Question

A simple pendulum with a bob of mass $m = 2$ kg, charge $q = 5\mu c$ and string of length $l = 1$ m is given a horizontal velocity $u$ in uniform electric field $E = 2 \times 10^6$ V/m at its bottom most point A, as shown in figure. If particle leaves the circle at point C. Find $\frac{u^2}{5}$. $(g = 10$ m/s$^2)$

Answer 1

5.2

Answer 2

The problem describes a simple pendulum with a charged bob in a uniform electric field. The bob is given a horizontal velocity at the bottommost point and leaves the circular path at point C. We are given the mass of the bob $m=2$ kg, charge $q=5\mu c = 5 \times 10^{-6}$ C, string length $l=1$ m, electric field $E=2 \times 10^6$ V/m, and gravitational acceleration $g=10$ m/s$^2$. The angle of point C from the vertical downwards direction is given as $53^\circ$. Let $\theta = 53^\circ$.

At point C, the bob leaves the circular path, which means the tension in the string is zero. The forces acting on the bob at point C are gravity $mg$ downwards, electric force $qE$ horizontally to the right, and tension $T_C$ along the string towards the pivot. The angle of the string with the vertical downwards direction is $\theta$. Applying Newton's second law in the radial direction (towards the pivot): $T_C - mg \cos\theta + qE \sin\theta = \frac{m v_C^2}{l}$ Since $T_C = 0$ at point C: $-mg \cos\theta + qE \sin\theta = \frac{m v_C^2}{l}$ $v_C^2 = \frac{l}{m} (qE \sin\theta - mg \cos\theta)$

We are given $\theta = 53^\circ$. We use the values $\sin 53^\circ \approx 0.8$ and $\cos 53^\circ \approx 0.6$. $qE = (5 \times 10^{-6} \text{ C}) \times (2 \times 10^6 \text{ V/m}) = 10$ N $mg = (2 \text{ kg}) \times (10 \text{ m/s}^2) = 20$ N $v_C^2 = \frac{1}{2} (10 \times 0.8 - 20 \times 0.6) = \frac{1}{2} (8 - 12) = -2$. A negative value for $v_C^2$ is not possible. This indicates that the angle $53^\circ$ is not measured from the vertical downwards direction in the usual sense for circular motion problems where the tension becomes zero.

Let's reconsider the angle from the figure. The angle $53^\circ$ is shown above the horizontal line passing through the pivot. This suggests that the angle is measured from the vertical upwards direction. Let's assume the angle from the vertical upwards direction is $53^\circ$. Let the angle from the vertical downwards direction be $\alpha$. Then $\alpha = 180^\circ - 53^\circ = 127^\circ$. At point C, the angle from the vertical downwards direction is $\alpha = 127^\circ$. The forces in the radial direction (towards the pivot) are: Tension: $T_C$ Component of gravity: $-mg \cos\alpha$ Component of electric force: $-qE \sin\alpha$ So, $T_C - mg \cos\alpha - qE \sin\alpha = \frac{m v_C^2}{l}$. At $T_C = 0$: $-mg \cos\alpha - qE \sin\alpha = \frac{m v_C^2}{l}$. $v_C^2 = -\frac{l}{m} (mg \cos\alpha + qE \sin\alpha)$. With $\alpha = 127^\circ$, $\cos 127^\circ = -\cos 53^\circ \approx -0.6$ and $\sin 127^\circ = \sin 53^\circ \approx 0.8$. $v_C^2 = -\frac{1}{2} (20 \times (-0.6) + 10 \times 0.8) = -\frac{1}{2} (-12 + 8) = -\frac{1}{2} (-4) = 2$. This gives a valid value for $v_C^2$. So, $v_C^2 = 2$ m$^2$/s$^2$.

Now, we use conservation of energy between point A and point C. Let the potential energy at point A be zero. The height of point C above A is $h = l(1 + \cos 53^\circ) = l(1 + 0.6) = 1.6l$ if the angle $53^\circ$ is from the vertical upwards. Let's use the angle $\alpha = 127^\circ$ from the vertical downwards direction. The height of C above A is $h = l(1 - \cos \alpha) = l(1 - \cos 127^\circ) = l(1 - (-0.6)) = 1.6l$. The electric potential energy is $PE_e = -qEx$, where x is the horizontal displacement from a reference point. Let's take the reference point at A, so $x_A = 0$. The x-coordinate of C relative to A is $x_C = l \sin \alpha = l \sin 127^\circ = l \sin 53^\circ \approx 0.8l$. $PE_g(A) = 0$, $PE_e(A) = 0$. $PE_g(C) = mgh = mg l(1 - \cos \alpha) = 20 \times 1 \times (1 - (-0.6)) = 20 \times 1.6 = 32$ J. $PE_e(C) = -qEx_C = -qE l \sin \alpha = -10 \times 1 \times 0.8 = -8$ J. $KE(A) = \frac{1}{2} m u^2 = \frac{1}{2} \times 2 \times u^2 = u^2$. $KE(C) = \frac{1}{2} m v_C^2 = \frac{1}{2} \times 2 \times 2 = 2$ J.

