Chapter 3, Problem 3.36P

To determine

The velocities of the two masses just after the impulse has been applied.

Answer to Problem 3.36P

The total velocity of the left mass is $\underset{\_}{\frac{F\Delta t}{2m}\cos \gamma \widehat{x}+\frac{F\Delta t}{m}\sin \gamma \widehat{y}}$ and the total velocity for the right mass of the dumbbell is $\underset{\_}{\frac{F\Delta t}{2m}\cos \gamma \widehat{x}}$.

Explanation of Solution

Write the expression for the $x$ component of the force acting on both dumbbells.

    $F_x=F\cos \gamma \widehat{x}$        (I)

Here, $F$ is the force acts on the left dumbbell, $\gamma$ is the angle from the $x$ axis.

Write the expression for the $y$ component of the force exerts on the dumbbell.

    $F_y=F\sin \gamma \widehat{y}$        (II)

Use equation (II) and (I) to write the expression for the net force on the system.

    $\begin{array}{c}F=F_x+F_y\\ =F\cos \gamma \widehat{x}++F\sin \gamma \widehat{y}\end{array}$        (III)

Here, $F$ is the net force on the system.

Figure $1$ shows the force $F$ acting on the dumbbell in the north-easterly direction at an angle of $\gamma$ from the $x$ axis.

The force exerted on the dumbbell for a short time is called as impulse exerts on the dumbbell. The change in momentum to the dumbbell is equal to the impulse exerts on the dumbbell.

    $\Delta P=F\Delta t$        (IV)

Here, $\Delta P$ is the change in momentum, $\Delta t$ is the time interval.

The initial momentum is zero because initially the dumbbell is at rest.

Write the expression for $\Delta P$.

    $\begin{array}{c}\Delta P=P_{\text{f}}-P_{\text{i}}\\ =P_{\text{f}}-0\\ =P_{\text{f}}\end{array}$        (V)

Use equation (V) and (III) in (IV) to solve for the momentum.

    $\begin{array}{c}{P}_{\text{f}}=\left(F\cos \gamma \widehat{x}+F\sin \gamma \widehat{y}\right)\Delta t\\ =\left(F\cos \gamma \Delta t\right)\widehat{x}+\left(F\sin \gamma \Delta t\right)\widehat{y}\end{array}$        (VI)

Write the expression for the final linear momentum of the system consists of mass $2m$ located at the center of the mass.

    $P_{\text{f}}=2m{v}_{\text{cm}}$        (VII)

Compare the equation (VI) and (VII) to solve for $v_{\text{cm}}$.

    $\begin{array}{c}2m{v}_{\text{cm}}=\left(F\cos \gamma \Delta t\right)\widehat{x}+\left(F\sin \gamma \Delta t\right)\widehat{y}\\ v_{\text{cm}}=\frac{1}{2m}\left[\left(F\cos \gamma \Delta t\right)\widehat{x}+\left(F\sin \gamma \Delta t\right)\widehat{y}\right]\\ =\frac{F\cos \gamma \Delta t}{2m}\widehat{x}+\frac{F\sin \gamma \Delta t}{2m}\widehat{y}\end{array}$        (VIII)

Write the expression for the torque acting on the dumbbell.

    $\tau =b\times F$        (IX)

Here, $\tau$ is the torque acting on the dumbbell, $b$ is the distance from one end of the dumbbell to its center of mass.

The torque about the center of mass is pointed into the page or in the $-z$ direction.

    $\left|F\right|=F\sin \gamma$        (X)

Use equation (X) in (IX) to solve for the torque.

    $\tau =bF\sin \gamma \left(-\widehat{z}\right)$        (XI)

Write the expression for the total angular momentum transferred to the dumbbell during time interval $\Delta t$.

    $L=\tau \Delta t$        (XII)

Here, $L$ is the total angular momentum.

Use equation (XI) in (XII) to solve for $L$.

    $\begin{array}{c}L=bF\sin \gamma \left(-\widehat{z}\right)\Delta t\\ =-bF\sin \gamma \Delta t\left(\widehat{z}\right)\end{array}$        (XIII)

Write the general expression for the angular momentum.

