Question

A small particle P starts from point O with a negligible speed and increases its speed to a value $v = \sqrt{2gy}$, where y is the vertical drop from O. When x = 50 ft, determine the n-component of acceleration of the particle.

Answer 1

1.8376 ft/s^2

Answer 2

The n-component (normal component) of acceleration, $a_n$, for a particle moving along a curved path is given by:

$a_n = \frac{v^2}{\rho}$

where $v$ is the speed of the particle and $\rho$ is the radius of curvature of the path.

1. Determine the vertical drop (y) at x = 50 ft: The path of the particle is given by the equation $y = \left(\frac{x}{20}\right)^2$ ft. Substitute $x = 50$ ft into the equation: $y = \left(\frac{50}{20}\right)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25 \text{ ft}$

2. Determine the speed (v) at x = 50 ft: The speed of the particle is given by $v = \sqrt{2gy}$. We will use the standard acceleration due to gravity $g = 32.2 \text{ ft/s}^2$. First, calculate $v^2$: $v^2 = 2gy = 2 \times 32.2 \times 6.25 = 64.4 \times 6.25 = 402.5 \text{ (ft/s)}^2$

3. Determine the radius of curvature ($\rho$) at x = 50 ft: The formula for the radius of curvature for a curve $y = f(x)$ is: $\rho = \frac{[1 + (y')^2]^{3/2}}{|y''|}$ First, find the first and second derivatives of $y$ with respect to $x$: Given $y = \frac{x^2}{400}$. The first derivative is: $y' = \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^2}{400}\right) = \frac{2x}{400} = \frac{x}{200}$ The second derivative is: $y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{x}{200}\right) = \frac{1}{200}$ Now, evaluate $y'$ at $x = 50$ ft: $y'(50) = \frac{50}{200} = \frac{1}{4}$ Substitute $y'(50)$ and $y''$ into the formula for $\rho$: $\rho = \frac{\left[1 + \left(\frac{1}{4}\right)^2\right]^{3/2}}{\left|\frac{1}{200}\right|} = \frac{\left[1 + \frac{1}{16}\right]^{3/2}}{\frac{1}{200}}$ $\rho = \frac{\left[\frac{17}{16}\right]^{3/2}}{\frac{1}{200}} = 200 \times \left(\frac{17}{16}\right)^{3/2}$ $\rho = 200 \times \frac{17\sqrt{17}}{16^{3/2}} = 200 \times \frac{17\sqrt{17}}{64} = \frac{25 \times 17\sqrt{17}}{8} = \frac{425\sqrt{17}}{8} \text{ ft}$

4. Calculate the n-component of acceleration ($a_n$): Using the formula $a_n = \frac{v^2}{\rho}$: $a_n = \frac{402.5}{\frac{425\sqrt{17}}{8}} = \frac{402.5 \times 8}{425\sqrt{17}}$ $a_n = \frac{3220}{425\sqrt{17}}$ To simplify the fraction, divide the numerator and denominator by 5: $a_n = \frac{644}{85\sqrt{17}}$ Now, calculate the numerical value using $\sqrt{17} \approx 4.1231$: $a_n \approx \frac{644}{85 \times 4.1231} = \frac{644}{350.4635} \approx 1.8376 \text{ ft/s}^2$

The final answer is $1.8376 \text{ ft/s}^2$.

Explanation of the solution:

1. Calculate the vertical position ($y$) of the particle at the given horizontal position ($x = 50$ ft) using the path equation $y = (x/20)^2$.
2. Calculate the square of the particle's speed ($v^2$) at this vertical position using the given speed relation $v = \sqrt{2gy}$.
3. Calculate the radius of curvature ($\rho$) of the path at $x = 50$ ft. This involves finding the first and second derivatives of $y$ with respect to $x$ and applying the formula $\rho = \frac{[1 + (y')^2]^{3/2}}{|y''|}$.
4. Finally, determine the n-component of acceleration ($a_n$) using the kinematic formula $a_n = v^2/\rho$.

Answer:

The n-component of acceleration of the particle is approximately $1.8376 \text{ ft/s}^2$.