Chapter 19.1, Problem 19.15P

A 5-kg collar C is released from rest in the position shown and slides without friction on a vertical rod until it hits a spring with a constant of k = 720 N/m that it compresses. The velocity of the collar is reduced to zero, and the collar reverses the direction of its motion and returns to its initial position. The cycle is then repeated. Determine (a) the period of the motion of the collar, (b) the velocity of the collar 0.4 s after it was released. (Note: This is a periodic motion, but it is not simple harmonic motion.)

Fig. P19.15

To determine

The period $\left({\tau }_n\right)$ of the motion of the collar.

Answer to Problem 19.15P

The period $\left({\tau }_n\right)$ of the motion of the collar is $\underset{\_}{0.943\sec }$.

Explanation of Solution

Given Information:

The mass (m) of the collar C is 5 kg.

The spring constant (k) is $720\text{N/m}$.

The value of acceleration due to gravity (g) is $9.81{\text{m/s}}^{\text{2}}$.

The vertical distance (h) between the collar and the spring is 0.5 m.

Calculation:

Calculate the natural circular frequency $\left({\omega }_n\right)$ using the relation:

${\omega }_n=\sqrt{\frac{k}{m}}$

Substitute $720\text{N/m}$ for k and 5 kg for m.

$\begin{array}{c}{\omega }_n=\sqrt{\frac{\text{720}\text{N/m}}{5}}\\ =\sqrt{144}\\ =\text{12}\text{rad}/\text{s}\end{array}$

Calculate the initial velocity $\left(v_0\right)$ using the relation:

$v_0=\sqrt{2gh}$

Substitute $9.81{\text{m/s}}^{\text{2}}$ for g and 0.5 m for h.

$\begin{array}{c}{v}_0=\sqrt{2\times 9.81\times 0.5}\\ =3.132\text{m/s}\end{array}$

Calculate the free fall time $\left(t_1\right)$ using the relation:

$t_1=\sqrt{\frac{2h}{g}}$

Substitute $9.81{\text{m/s}}^{\text{2}}$ for g and 0.5 m for h.

$\begin{array}{c}{t}_1=\sqrt{\frac{2\left(0.5\right)}{9.81}}\\ =0.3193\sec \end{array}$

Measure the displacement (x) from the position of static displacement of the spring.

Calculate the weight (W) of the collar C using the relation:

$W=mg$

Substitute 5kg for m and $9.81{\text{m/s}}^{\text{2}}$ for g:

$\begin{array}{c}W=5\times 9.81\\ =49.05\text{kg}\cdot {\text{ft/s}}^{\text{2}}\end{array}$

Calculate the static displacement $\left({\delta }_{st}\right)$ using the relation:

${\delta }_{st}=\frac{W}{k}$

Substitute $49.05\text{kg}\cdot {\text{ft/s}}^{\text{2}}$ for W and $720\text{N/m}$ for k.

$\begin{array}{c}{\delta }_{st}=\frac{49.05}{720}\\ =0.068125\text{m}\end{array}$

Show the displacement of the collar and spring while impact as in Figure (1).

Write the equation of motion for simple harmonic motion as below:

$m\ddot{x}+kx=0$

Substitute 5 kg for m and $720\text{N/m}$ for k.

$5\ddot{x}+\left(720\right)k=0$

Write the expression for displacement (x):

$x=x_m\sin \left({\omega }_{n}t+\varphi \right)$ (1)

Differentiate the above equation.

$\dot{x}=x_m{\omega }_n\cos \left({\omega }_{n}t+\varphi \right)$ (2)

When time (t) is 0 the displacement (x) is equal to ${\delta }_{st}$ and the velocity $\left(\dot{x}\right)$ is $3.132\text{m/s}$.

Substitute 0 for t and $0.068125\text{m}$ for x in equation (1).

$\begin{array}{l}0.068125=x_m\sin \left({\omega }_n\left(0\right)+\varphi \right)\\ 0.068125=x_m\sin \left(\varphi \right)\end{array}$

Rewrite the above equation,

$\begin{array}{l}0.068125=x_m\sin \left(\varphi \right)\\ x_m=\frac{0.068125}{\sin \left(\varphi \right)}\end{array}$ (3)

Substitute 0 for t, $-3.132\text{m/s}$ for $\dot{x}$, and $\text{12}\text{rad}/\text{s}$ for ${\omega }_n$ in Equation (2).

