Chapter 17.2, Problem 17.91P

To determine

(a)

Calculate the angular velocity of the ring when collar passes through ${90}^0$ ? Collar is free to slide on a ring and ring is attached to a vertical shaft. Which rotates in a fixed bearing. Initially, collar is at top of the ring $\theta =0$ at $35$ rad/sec of angular velocity. Refer figure.

Answer to Problem 17.91P

The angular velocity of the ring when coller is at ${90}^0$ from the top is 15 rad/sec.

Explanation of Solution

Given:

The weight of collar = ${\omega }_c=4lb=1.814kg$

Mass of ring = ${\omega }_r=6lb=2.721kg$

The radius of the ring,

$\begin{array}{l}{R}_r=10in\\ =0.254m\\ \theta =0\\ v_c=0\end{array}$

The angular velocity $({\omega }_c)$ of the collar = $35$ rad/sec.

Concept used:

Conservation of angular momentum.

Conservation of energy.

Calculation:

Position (1): $\theta =0v_c=0$

Position (2): $\theta ={90}^0{(v_c)}_4=v_y=R{\omega }_2$

By conservation of angular momentum,

$\begin{array}{l}{I}_R{\omega }_1=I_R{\omega }_2+m_c{v}_{y}R\\ m_R{R}^2{\omega }_1=(m_R+2m_c)R^2{\omega }_2\\ {\omega }_2=\frac{m_R{\omega }_1}{(m_R+2m_c)}\end{array}$

Potential energy at position 1 and position 2

$\begin{array}{l}{v}_1=m_{c}gR\\ v_2=0\end{array}$

The kinetic energy of position 1 and position 2

$\begin{array}{l}{T}_1=\frac{I_R{\omega }_1{}^2}{2}=\frac{1}{2}{m}_R{R}^2{\omega }_1{}^2\\ T_2=\frac{I_R{\omega }_1{}^2}{2}+\frac{1}{2}{m}_c(v_x{}^2+v_Y{}^2)\\ =\frac{m_R{R}^2{\omega }_2{}^2}{4}+\frac{m_c{R}^2{\omega }_2{}^2}{2}+\frac{m_c{v}_y{}^2}{2}\end{array}$

Applying conservation of energy principle,

$\begin{array}{l}{T}_1+v_1=T_2+v_2\\ \frac{m_R{R}^2{\omega }_1{}^2}{4}+m_{c}gR=\frac{m_R{R}^2{\omega }_2{}^2}{4}+\frac{m_c{R}^2{\omega }_2{}^2}{2}+\frac{m_c{v}_y{}^2}{2}\end{array}$

But,

Mass of collar, $m_c=\frac{w_c}{g}=0.1849kg$

Mass of ring, $m_R=\frac{w_R}{g}=0.277kg$

$\begin{array}{l}{\omega }_2=\frac{m_R{\omega }_1}{(m_R+2m_c)}=\frac{0.277\times 35}{(0.277+2\times 0.1849)}\\ =14.98\\ {\omega }_2=15rad/\sec \end{array}$

Conclusion:

Using conservation of angular momentum we are able to find, the angular velocity of the ring for $\theta ={90}^0$ is $15$ a rad/sec.

To determine

(b)

What will be the velocity of collar with respect to ring? Collar is free to slide on a ring and ring is attached to a vertical shaft. Which rotates in a fixed bearing. Initially, collar is at top of the ring $\theta =0$ at $35$ rad/sec of angular velocity. Refer figure.

Answer to Problem 17.91P

The corresponding velocity of collar relative to ring is $6.242$ m/s.

Explanation of Solution

Given:

The weight of color = ${\omega }_c=4lb=1.814kg$

Mass of ring = ${\omega }_r=6lb=2.721kg$

The radius of the ring,

$\begin{array}{l}{R}_r=10in\\ =0.254m\\ \theta =0\\ v_c=0\end{array}$

The angular velocity $({\omega }_c)$ of the collar = $35$ rad/sec.

Concept used:

Conservation of angular momentum.

Conservation of energy.

Calculation:

Potential energy at position 1 and position 2

$\begin{array}{l}{v}_1=m_{c}gR\\ v_2=0\end{array}$

The kinetic energy of position 1 and position 2

$\begin{array}{l}{T}_1=\frac{I_R{\omega }_1{}^2}{2}=\frac{1}{2}{m}_R{R}^2{\omega }_1{}^2\\ T_2=\frac{I_R{\omega }_1{}^2}{2}+\frac{1}{2}{m}_c(v_x{}^2+v_Y{}^2)\\ =\frac{m_R{R}^2{\omega }_2{}^2}{4}+\frac{m_c{R}^2{\omega }_2{}^2}{2}+\frac{m_c{v}_y{}^2}{2}\end{array}$

Applying conservation of energy principle,

$\begin{array}{l}{T}_1+v_1=T_2+v_2\\ \frac{m_R{R}^2{\omega }_1{}^2}{4}+m_{c}gR=\frac{m_R{R}^2{\omega }_2{}^2}{4}+\frac{m_c{R}^2{\omega }_2{}^2}{2}+\frac{m_c{v}_y{}^2}{2}\end{array}$

$\begin{array}{l}\frac{m_R{R}^2{\omega }_1{}^2}{4}+m_{c}gR=(\frac{m_R}{4}+\frac{m_c}{2})R^2{\omega }_2{}^2+\frac{m_c{v}_y{}^2}{2}\\ 5.472+0.4607=2.347+0.092v{y}^2\\ v_y=6.242m/s.\end{array}$

Conclusion:

Using conservation of energy principle we are able to find the velocity of collar relative to ring at $\theta ={90}^0$ is $6.242m/s$