Chapter 12, Problem 12.133RP

Disk A rotates in a horizontal plane about a vertical axis at the ${\stackrel{\dot{}}{\theta }}_0=10\text{ rad/s}$. 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

Find the position of the slider and horizontal force exerted on the slider by disk at $t=0.1\text{s}$ for the spring constant of $100\text{N}/\text{m}$.

Answer to Problem 12.133RP

The position of the slider at $t=0.1\text{s}$ for the spring constant of $100\text{N}/\text{m}$ is $\underset{\_}{0.5\text{m}}$.

The horizontal force exerted on the slider by disk at $t=0.1\text{s}$ for the spring constant $100\text{N}/\text{m}$ is $\underset{\_}{0\text{N}}$.

Explanation of Solution

Given information:

The polar coordinate $\left(\dot{\theta }\right)$ of velocity is $10\text{rad/s}$.

The mass $\left(m_B\right)$ of slider B is 1 kg.

The distance $\left(r\right)$ of the slider from the point O is 500 mm.

The spring constant (k) is 100 N/m.

Calculation:

Consider the Position of the slider is in point O

Find the displacement of spring when $r=0$.

${\left(x_{spr}\right)}_0=0$

Consider distance of the slider (r) from the point O is 500 mm.

Find the displacement of spring when $r=500\text{mm}$.

${\left(x_{spr}\right)}_r=r$

Substitute 500 mm for r.

${\left(x_{spr}\right)}_r=500\text{mm}$

Find the restoring force (F) of spring when $r=500\text{mm}$.

$-F=k{\left(x_{spr}\right)}_r$

Substitute 100 N/m for k and 500 mm for ${\left(x_{spr}\right)}_r$.

$\begin{array}{l}-F=\left(100\right)\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\\ F=-50\text{N}\end{array}$

Sketch the free body diagram and kinetic diagram of forces on disk A and spring as shown in in Figure (1).

Refer Figure (1).

Write the equation of radial component of acceleration $\left(a_r\right)$.

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

Apply Newton’s law of equation along radial direction.

The radial force is equal to the restoring force.

Find the equation of restoring force (F).

$\begin{array}{l}{F}_r=F\\ F=m_B{a}_r\end{array}$

Substitute $\ddot{r}-r{\dot{\theta }}^2$ for $a_r$.

$F=m_B\left(\ddot{r}-r{\dot{\theta }}^2\right)$

Substitute $-50$ for F, 1 kg for $m_B$, 500 mm for r, and 10 rad/s for $\dot{\theta }$.

$\begin{array}{l}-50=1\left[\ddot{r}-\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times {10}^2\right)\right]\\ -50=\ddot{r}-50\\ \ddot{r}=0\end{array}$

Write the equation of $\ddot{r}$ in terms of derivative of $\dot{r}$.

$\frac{d}{dt}\left(\dot{r}\right)=\ddot{r}$ (1)

Integrate Equation (1) to find $\dot{r}$.

${\displaystyle \underset{0}{\overset{\dot{r}}{\int }}d\left(\dot{r}\right)}={\displaystyle \underset{0}{\overset{0.1}{\int }}\ddot{r}dt}$ (2)

Use Equation (1) to substitute for $\ddot{r}$ in Equation (2).

$\begin{array}{l}{\displaystyle \underset{0}{\overset{\dot{r}}{\int }}d\left(\dot{r}\right)}={\displaystyle \underset{0}{\overset{0.1}{\int }}0dt}\\ \dot{r}-0=0\\ \dot{r}=0\end{array}$ (3)

Slider B is at initial position when $\dot{r}=0$.

Write $\dot{r}$ in terms of derivative of r.

$\frac{d}{dt}\left(r\right)=\dot{r}$ (4)

Integrate Equation (4) to find $r$.

${\displaystyle \underset{r_0}{\overset{r}{\int }}d\left(r\right)}={\displaystyle \underset{0}{\overset{0.1}{\int }}\dot{r}dt}$ (5)

Use Equation (3) to substitute for $\dot{r}$ in Equation (5).

$\begin{array}{c}{\displaystyle \underset{r_0}{\overset{r}{\int }}d\left(r\right)}={\displaystyle \underset{0}{\overset{0.1}{\int }}0dt}\\ r-r_0=0\\ r=r_0\end{array}$ (6)

Find the position of the slider at $t=0.1\text{s}$ for the spring constant of $100\text{N}/\text{m}$.

Use Equation (4) to substitute for $r_0$ in Equation (6).

$r=0.5\text{m}$

Thus, the position of the slider at $t=0.1\text{s}$ for the spring constant $100\text{N}/\text{m}$ is $\underset{\_}{0.5\text{m}}$.

