Question

A particle of mass $m$ is initially situated at the point $P$ inside a hemispherical surface of radius $r$ as shown in figure. A horizontal acceleration of magnitude $a_0$ is suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is

• A. $\sqrt{\frac{4r \sin \alpha}{a_0}}$
• B. $\sqrt{\frac{4r \tan \alpha}{a_0}}$
• C. $\sqrt{\frac{4r \cos \alpha}{a_0}}$
• D. None of these

Answer 1

Answer: (A)

$\sqrt{\frac{4r \sin \alpha}{a_0}}$

Answer 2

Let's assume the center of the hemispherical surface is at the origin $(0,0)$. The equation of the hemispherical surface is $x^2 + y^2 = r^2$ for $y \ge 0$.

From the figure, the particle is initially at point $P$. Let's assume the angle $\alpha$ is measured from the vertical axis to the line segment from the origin to $P$. Since $P$ is on the left side, the coordinates of $P$ are $(-r \sin \alpha, r \cos \alpha)$. The particle is initially at rest, so the initial velocity is $\vec{v}_0 = 0$. A horizontal acceleration of magnitude $a_0$ is applied in the horizontal direction, which is the positive x-direction as shown in the figure. So the acceleration is $\vec{a} = a_0 \hat{i}$.

The initial position vector is $\vec{r}_0 = -r \sin \alpha \hat{i} + r \cos \alpha \hat{j}$. The position vector of the particle at time $t$ is given by $\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2$ $\vec{r}(t) = (-r \sin \alpha \hat{i} + r \cos \alpha \hat{j}) + 0 \cdot t + \frac{1}{2} a_0 t^2 \hat{i}$ $\vec{r}(t) = (-r \sin \alpha + \frac{1}{2} a_0 t^2) \hat{i} + r \cos \alpha \hat{j}$

Let the position of the particle at time $t$ be $(x(t), y(t))$. Then $x(t) = -r \sin \alpha + \frac{1}{2} a_0 t^2$ and $y(t) = r \cos \alpha$. The particle touches the sphere again when its position $(x(t), y(t))$ satisfies the equation of the hemisphere $x^2 + y^2 = r^2$ and $y \ge 0$. Since the acceleration is only in the x-direction, the y-coordinate of the particle remains constant, $y(t) = r \cos \alpha$. For the particle to be on the hemisphere, we must have $y(t) \ge 0$, which means $r \cos \alpha \ge 0$. Since $r > 0$, we have $\cos \alpha \ge 0$. From the figure, $\alpha$ is an acute angle, so $\cos \alpha > 0$.

Substitute $x(t)$ and $y(t)$ into the equation of the sphere: $x(t)^2 + y(t)^2 = r^2$ $(-r \sin \alpha + \frac{1}{2} a_0 t^2)^2 + (r \cos \alpha)^2 = r^2$ $(-r \sin \alpha + \frac{1}{2} a_0 t^2)^2 + r^2 \cos^2 \alpha = r^2$ $(-r \sin \alpha + \frac{1}{2} a_0 t^2)^2 = r^2 - r^2 \cos^2 \alpha$ $(-r \sin \alpha + \frac{1}{2} a_0 t^2)^2 = r^2 (1 - \cos^2 \alpha)$ $(-r \sin \alpha + \frac{1}{2} a_0 t^2)^2 = r^2 \sin^2 \alpha$ Taking the square root of both sides: $-r \sin \alpha + \frac{1}{2} a_0 t^2 = \pm \sqrt{r^2 \sin^2 \alpha}$ $-r \sin \alpha + \frac{1}{2} a_0 t^2 = \pm r \sin \alpha$

Case 1: $-r \sin \alpha + \frac{1}{2} a_0 t^2 = -r \sin \alpha$. $\frac{1}{2} a_0 t^2 = 0$ $t^2 = 0 \implies t = 0$. This corresponds to the initial position.

Case 2: $-r \sin \alpha + \frac{1}{2} a_0 t^2 = r \sin \alpha$. $\frac{1}{2} a_0 t^2 = r \sin \alpha + r \sin \alpha$ $\frac{1}{2} a_0 t^2 = 2r \sin \alpha$ $t^2 = \frac{4r \sin \alpha}{a_0}$ $t = \sqrt{\frac{4r \sin \alpha}{a_0}}$ (since time $t$ must be non-negative).

This is the time taken for the particle to touch the sphere again.