Chapter 12, Problem 12.133RP

Disk A rotates in a horizontal plane about a vertical axis at the constant rate ${\stackrel{.}{\theta }}_0=10$ rad/s. Slider B has mass 1 kg and moves in a frictionless slot cut in the disk. The slider is attached to a spring of constant k, which is undeformed when $r=0$ . Knowing that the slider is released with no radial velocity in the position $r=500$ mm, determine the position of the slider and the horizontal force exerted on it by the disk at $t=0.1$ s for (a) $k=100$ N/m, (b) $k=200$ N/m.

  

To determine

(a)

The position of slider and horizontal force exerted on at $t=0.1s$ for $k=100N/m$.

Answer to Problem 12.133RP

$\begin{array}{l}r=0.5\\ F_A=0\end{array}$

Explanation of Solution

Given information:

Disk A rotates at constant rate ${\dot{\theta }}_0=10rad/s$

Mass of slider B is $1kg$

The spring constant is $k$

Slider is released at $r=500mm$

Newton’s second law of motion is denoted as,

$\to \sum F=ma$

The radial component of acceleration is defined as,

$a_r=\ddot{r}-r{\dot{\theta }}^2$

The spring force is defined as,

$F_{spring}=kx$

Where,

$x$ - Elongation

Calculation:

According to given information, we know that

$\begin{array}{l}r=0\\ r_0=0.5m\\ \dot{\theta }=10rad/s\\ \ddot{\theta }=0\end{array}$

For block B,

$\leftarrow \sum F_r=m_B{a}_r$

Substitute,

$\begin{array}{l}-F_{spring}=m_B{a}_r\\ -kr=m_B\left(\ddot{r}-r{\dot{\theta }}^2\right)\to \left(1\right)\end{array}$

Rearrange,

$\ddot{r}+\left(\frac{k}{m_B}-{\dot{\theta }}^2\right)r=0$

Then,

$\begin{array}{l}\downarrow \sum F_{\theta }=m_B{a}_{\theta }\\ F_A=m_B\left(0+2\dot{r}\dot{\theta }\right)\to \left(2\right)\end{array}$

Therefore, for $k=100N/m$

$\begin{array}{l}\ddot{r}+\left(\frac{k}{m_B}-{\dot{\theta }}^2\right)r=0\\ \ddot{r}+\left(\frac{100N/m}{1kg}-{\left(10rad/s\right)}^2\right)r=0\\ \ddot{r}=0\end{array}$

Therefore,

$\begin{array}{l}\frac{d\dot{r}}{dt}=\ddot{r}=0\\ \underset{0}{\overset{\dot{r}}{\int }}d\dot{r}=\underset{0}{\overset{0.1}{\int }}0dt\\ \dot{r}=0\end{array}$

Then,

$\frac{dr}{dt}=\dot{r}=0$

At $t=0,r_0=0.5m$

$\begin{array}{l}\underset{r_0}{\overset{r}{\int }}dr=\underset{0}{\overset{0.1}{\int }}0dt\\ r=r_0=0.5\end{array}$

Now, according to equation 2,

$F_A=0$

Because of $\dot{r}=0$

Conclusion:

The position of slider and horizontal force exerted on at $t=0.1s$ for $k=100N/m$ is equal to,

$\begin{array}{l}r=0.5\\ F_A=0\end{array}$

To determine

(b)

The position of slider and horizontal force exerted on at $t=0.1s$ for $k=200N/m$

Answer to Problem 12.133RP

$\begin{array}{l}r=0.27m\\ F_A=-84.147N\end{array}$

Explanation of Solution

Given information:

Disk A rotates at constant rate ${\dot{\theta }}_0=10rad/s$

Mass of slider B is $1kg$

The spring constant is $k$

Slider is released at $r=500mm$

Newton’s second law of motion is denoted as,

$\to \sum F=ma$

The radial component of acceleration is defined as,

$a_r=\ddot{r}-r{\dot{\theta }}^2$

The spring force is defined as,

$F_{spring}=kx$

Where,

$x$ - Elongation

Calculation:

According to sub-part a,

We have found,

$\ddot{r}+\left(\frac{k}{m_B}-{\dot{\theta }}^2\right)r=0$

At $k=200N/m$, it becomes,

$\begin{array}{l}\ddot{r}+\left(\frac{200N/m}{1kg}-{\left(10rad/s\right)}^2\right)r=0\\ \ddot{r}+100r=0\end{array}$

But we can define $\ddot{r}$ as,

$\ddot{r}=v_r\frac{d{v}_r}{dr}$

Then,

$v_r\frac{d{v}_r}{dr}+100r=0$

At $t=0,v_r=0,r=r_0$

$\begin{array}{l}\underset{0}{\overset{v_r}{\int }}{v}_{r}d{v}_r=-100\underset{r_0}{\overset{r}{\int }}rdr\\ v_r=10\sqrt{r_0{}^2-r^2}\end{array}$

But we know that,

$v_r=\frac{dr}{dt}$

Therefore,

At $t=0,r=r_0$

$\begin{array}{l}\frac{dr}{dt}=10\sqrt{r_0{}^2-r^2}\\ \underset{r_0}{\overset{r}{\int }}\frac{1}{\sqrt{r_0{}^2-r^2}}dr=\underset{0}{\overset{t}{\int }}10dt\end{array}$

To solve this, assume

$\begin{array}{l}r=r_0\sin \varphi \\ dr=r_0\cos \varphi d\varphi \end{array}$

Therefore,

$\begin{array}{l}\underset{\pi /2}{\overset{{\sin }^{-1}\left(r/r_0\right)}{\int }}\frac{r_0\cos \varphi d\varphi }{\sqrt{r_0{}^2-{\left(r_0\sin \varphi \right)}^2}}=\underset{0}{\overset{t}{\int }}10dt\\ \underset{\pi /2}{\overset{{\sin }^{-1}\left(r/r_0\right)}{\int }}d\varphi =10t\end{array}$

By solving further,

${\sin }^{-1}\left(r/r_0\right)-\pi /2=10t$

Rearrange,

$\begin{array}{l}r=r_0\sin \left(\pi /2+10t\right)\\ r=r_0\cos 10t\to \left(1\right)\end{array}$

Substitute,

$\begin{array}{l}r=0.5\cos 10\left(0.1\right)\\ r=0.27m\end{array}$

Differentiate equation 1,

$\dot{r}=-5\sin 10t$

Now find the force exerted,

$\begin{array}{l}{F}_A=m_B\left(0+2\dot{r}\dot{\theta }\right)\\ F_A=1kg\left(2\left(\left(-5\sin 10\times 0.1\right)\right)10rad/s\right)=-84.147N\end{array}$

Conclusion:

The position of slider and horizontal force exerted on at $t=0.1s$ for $k=200N/m$ is equal to,

$\begin{array}{l}r=0.27m\\ F_A=-84.147N\end{array}$