Refer Figure 1.

Apply Newton’s law of Equation along transverse direction.

Write the transverse component of acceleration $\left(a_{\theta }\right)$ of the disk A.

$a_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }$

Here, $\ddot{\theta }$ is the polar coordinate of transverse acceleration.

The transverse force is the horizontal force exerted on the slider by disk.

The disk is rotating at constant rate. Therefore, the polar coordinate of transverse acceleration, $\ddot{\theta }$ is zero.

Find the horizontal force exerted on the slider by disk at $t=0.1\text{s}$ for the spring constant 100 N/m.

Write the equation of transverse force $\left(F_{\theta }\right)$.

$F_{\theta }=m_B{a}_{\theta }$

Substitute $F_H$ for $F_{\theta }$, 1 kg for m, and $r\ddot{\theta }-2\dot{r}\dot{\theta }$ for $a_{\theta }$.

$\begin{array}{c}{F}_H=\left(1\right)\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)\\ =r\ddot{\theta }+2\dot{r}\dot{\theta }\end{array}$

Substitute 0 for $\ddot{\theta }$.

$\begin{array}{c}{F}_H=r\left(0\right)+2\dot{r}\dot{\theta }\\ =2\dot{r}\dot{\theta }\end{array}$ (7)

Substitute Equation (3) in Equation (7).

$\begin{array}{c}{F}_H=-2\left(0\right)\dot{\theta }\\ =0\end{array}$

Thus, the horizontal force exerted on the slider by disk at $t=0.1\text{s}$ for the spring constant $100\text{N}/\text{m}$ is $\underset{\_}{0\text{N}}$.

To determine

Find the position of the slider and horizontal force exerted on the slider by disk at $t=0.1\text{s}$ for the spring constant of $200\text{N}/\text{m}$.

Answer to Problem 12.133RP

The position of the slider at $t=0.1\text{s}$ for the spring constant of $200\text{N}/\text{m}$ is $\underset{\_}{0.270\text{m}}$.

The horizontal force exerted on the slider by disk at $t=0.1\text{s}$ for the spring constant $200\text{N}/\text{m}$ is $\underset{\_}{-84.1\text{N}}$.

Explanation of Solution

Calculation:

Consider the Position of the slider is in point O

Find the displacement of spring when $r=0$.

${\left(x_{spr}\right)}_0=0$

Consider distance of the slider (r) from the point O is 500 mm.

Find the displacement of spring when $r=500\text{mm}$.

${\left(x_{spr}\right)}_r=r$

Substitute 500 mm for r.

${\left(x_{spr}\right)}_r=500\text{mm}$

Find the restoring force (F) of spring when $r=500\text{mm}$.

$-F=k{\left(x_{spr}\right)}_r$

Substitute 200 N/m for k and 500 mm for ${\left(x_{spr}\right)}_r$.

$\begin{array}{l}-F=\left(200\right)\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\\ F=-100\text{N}\end{array}$

Refer Figure (1).

Write the equation of radial component of acceleration $\left(a_r\right)$.

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

Apply Newton’s law of equation along radial direction.

The radial force is equal to the restoring force.

Find the equation of restoring force (F).

$\begin{array}{l}{F}_r=F\\ F=m_B{a}_r\end{array}$

Substitute $\ddot{r}-r{\dot{\theta }}^2$ for $a_r$.

$F=m_B\left(\ddot{r}-r{\dot{\theta }}^2\right)$

Substitute $-100$ for F, 1 kg for $m_B$, 500 mm for r, and 10 rad/s for $\dot{\theta }$.

$\begin{array}{l}-100=1\left[\ddot{r}-\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times {10}^2\right)\right]\\ -50=\ddot{r}-50\\ \ddot{r}=-50{\text{m/s}}^2\end{array}$

Write the equation of radial velocity of the slider in terms of r.

$\begin{array}{c}{v}_r=\frac{dr}{dt}\\ =\dot{r}\end{array}$ (8)

Here, $v_r$ is the radial velocity of slider.

Write equation of the rate of change of position coordinate in terms of differential equation.

$\dot{r}=\frac{dr}{dt}$ (9)

Apply differentiation to Equation (8)

$\begin{array}{c}\ddot{r}=\frac{d}{dt}\left(\frac{dr}{dt}\right)\\ =\frac{d\dot{r}}{dt}\end{array}$ (10)

Rewrite Equation (10) by multiplying and dividing the right-hand side by dr.

$\ddot{r}=\frac{dr}{dt}\frac{d}{dr}\left(\dot{r}\right)$ (11)

Substitute Equation (10) to rewrite Equation (11).

