Chapter 17.2, Problem 17.91P

A small 4-lb collar C can slide freely on a thin ring of weight 6 lb and radius 10 in. The ring is welded to a short vertical shaft, which can rotate freely in a fixed bearing. Initially, the ring has an angular velocity of 35 rad/s and the collar is at the top of the ring (θ = 0) when it is given a slight nudge. Neglecting the effect of friction, determine (a) the angular velocity of the ring as the collar passes through the position θ = 90°, (b) the corresponding velocity of the collar relative to the ring.

Fig. P17.91

To determine

Find the angular velocity of the ring as the collar passes through the position $\theta =90°$.

Answer to Problem 17.91P

The angular velocity of the ring as the collar passes through the position $\theta =90°$ is $\underset{\_}{15.00\text{rad/s}}$.

Explanation of Solution

Given information:

The weight $\left(W_C\right)$ of collar C is 4 lb.

The weight $\left(W_R\right)$ of thin ring is 6 lb.

The initial angular velocity $\left({\omega }_1\right)$ of the ring is 35 rad/s.

Calculation:

Find the mass $\left(m_C\right)$ of the collar C using the equation:

$m_C=\frac{W_C}{g}$

Substitute 4 lb for $W_C$ and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}{m}_C=\frac{4}{32.2}\\ =0.12422\text{lb}\cdot {\text{s}}^2\text{/ft}\end{array}$

Find the mass $\left(m_C\right)$ of the ring using the equation:

$m_R=\frac{W_R}{g}$

Substitute 6 lb for $W_C$ and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}{m}_R=\frac{6}{32.2}\\ =0.18634\text{lb}\cdot {\text{s}}^2\text{/ft}\end{array}$

Sketch the component of velocity of the given system at initial position as shown in Figure (1).

Find the centroidal moment of inertia $\left({\overline{I}}_R\right)$ of the ring using the equation:

${\overline{I}}_R=\frac{1}{2}{m}_R{R}^2$

Here, R is the radius of the ring.

The collar C is located at top of the ring at initial position. Therefore, the angle $\left(\theta \right)$ is zero at initial position.

Sketch the component of velocity of the given system at final position as shown in Figure (2).

Refer Figure (2).

Write the equation of the velocity ${\left(v_C\right)}_y$ of the collar C in y-direction.

${\left(v_C\right)}_y=R{\omega }_2$

Here, ${\omega }_2$ is the angular velocity of the system at final position.

Consider the conservation of angular momentum.

${\overline{I}}_R{\omega }_1={\overline{I}}_R{\omega }_2+m_C{\left(v_C\right)}_{y}R$

Substitute $\frac{1}{2}{m}_R{R}^2$ for ${\overline{I}}_R$, and $R{\omega }_2$ for ${\left(v_C\right)}_y$.

$\begin{array}{l}\left(\frac{1}{2}{m}_R{R}^2\right){\omega }_1=\left(\frac{1}{2}{m}_R{R}^2\right){\omega }_2+m_C\left(R{\omega }_2\right)R\\ \frac{1}{2}{m}_R{R}^2{\omega }_1=\frac{1}{2}{m}_R{R}^2{\omega }_2+m_C{R}^2{\omega }_2\\ \frac{1}{2}{m}_R{R}^2{\omega }_1=\frac{\left(m_R+2m_C\right)R^2}{2}{\omega }_2\end{array}$

Simplify the Equation.

$\begin{array}{l}{m}_R{R}^2{\omega }_1=\left(m_R+2m_C\right)R^2{\omega }_2\\ {\omega }_2=\frac{m_R{R}^2}{\left(m_R+2m_C\right)R^2}{\omega }_1\\ {\omega }_2=\frac{m_R}{\left(m_R+2m_C\right)}{\omega }_1\end{array}$

Substitute $\frac{6\text{lb}}{32.2\text{ft}/\text{s}}$ for $m_R$, $\frac{4\text{lb}}{32.2\text{ft}/\text{s}}$ for $m_C$, and $35\text{rad}/\text{s}$ for ${\omega }_1$.

$\begin{array}{c}{\omega }_2=\frac{\frac{6\text{lb}}{32.2\text{ft}/\text{s}}}{\frac{6\text{lb}}{32.2\text{ft}/\text{s}}+2\left(\frac{4\text{lb}}{32.2\text{ft}/\text{s}}\right)}\left(35\text{rad}/\text{s}\right)\\ =\frac{6}{14}\left(35\text{rad}/\text{s}\right)\\ =15\text{rad}/\text{s}\end{array}$

Thus, the angular velocity of the ring as the collar passes through the position $\theta =90°$ is $\underset{\_}{15.00\text{rad/s}}$.

To determine

Find the corresponding velocity of the collar relative to the ring.

Answer to Problem 17.91P

The corresponding velocity of the collar relative to the ring is $\underset{\_}{20.5\text{ft/s}}$.

Explanation of Solution

Calculation:

Find the equation of the kinetic energy $\left(T_1\right)$ of the system at initial position.

$T_1=\frac{1}{2}{\overline{I}}_R{\omega }_1{}^2$

Substitute $\frac{1}{2}{m}_R{R}^2$ for ${\overline{I}}_R$.

