Question

A particle of mass $m$ is moving with constant speed in a vertical circle in $x-z$ plane. There is a small bulb at some distance on $z$-axis. The maximum distance of the shadow of the particle on $x$-axis from origin equal to

• A. $\frac{175}{24}$ m
• B. $\frac{125}{24}$ m
• C. 25 m
• D. 24 m

Answer 1

Answer: (A)

175/24 m

Answer 2

The problem asks for the maximum distance of the shadow of a particle on the x-axis from the origin. The particle moves in a vertical circle in the x-z plane.

1. Identify the coordinates of the relevant points:

   • The circle is centered at the origin (0,0) and has a radius $R = 7$ m.
   • The position of the particle P on the circle can be parameterized as $(x_p, z_p) = (R\cos\theta, R\sin\theta) = (7\cos\theta, 7\sin\theta)$, where $\theta$ is the angle measured from the positive x-axis.
   • The light source (bulb) is on the z-axis. From the diagram, it is 18 m above the top of the circle. The top of the circle is at $z = R = 7$ m. Therefore, the z-coordinate of the light source L is $Z_L = 7 + 18 = 25$ m. So, $L = (0, 25)$.
   • The shadow S is cast on the x-axis, so its z-coordinate is 0. Let its position be $S = (x_s, 0)$.

2. Establish collinearity:

   The light ray passes through the light source L, the particle P, and the shadow S. This means the points L, P, and S are collinear. We can use the property that the slope of the line LP is equal to the slope of the line LS.

   Slope of LP = $\frac{z_p - Z_L}{x_p - 0} = \frac{z_p - Z_L}{x_p}$

   Slope of LS = $\frac{0 - Z_L}{x_s - 0} = \frac{-Z_L}{x_s}$

   Equating the slopes: $\frac{z_p - Z_L}{x_p} = \frac{-Z_L}{x_s}$ Rearranging to solve for $x_s$: $x_s (z_p - Z_L) = -Z_L x_p$ $x_s = \frac{-Z_L x_p}{z_p - Z_L} = \frac{Z_L x_p}{Z_L - z_p}$

3. Substitute the coordinates and parameters:

   Substitute $Z_L = 25$, $x_p = 7\cos\theta$, and $z_p = 7\sin\theta$ into the equation for $x_s$: $x_s = \frac{25 (7\cos\theta)}{25 - 7\sin\theta} = \frac{175\cos\theta}{25 - 7\sin\theta}$

4. Find the maximum value of $|x_s|$ using calculus:

   To find the maximum distance, we need to find the maximum value of $|x_s|$. This involves finding the critical points by differentiating $x_s$ with respect to $\theta$ and setting the derivative to zero.

   Let $f(\theta) = x_s = \frac{175\cos\theta}{25 - 7\sin\theta}$.

   Using the quotient rule $\frac{d}{d\theta}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:

   $u = 175\cos\theta \implies u' = -175\sin\theta$

   $v = 25 - 7\sin\theta \implies v' = -7\cos\theta$

   $\frac{dx_s}{d\theta} = \frac{(-175\sin\theta)(25 - 7\sin\theta) - (175\cos\theta)(-7\cos\theta)}{(25 - 7\sin\theta)^2}$ $\frac{dx_s}{d\theta} = \frac{-175 \times 25\sin\theta + 175 \times 7\sin^2\theta + 175 \times 7\cos^2\theta}{(25 - 7\sin\theta)^2}$ $\frac{dx_s}{d\theta} = \frac{175(-25\sin\theta + 7(\sin^2\theta + \cos^2\theta))}{(25 - 7\sin\theta)^2}$ Since $\sin^2\theta + \cos^2\theta = 1$: $\frac{dx_s}{d\theta} = \frac{175(7 - 25\sin\theta)}{(25 - 7\sin\theta)^2}$ Set $\frac{dx_s}{d\theta} = 0$ to find critical points: $175(7 - 25\sin\theta) = 0$ $7 - 25\sin\theta = 0$ $\sin\theta = \frac{7}{25}$

5. Calculate $\cos\theta$ and the corresponding $x_s$ values:

   Using the identity $\sin^2\theta + \cos^2\theta = 1$: $\cos^2\theta = 1 - \left(\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625}$ $\cos\theta = \pm\sqrt{\frac{576}{625}} = \pm\frac{24}{25}$

   • Case 1: $\sin\theta = \frac{7}{25}$ and $\cos\theta = \frac{24}{25}$ (Particle in the first quadrant) $x_s = \frac{175 \left(\frac{24}{25}\right)}{25 - 7 \left(\frac{7}{25}\right)} = \frac{7 \times 25 \times \frac{24}{25}}{25 - \frac{49}{25}} = \frac{7 \times 24}{\frac{625 - 49}{25}} = \frac{168}{\frac{576}{25}}$ $x_s = \frac{168 \times 25}{576} = \frac{(7 \times 24) \times 25}{24 \times 24} = \frac{7 \times 25}{24} = \frac{175}{24} \text{ m}$

   • Case 2: $\sin\theta = \frac{7}{25}$ and $\cos\theta = -\frac{24}{25}$ (Particle in the second quadrant) $x_s = \frac{175 \left(-\frac{24}{25}\right)}{25 - 7 \left(\frac{7}{25}\right)} = \frac{-175 \times \frac{24}{25}}{25 - \frac{49}{25}} = -\frac{175}{24} \text{ m}$

6. Determine the maximum distance:

   The distance of the shadow from the origin is $|x_s|$.

   From Case 1, $|x_s| = \left|\frac{175}{24}\right| = \frac{175}{24}$ m.

   From Case 2, $|x_s| = \left|-\frac{175}{24}\right| = \frac{175}{24}$ m.

   Both critical points yield the same maximum distance from the origin.

The maximum distance of the shadow of the particle on the x-axis from the origin is $\frac{175}{24}$ m.