Chapter 9, Problem 9.24P

(a)

To determine

Plot the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,1\right)$.

Answer to Problem 9.24P

The orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,1\right)$ are plotted.

Explanation of Solution

Write the expression for the general equation for the position of the puck,

    $\begin{array}{c}x\left(t\right)=\left(x_0+v_{x0}t\right)\cos \Omega t+\left(v_{y0}+\Omega x_0\right)t\sin \Omega t\\ y\left(t\right)=-\left(x_0+v_{x0}t\right)\sin \Omega t+\left(v_{y0}+\Omega x_0\right)t\cos \Omega t\end{array}$        (I)

Here, $x\left(t\right),y\left(t\right)$ are the position of the puck in the x and y direction, $v_{x0},v_{y0}$ are the initial velocity in the x and y-direction, $x_0$ is the initial position in x direction, $t$ is the time and $\Omega$ is the angular velocity.

In this case, the initial velocity of the puck.

    $v_0=\left(0,1\right)$        (II)

So,

    $v_{x0}=0$        (III)

    $v_{y0}=1$        (IV)

Conclusion:

Substitute $1$ for $x_0$, $1$ for $\Omega$, $0$ for $v_{x0}$ and $1$ for $v_{y0}$ in (I),

    $\begin{array}{c}x\left(t\right)=\left(1+0\right)\cos \left(1\right)t+\left(1+\left(1\right)\left(1\right)\right)t\sin \left(1\right)t\\ =\cos t+2t\sin t\end{array}$

    $\begin{array}{c}y\left(t\right)=-\left(1+0\right)\sin \left(1\right)t+\left(1+\left(1\right)\left(1\right)\right)t\cos \left(1\right)t\\ =-\sin t+2t\cos t\end{array}$

Using plotting program of Mathematics parametric plot to plot of the orbit of the above parametricvalue.

The figure 1 shows the orbits of the puck in the turntable.

Therefore, the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,1\right)$ are plotted.

(b)

To determine

Plot the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,0\right)$.

Answer to Problem 9.24P

The orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,0\right)$ are plotted.

Explanation of Solution

In this case, the initial velocity of the puck.

    $v_0=\left(0,0\right)$        (V)

So,

    $v_{x0}=0$

    $v_{y0}=0$

Conclusion:

Substitute $1$ for $x_0$, $1$ for $\Omega$, $0$ for $v_{x0}$ and $0$ for $v_{y0}$ in (I),

    $\begin{array}{c}x\left(t\right)=\left(1+0\right)\cos \left(1\right)t+\left(0+\left(1\right)\left(1\right)\right)t\sin \left(1\right)t\\ =\cos t+t\sin t\end{array}$

    $\begin{array}{c}y\left(t\right)=-\left(1+0\right)\sin \left(1\right)t+\left(0+\left(1\right)\left(1\right)\right)t\cos \left(1\right)t\\ =-\sin t+t\cos t\end{array}$

Using plotting program of Mathematics parametric plot to plot of the orbit of the above parametricvalue.

The figure 2 shows the orbits of the puck in the turntable.

Therefore, the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,0\right)$ are plotted.

(c)

To determine

Plot the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,-1\right)$.

Answer to Problem 9.24P

The orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,-1\right)$ are plotted.

Explanation of Solution

In this case, the initial velocity of the puck.

    $v_0=\left(0,0\right)$        (VI)

So,

    $v_{x0}=0$

    $v_{y0}=-1$

Conclusion:

Substitute $1$ for $x_0$, $1$ for $\Omega$, $0$ for $v_{x0}$ and $-1$ for $v_{y0}$ in (I),

    $\begin{array}{c}x\left(t\right)=\left(1+0\right)\cos \left(1\right)t+\left(\left(-1\right)+\left(1\right)\left(1\right)\right)t\sin \left(1\right)t\\ =\cos t\end{array}$

    $\begin{array}{c}y\left(t\right)=-\left(1+0\right)\sin \left(1\right)t+\left(\left(-1\right)+\left(1\right)\left(1\right)\right)t\cos \left(1\right)t\\ =-\sin t\end{array}$

Using plotting program of Mathematics parametric plot to plot of the orbit of the above parametricvalue.

The figure 3 shows the orbits of the puck in the turntable.