    $L=I\omega$        (XIV)

Here, $I$ is the moment of inertia, $\omega$ is the angular frequency.

Write the expression for $I$.

    $I=2m{b}^2$        (XV)

Use equation (XIII) and (XV) in (XIV) to solve for $\omega$.

    $\begin{array}{c}\omega =\frac{-\left(bF\sin \gamma \Delta t\right)\widehat{z}}{2m{b}^2}\\ =-\frac{F\Delta t}{2mb}\sin \gamma \widehat{z}\\ \left|\omega \right|=\frac{F\Delta t}{2mb}\sin \gamma \end{array}$        (XVI)

Write the expression for the total velocity of the left mass of the dumbbell.

    $v_{\text{left}}=v_{\text{cm}}+\omega b\widehat{y}$        (XVII)

Use equation (VIII) and (XVI) in (XVII) to solve for $v_{\text{left}}$.

    $\begin{array}{c}{v}_{\text{left}}=\frac{F\cos \gamma \Delta t}{2m}\widehat{x}+\frac{F\sin \gamma \Delta t}{2m}\widehat{y}+\left(\frac{F\Delta t}{2mb}\sin \gamma \right)b\widehat{y}\\ =\frac{F\cos \gamma \Delta t}{2m}\widehat{x}+\frac{2F\sin \gamma \Delta t}{2m}\widehat{y}\\ =\frac{F\Delta t}{2m}\cos \gamma \widehat{x}+\frac{F\Delta t}{m}\sin \gamma \widehat{y}\end{array}$        (XVIII)

Write the expression for the total velocity of the right mass of the dumbbell.

    $v_{\text{right}}=v_{\text{cm}}-b\omega \widehat{y}$        (XIX)

Use equation (VIII) and (XVI) in (XVIII) to solve for $v_{\text{right}}$.

    $\begin{array}{c}{v}_{\text{right}}=\frac{F\cos \gamma \Delta t}{2m}\widehat{x}+\frac{F\sin \gamma \Delta t}{2m}\widehat{y}-\left(\frac{F\Delta t}{2mb}\sin \gamma \right)b\widehat{y}\\ =\frac{F\Delta t}{2m}\cos \gamma \widehat{x}\end{array}$        (XX)

If $\gamma =0°$, then the velocity of the left mass.

    $\begin{array}{c}{v}_{\text{left}}=\frac{F\Delta t}{2m}\cos 0°\widehat{x}+\frac{F\Delta t}{m}\sin 0°\widehat{y}\\ =\frac{F\Delta t}{2m}\widehat{x}\end{array}$        (XXI)

If $\gamma =0°$, then the velocity of the right mass.

    $\begin{array}{c}{v}_{\text{right}}=\frac{F\Delta t}{2m}\cos 0°\widehat{x}\\ =\frac{F\Delta t}{2m}\widehat{x}\end{array}$        (XXII)

The velocity of the left mass is equal to the velocity of the right mass when $\gamma =0°$.

If $\gamma =90°$, then the velocity of the left mass.

    $\begin{array}{c}{v}_{\text{right}}=\frac{F\Delta t}{2m}\cos 90°\widehat{x}+\frac{F\Delta t}{m}\sin 90°\widehat{y}\\ =\frac{F\Delta t}{m}\widehat{y}\end{array}$        (XXIII)

If $\gamma =90°$, then the velocity of the right mass.

    $\begin{array}{c}{v}_{\text{right}}=\frac{F\Delta t}{2m}\cos 90°\widehat{x}\\ =0\end{array}$        (XXIV

The velocity of the right mass is zero.

Conclusion:

Therefore, the total velocity of the left mass is $\underset{\_}{\frac{F\Delta t}{2m}\cos \gamma \widehat{x}+\frac{F\Delta t}{m}\sin \gamma \widehat{y}}$ and the total velocity for the right mass of the dumbbell is $\underset{\_}{\frac{F\Delta t}{2m}\cos \gamma \widehat{x}}$.