Conservation of energy: $KE(A) + PE_g(A) + PE_e(A) = KE(C) + PE_g(C) + PE_e(C)$ $u^2 + 0 + 0 = 2 + 32 + (-8)$ $u^2 = 2 + 32 - 8 = 26$.

The question asks for $\frac{u^2}{5}$. $\frac{u^2}{5} = \frac{26}{5} = 5.2$.

Let's verify the angle interpretation from the figure. The angle $53^\circ$ is shown between the vertical line (downwards) and the string. Point C is above the horizontal line. If the angle is measured from the vertical downwards, then $\theta = 53^\circ$. Then the height above A is $l(1 - \cos 53^\circ) = 1(1 - 0.6) = 0.4$ m. The horizontal displacement from A is $l \sin 53^\circ = 1 \times 0.8 = 0.8$ m to the right. If $\theta = 53^\circ$ from the vertical downwards, then $v_C^2 = -2$, which is not possible.

Let's assume the angle $53^\circ$ is measured from the vertical upwards direction. Then the angle from the vertical downwards is $180^\circ - 53^\circ = 127^\circ$. Height of C above A is $l(1 + \cos 53^\circ) = 1(1 + 0.6) = 1.6$ m. Horizontal displacement of C from A is $l \sin 53^\circ = 1 \times 0.8 = 0.8$ m to the right. Using $\alpha = 127^\circ$ from vertical downwards, $v_C^2 = 2$. Energy conservation: $KE(A) = u^2$. $PE_g(A) = 0$. $PE_e(A) = 0$. $KE(C) = \frac{1}{2} m v_C^2 = \frac{1}{2} \times 2 \times 2 = 2$. $PE_g(C) = mgh = mg l(1 - \cos 127^\circ) = 20 \times 1 \times (1 - (-0.6)) = 20 \times 1.6 = 32$. $PE_e(C) = -qE x_C$. The x-coordinate of C relative to A is $l \sin 127^\circ = l \sin 53^\circ = 1 \times 0.8 = 0.8$. $PE_e(C) = -(5 \times 10^{-6}) \times (2 \times 10^6) \times 0.8 = -10 \times 0.8 = -8$. $u^2 + 0 + 0 = 2 + 32 - 8 = 26$. $\frac{u^2}{5} = \frac{26}{5} = 5.2$.

Let's verify the angle interpretation from the figure again. The angle $53^\circ$ is shown between the vertical line (downwards) and the string. However, point C is in the second quadrant relative to the pivot, if the vertical is y-axis and horizontal is x-axis. If the angle is from the vertical downwards, and it's $53^\circ$, then the point should be in the first or fourth quadrant. The figure shows point C at an angle of $53^\circ$ from the vertical line passing through the pivot. The vertical line is drawn downwards. So the angle is measured from the vertical downwards direction. But point C is above the horizontal line. This means the angle from the vertical downwards direction is greater than $90^\circ$. It is possible that the angle $53^\circ$ is the angle between the string and the horizontal line at point C. If the angle with the horizontal is $53^\circ$, then the angle with the vertical is $90^\circ - 53^\circ = 37^\circ$. Let's assume the angle with the vertical downwards is $\theta$. From the figure, it looks like the angle between the vertical downwards and the string is $53^\circ$. But this is inconsistent with point C being above the horizontal line. Let's assume the angle $53^\circ$ is the angle from the vertical line passing through the pivot, measured from the horizontal line passing through the pivot. This is also not standard.

Let's assume the angle $53^\circ$ is the angle from the vertical upwards direction, as this gives a consistent result. Angle from vertical upwards = $53^\circ$. Angle from vertical downwards $\alpha = 180^\circ - 53^\circ = 127^\circ$. $v_C^2 = 2$. $u^2 = 26$. $\frac{u^2}{5} = 5.2$.

Final check of calculations: $qE = 10$ N, $mg = 20$ N, $l=1$ m, $m=2$ kg. Angle from vertical downwards $\alpha = 127^\circ$. $\cos 127^\circ = -0.6$, $\sin 127^\circ = 0.8$. $v_C^2 = -\frac{l}{m} (mg \cos\alpha + qE \sin\alpha) = -\frac{1}{2} (20 \times (-0.6) + 10 \times 0.8) = -\frac{1}{2} (-12 + 8) = -\frac{1}{2} (-4) = 2$. Height of C above A = $l(1 - \cos\alpha) = 1(1 - (-0.6)) = 1.6$. Horizontal displacement of C from A = $l \sin\alpha = 1 \times 0.8 = 0.8$. Energy conservation: $\frac{1}{2} m u^2 = \frac{1}{2} m v_C^2 + mg l(1 - \cos\alpha) - qE l \sin\alpha$. $\frac{1}{2} \times 2 \times u^2 = \frac{1}{2} \times 2 \times 2 + 20 \times 1 \times 1.6 - 10 \times 1 \times 0.8$. $u^2 = 2 + 32 - 8 = 26$. $\frac{u^2}{5} = \frac{26}{5} = 5.2$.

Assuming the angle $53^\circ$ is measured from the vertical upwards direction.

The final answer is $\frac{u^2}{5} = 5.2$.