Therefore, the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,-1\right)$ are plotted.

(d)

To determine

Plot the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(-0.5,-0.5\right)$.

Answer to Problem 9.24P

The orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(-0.5,-0.5\right)$ are plotted.

Explanation of Solution

In this case, the initial velocity of the puck.

    $v_0=\left(0,0\right)$        (VII)

So,

    $v_{x0}=0$

    $v_{y0}=0$

Conclusion:

Substitute $1$ for $x_0$, $1$ for $\Omega$, $-0.5$ for $v_{x0}$ and $-0.5$ for $v_{y0}$ in (I),

    $\begin{array}{c}x\left(t\right)=\left(1+\left(-0.5\right)t\right)\cos \left(1\right)t+\left(\left(-0.5\right)+\left(1\right)\left(1\right)\right)t\sin \left(1\right)t\\ =\left(1-0.5t\right)\cos t+0.5t\sin t\end{array}$

    $\begin{array}{c}y\left(t\right)=-\left(1+\left(-0.5\right)t\right)\sin \left(1\right)t+\left(-0.5+\left(1\right)\left(1\right)\right)t\cos \left(1\right)t\\ =-\left(1-0.5t\right)\sin t+0.5t\cos t\end{array}$

Using plotting program of Mathematics parametric plot to plot of the orbit of the above parametricvalue.

The figure 4 shows the orbits of the puck in the turntable.

Therefore, the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(-0.5,-0.5\right)$ are plotted.

(e)

To determine

Plot the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(-0.7,-0.7\right)$.

Answer to Problem 9.24P

The orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(-0.7,-0.7\right)$ are plotted.

Explanation of Solution

In this case, the initial velocity of the puck.

    $v_0=\left(0,0\right)$        (VIII)

So,

    $v_{x0}=0$

    $v_{y0}=0$

Conclusion:

Substitute $1$ for $x_0$ , $1$ for $\Omega$ , $-0.7$ for $v_{x0}$ and $-0.7$ for $v_{y0}$ in (I),

    $\begin{array}{c}x\left(t\right)=\left(1+\left(-0.7\right)t\right)\cos \left(1\right)t+\left(-0.7+\left(1\right)\left(1\right)\right)t\sin \left(1\right)t\\ =\left(1-0.7t\right)\cos t+0.3t\sin t\end{array}$

    $\begin{array}{c}y\left(t\right)=-\left(1+\left(-0.7\right)t\right)\sin \left(1\right)t+\left(-0.7+\left(1\right)\left(1\right)\right)t\cos \left(1\right)t\\ =-\left(1-0.7t\right)\sin t+0.3t\cos t\end{array}$

Using plotting program of Mathematics parametric plot to plot of the orbit of the above parametricvalue.

The figure 5 shows the orbits of the puck in the turntable.

Therefore, the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(-0.7,-0.7\right)$ are plotted.

(f)

To determine

Plot the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,-0.1\right)$.

Answer to Problem 9.24P

The orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,-0.1\right)$ are plotted.

Explanation of Solution

In this case, the initial velocity of the puck.

    $v_0=\left(0,0\right)$        (IX)

So,

    $v_{x0}=0$

    $v_{y0}=0$

Conclusion:

Substitute $1$ for $x_0$ , $1$ for $\Omega$ , $0$ for $v_{x0}$ and $-0.1$ for $v_{y0}$ in (I),

    $\begin{array}{c}x\left(t\right)=\left(1+0\right)\cos \left(1\right)t+\left(-0.1+\left(1\right)\left(1\right)\right)t\sin \left(1\right)t\\ =\cos t+0.9t\sin t\end{array}$

    $\begin{array}{c}y\left(t\right)=-\left(1+0\right)\sin \left(1\right)t+\left(-0.1+\left(1\right)\left(1\right)\right)t\cos \left(1\right)t\\ =-\sin t+0.9t\cos t\end{array}$

Using the plotting program of Mathematics parametric plot to plot of the orbit of the above parametricvalue.

The figure 2 shows the orbits of the puck in the turntable.

Therefore, the orbits of the puck on a rotating turntable with $x_0=\Omega =1$ and the initial velocity $v_0=\left(0,-0.1\right)$ are plotted.