However, this does not guarantee uniform circular motion throughout the path. Assuming this is the intended calculation despite the flawed premise.

The final answer is $\boxed{\sqrt{\frac{L}{m}\sqrt{(mg)^2 + (qE)^2}}}$.