Chapter 11, Problem 11.192RP

To determine

(a)

The velocity of point B at $t=2s$

Answer to Problem 11.192RP

$v=3.0655m/s$ At $59°$

Explanation of Solution

Given information:

Distance between A and B is equal to $5m$

$\ddot{r}=-1.0m/s^2$

$\ddot{\theta }=-0.5rad/s^2$

Velocity of the skier is denoted as,

$v=v_r{e}_r+v_{\theta }{e}_{\theta }=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }$

For a uniformly accelerated motion,

$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

In the above equation,

$x$ - End distance

$x_0$ - Start distance

$v_0$ - Initial velocity

$t$ - Elapsed time

$a$ -Acceleration

Calculation:

For radial motion,

$r=r_0+{\dot{r}}_{0}t+\frac{1}{2}\ddot{r}{t}^2$

We know that,

$\begin{array}{l}{r}_0=5m\\ {\dot{r}}_0=0\\ \ddot{r}=-1.0m/s^2\end{array}$

Therefore,

At $t=2s$

$\begin{array}{l}r=5+0+\frac{1}{2}\left(-1.0m/s^2\right){\left(2s\right)}^2\\ r=3m\end{array}$

$\begin{array}{l}\dot{r}={\dot{r}}_0+\ddot{r}t\\ \dot{r}=0+\left(-1.0m/s^2\right)\left(2s\right)\\ \dot{r}=-2m/s\end{array}$

For angular motion,

At $t=2s$

$\begin{array}{l}\theta ={\theta }_0+{\dot{\theta }}_{0}t+\frac{1}{2}\ddot{\theta }{t}^2\\ \theta =\frac{\pi }{3}+0+\frac{1}{2}\left(-0.5rad/s^2\right){\left(2s\right)}^2\\ \theta =0.0472rad=2.70°\end{array}$

$\begin{array}{l}\dot{\theta }={\dot{\theta }}_0+\ddot{\theta }t\\ \dot{\theta }=0+\left(-0.5rad/s^2\right)\left(2s\right)\\ \dot{\theta }=-1.0rad/s\end{array}$

Now, find the velocity at point B at $t=2s$

$\begin{array}{l}v=v_r{e}_r+v_{\theta }{e}_{\theta }=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }\\ v=\left(-2m/s\right)e_r+\left(3m\right)\left(-1.0rad/s\right)e_{\theta }\\ v=\left(-2m/s\right)e_r+\left(-3m/s\right)e_{\theta }\end{array}$

Therefore,

$\begin{array}{l}\tan \alpha =\frac{v_{\theta }}{v_r}=\frac{-3}{-2}=1.5\\ \alpha =56.31°\end{array}$

To find the magnitude of the velocity,

$v=\sqrt{v_{\theta }{}^2+v_r{}^2}=\sqrt{{\left(-2\right)}^2+{\left(-3\right)}^2}=3.0655m/s$

Direction of the velocity is equal to,

$\theta +\alpha =2.70°+56.31°=59°$

Conclusion:

Therefore velocity at point B is equal to,

$v=3.0655m/s$ At $59°$

To determine

(b)

The acceleration of point B

Answer to Problem 11.192RP

$4.717m/s^2$ At $29.3°$

Explanation of Solution

Given information:

Distance between A and B is equal to $5m$

$\ddot{r}=-1.0m/s^2$

$\ddot{\theta }=-0.5rad/s^2$

Acceleration of the skier is denoted as,

$a=a_r{e}_r+a_{\theta }{e}_{\theta }=\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }$

Calculation:

To find the acceleration of point B,

$a=a_r{e}_r+a_{\theta }{e}_{\theta }=\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }$

Then,

$\begin{array}{l}a=a_r{e}_r+a_{\theta }{e}_{\theta }=\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }\\ a=\left[\left(-1.0m/s^2\right)-\left(3m\right){\left(-1.0rad/s\right)}^2\right]e_r+\left[\left(3m\right)\left(-0.5rad/s^2\right)+2\left(-2m/s\right)\left(-1.0rad/s\right)\right]e_{\theta }\\ a=\left(-4m/s\right)e_r+\left(2.5m/s\right)e_{\theta }\end{array}$

Therefore,

$\begin{array}{l}\tan \beta =\frac{a_{\theta }}{a_r}=\frac{2.5}{-4}=-0.625\\ \beta =-32°\end{array}$

The magnitude is equal to,

$a=\sqrt{{\left(-4\right)}^2+{2.5}^2}=4.717m/s^2$

Then,

$\theta +\beta =2.70°-32°=29.3°$

Conclusion:

The acceleration of point B is equal to $4.717m/s^2$ at $29.3°$

To determine

(c)

The radius of curvature of the path

Answer to Problem 11.192RP

Radius of curvature of the path is $2.757m$

Explanation of Solution

Given information:

Distance between A and B is equal to $5m$

$\ddot{r}=-1.0m/s^2$

$\ddot{\theta }=-0.5rad/s^2$

Radius of curvature is defined as,

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

The tangential component is denoted as,

$a_t=\left(a.e_t\right)e_t$

The normal component is denoted as,

$a_n=a-a_t$

$e_t$ is defined as,

$e_t=\frac{V}{v}$

Calculation:

Find $e_t$

$e_t=\frac{V}{v}=\frac{-2e_r-3e_{\theta }}{3.0655m/s}=-0.5547e_r-0.8320e_{\theta }$

Find the tangential component $a_t$

$\begin{array}{l}{a}_t=\left(a.e_t\right)e_t\\ a_t=\left(-4e_r+2.5e_{\theta }\right).\left(-0.5547e_r-0.8320e_{\theta }\right)e_t\\ a_t=0.1387e_t\end{array}$

Find the normal component $a_n$

$\begin{array}{l}{a}_n=\left(-4e_r+2.5e_{\theta }\right).\left(0.1387\right)\left(-0.5547e_r-0.8320e_{\theta }\right)\\ a_n=-3.923e_r+2.615e_{\theta }\end{array}$

Therefore the magnitude will be,

$a_n=\sqrt{-{3.923}^2+{2.615}^2}=4.7149m/s^2$

To find the radius of curvature,

$\rho =\frac{v^2}{a_n}=\frac{{\left(3.0655m/s\right)}^2}{4.7149m/s^2}=2.757m$

Conclusion:

The radius of curvature is equal to $2.757m$