$\ddot{r}=\frac{dr}{dt}\frac{d}{dr}\left(\frac{dr}{dt}\right)$ (12

Substitute Equation (8) to rewrite Equation (12).

$\ddot{r}=v_r\frac{d{v}_r}{dr}$ (13)

Substitute $-50{\text{m/s}}^2$ for $\ddot{r}$.

$\begin{array}{c}-50=v_r\frac{d{v}_r}{dr}\\ v_{r}dr=-50dr\end{array}$ (14)

Apply the limits to integrate the Equation (14).

At the time of instant $t=0$ the slider is released with no radial velocity in the position $r=r_0$. Thus, at $t=0$, $v_r=0$ and $r=r_0$.

$\begin{array}{l}{\displaystyle \underset{0}{\overset{v_r}{\int }}{v}_{r}d{v}_r}=-100{\displaystyle \underset{r_0}{\overset{r}{\int }}rdr}\\ \frac{v_r^2}{2}-0=-100\left[\frac{r^2}{2}-\frac{r_0^2}{2}\right]\\ v_r^2=-100\left(r^2-r_0^2\right)\\ v_r=10\sqrt{r_0^2-r^2}\end{array}$ (15)

Substitute Equation (8) in Equation (15).

$\frac{dr}{dt}=10\sqrt{r_0^2-r^2}$ (16)

Integrate Equation (16).

$\begin{array}{c}{\displaystyle \underset{r_o}{\overset{r}{\int }}\frac{dr}{\sqrt{r_0^2-r^2}}}={\displaystyle \underset{0}{\overset{t}{\int }}10dt}\\ =10t\end{array}$ (17)

Use spherical polar coordinates and choose,

$r=r_0\sin \varphi$ (18)

Differentiate Equation (18).

$dr=r_0\cos \varphi d\varphi$ (19)

Rewrite Equation (18).

$\varphi ={\sin }^{-1}\left(\frac{r}{r_0}\right)$ (20)

Rewrite Equation (20) for $r=r_0$.

$\begin{array}{c}\varphi ={\sin }^{-1}\left(\frac{r_0}{r_0}\right)\\ ={\sin }^{-1}\left(1\right)\\ =\frac{\pi }{2}\end{array}$ (21)

Use Equation (20) and (21) to change the values of limit in Equation (17).

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

$\begin{array}{l}{\sin }^{-1}\left(r/r_0\right)-\frac{\pi }{2}=10t\\ {\sin }^{-1}\left(r/r_0\right)=\left(10t+\frac{\pi }{2}\right)\\ \sin {\sin }^{-1}\left(r/r_0\right)=\sin \left(10t+\frac{\pi }{2}\right)\\ r=r_0\sin \left(10t+\frac{\pi }{2}\right)\end{array}$ (22)

Apply the trigonometric formula of $\sin \left(\theta +\pi /2\right)$.

$\sin \left(\theta +\pi /2\right)=\cos \theta$ (23)

Use Equation (23) to rewrite Equation (22).

$r=r_0\cos 10t$ (24)

Substitute 0.5m for $r_0$ and 0.1s for t.

$\begin{array}{c}r=0.5\cos \left(10\times 0.1\right)\\ =0.5\times 0.540302\\ =0.27015\\ \approx 0.270\text{m}\end{array}$

Thus, the position of the slider at $t=0.1\text{s}$ for the spring constant $200\text{N}/\text{m}$ is $\underset{\_}{0.270\text{m}}$.

Find the radial polar coordinate of velocity using Equation (24).

Differentiate Equation (24) with respect to t.

$\dot{r}=-10r_0\sin 10t$

Substitute 500 mm for $r_0$

$\begin{array}{c}\dot{r}=-10\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\sin 10t\\ =-5\sin 10t\end{array}$

Find the horizontal force exerted on the slider by disk at $t=0.1\text{s}$ for the spring constant $200\text{N}/\text{m}$ using Equation (7).

$F_H=2\dot{r}\dot{\theta }$

Substitute $-5\sin 10t$ for $\dot{r}$.

$F_H=2\left(-5\sin 10t\right)\dot{\theta }$

Substitute 0.1 s for t and $10\text{rad}/\text{s}$ for $\dot{\theta }$.

$\begin{array}{c}{F}_H=2\left(-5\sin 10\times 0.1\right)\left(10\right)\\ =-84.1\text{N}\end{array}$

Thus, the horizontal force exerted on the slider by disk at $t=0.1\text{s}$ for the spring constant $200\text{N}/\text{m}$ is $\underset{\_}{-84.1\text{N}}$.