$\begin{array}{l}-3.132=x_m\left(12\right)\cos \left({\omega }_n\left(0\right)+\varphi \right)\\ -3.132=12x_m\cos \left(\varphi \right)\end{array}$ (4)

Calculate the phase angle $\left(\varphi \right)$:

Substitute $\frac{0.068125}{\sin \left(\varphi \right)}$ for $x_m$ in equation (4).

$\begin{array}{c}-3.132=12\times \frac{0.068125}{\sin \left(\varphi \right)}\times \cos \left(\varphi \right)\\ -3.132=12\times \left(0.068125\sec \left(\varphi \right)\right)\\ \frac{-3.132}{12}=\left(0.068125\sec \left(\varphi \right)\right)\left(\because \sec \varphi =\frac{1}{\tan \varphi }\right)\end{array}$

$\begin{array}{l}\frac{-0.261}{0.068125}=\frac{1}{\tan \varphi }\\ \tan \varphi =\frac{1}{-3.831}\\ \varphi ={\tan }^{-1}\left(-0.261\right)\\ \varphi =-0.25531\text{rad}\end{array}$

Calculate the amplitude $\left(x_m\right)$:

Substitute $-0.25531\text{rad}$ for $\varphi$ in Equation (3).

$\begin{array}{c}{x}_m=\frac{0.068125}{\sin \left(-0.25531\right)}\\ =-0.26975\text{m}\end{array}$

Hence, from time of impact, the ‘time of flight’ is the time necessary for the collar to come to rest on its downward motion. The time required for collar to rest return is $\left(t_2\right)$.

At time $\left(t_2\right)$ the velocity $\left(\dot{x}\right)$ is zero.

Calculate the time $\left(t_2\right)$:

Substitute 0 for $\dot{x}$ in equation (2).

$\begin{array}{l}0=x_m{\omega }_n\cos \left({\omega }_n{t}_2+\varphi \right)\\ 0=\cos \left({\omega }_n{t}_2+\varphi \right)\\ {\omega }_n{t}_2+\varphi ={\cos }^{-1}\left(0\right)\\ {\omega }_n{t}_2+\varphi =\frac{\pi }{2}\end{array}$

Substitute $\text{12}\text{rad}/\text{s}$ for ${\omega }_n$ and $-0.25531\text{rad}$ for $\varphi$.

$\begin{array}{l}12t_2+\left(-0.25531\right)=\frac{\pi }{2}\\ 12t_2=\frac{\pi }{2}+0.25531\\ t_2=\frac{1.826}{12}\\ t_2=0.15217\sec \end{array}$

Calculate the period of motion $\left({\tau }_n\right)$ using the relation:

${\tau }_n=2\left(t_1+t_2\right)$

Substitute $0.15217\sec$ for $t_2$ and $0.3193\sec$ for $t_1$.

$\begin{array}{c}{\tau }_n=2\left(0.3193+0.15217\right)\\ =0.9429\\ =0.943\sec \end{array}$

Therefore, the period $\left({\tau }_n\right)$ of the motion of the collar is $\underset{\_}{0.943\sec }$.

To determine

The velocity $\left(\dot{x}\right)$ of the collar 0.4 s after it was released.

Answer to Problem 19.15P

The velocity $\left(\dot{x}\right)$ of the collar 0.4 s after it was released is $\underset{\_}{2.45\text{m/s}(\downarrow )}$.

Explanation of Solution

Given Information:

The mass (m) of the collar C is 5 kg.

The spring constant (k) is $720\text{N/m}$.

The value of acceleration due to gravity (g) is $9.81{\text{m/s}}^{\text{2}}$.

The vertical distance (h) between the collar and the spring is 0.5 m.

Calculation:

Calculate the velocity $\left(\dot{x}\right)$ 0.4 sec after it was released.

Rewrite equation (2).

$\dot{x}=x_m{\omega }_n\cos \left({\omega }_n\left(0.4-t_1\right)+\varphi \right)$

Substitute $-0.26975\text{m}$ for $x_m$, $\text{12}\text{rad}/\text{s}$ for ${\omega }_n$, $-0.25531\text{rad}$ for $\varphi$, and $0.3193\sec$ for $t_1$.

$\begin{array}{c}\dot{x}\left(0.4\right)=\left(-0.26975\right)\left(12\right)\left(\cos \left(\left(12\left(0.4-0.3193\right)\right)+\left(-0.25531\right)\right)\right)\\ =-2.4483\text{m/s}\\ =2.45\text{m/s}(\downarrow )\end{array}$

Therefore, the velocity $\left(\dot{x}\right)$ of the collar 0.4 s after it was released is $\underset{\_}{2.45\text{m/s}(\downarrow )}$.