$\begin{array}{c}{T}_1=\frac{1}{2}\left(\frac{1}{2}{m}_R{R}^2\right){\omega }_1{}^2\\ =\frac{1}{4}{m}_R{R}^2{\omega }_1{}^2\end{array}$

Find the equation of the potential energy $\left(V_1\right)$ of the system at initial position.

$V_1=m_{C}gR$

Find the equation of the kinetic energy $\left(T_2\right)$ of the system at final position.

$T_2=\frac{1}{2}\left(\frac{1}{2}{m}_R{R}^2\right){\omega }_2{}^2+\frac{1}{2}{m}_C{\left(v_C\right)}_x{}^2+\frac{1}{2}{m}_C{\left(v_C\right)}_y{}^2$

Substitute $\frac{1}{2}{m}_R{R}^2$ for ${\overline{I}}_R$, and $R{\omega }_2$ for ${\left(v_C\right)}_x$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left(\frac{1}{2}{m}_R{R}^2\right){\omega }_2{}^2+\frac{1}{2}{m}_C{\left(R{\omega }_2\right)}^2+\frac{1}{2}{m}_C{\left(v_C\right)}_y{}^2\\ =\frac{1}{4}{m}_R{R}^2{\omega }_2{}^2+\frac{1}{2}{m}_C{R}^2{\omega }_2{}^2+\frac{1}{2}{m}_C{\left(v_C\right)}_y{}^2\end{array}$

The potential energy of the system at final position $\left(V_2\right)$ is zero.

Consider the conservation of energy.

$T_1+V_1=T_2+V_2$

Substitute $\frac{1}{4}{m}_R{R}^2{\omega }_1{}^2$ for $T_1$, $m_{C}gR$ for $V_1$, $\frac{1}{4}{m}_R{R}^2{\omega }_2{}^2+\frac{1}{2}{m}_C{R}^2{\omega }_2{}^2+\frac{1}{2}{m}_C{\left(v_C\right)}_y{}^2$ for $T_2$, and 0 for $V_2$.

$\begin{array}{l}\frac{1}{4}{m}_R{R}^2{\omega }_1{}^2+m_{C}gR=\frac{1}{4}{m}_R{R}^2{\omega }_2{}^2+\frac{1}{2}{m}_C{R}^2{\omega }_2{}^2+\frac{1}{2}{m}_C{\left(v_C\right)}_y{}^2+0\\ \frac{1}{4}{m}_R{R}^2{\omega }_1{}^2+m_{C}gR=\left(\frac{1}{4}{m}_R+\frac{1}{2}{m}_C\right)R^2{\omega }_2{}^2+\frac{1}{2}{m}_C{\left(v_C\right)}_y{}^2\end{array}$

Substitute $\frac{6\text{lb}}{32.2\text{ft}/\text{s}}$ for $m_R$, 10 in. for R, $\frac{4\text{lb}}{32.2\text{ft}/\text{s}}$ for $m_C$, $32.2\text{ft}/\text{s}$ for g, $35\text{rad}/\text{s}$ for ${\omega }_1$, and $15\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{l}\left\{\begin{array}{l}\frac{1}{4}\left(\frac{6\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right){\left(10\text{in}\text{.}\right)}^2{\left(35\text{rad}/\text{s}\right)}^2+\\ \left(\frac{4\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(10\text{in}\text{.}\right)\end{array}\right\}=\left\{\begin{array}{l}\left(\frac{1}{4}\left(\frac{6\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right)+\frac{1}{2}\left(\frac{4\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right)\right)\times \\ {\left(10\text{in}\text{.}\right)}^2{\left(15\text{rad}/\text{s}\right)}^2\\ +\frac{1}{2}\left(\frac{4\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right){\left(v_C\right)}_y{}^2\end{array}\right\}\\ \left\{\begin{array}{l}\frac{1}{4}\left(\frac{6\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right){\left(\frac{10}{12}\text{ft}\right)}^2{\left(35\text{rad}/\text{s}\right)}^2+\\ \left(\frac{4\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(\frac{10}{12}\text{ft}\right)\end{array}\right\}=\left\{\begin{array}{l}\left(\frac{1}{4}\left(\frac{6\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right)+\frac{1}{2}\left(\frac{4\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right)\right)\times \\ {\left(\frac{10}{12}\text{ft}\right)}^2{\left(15\text{rad}/\text{s}\right)}^2\\ +\frac{1}{2}\left(\frac{4\text{lb}}{32.2\text{ft}/{\text{s}}^2}\right){\left(v_C\right)}_y{}^2\end{array}\right\}\\ 39.6286+3.333=16.9836+0.0621{\left(v_C\right)}_y{}^2\end{array}$

$\begin{array}{l}0.0621{\left(v_C\right)}_y{}^2=42.9619-16.9836\\ {\left(v_C\right)}_y{}^2=\frac{25.9782}{0.0621}\\ {\left(v_C\right)}_y={\left(418.25\right)}^{0.5}\\ {\left(v_C\right)}_y=20.5\text{ft}/\text{s}\end{array}$

Thus, the corresponding velocity of the collar relative to the ring is $\underset{\_}{20.5\text{ft/s}}$.