Chapter 11, Problem 11.38P

Review. A thin, uniform, rectangular signboard hangs vertically above the door of a shop. The sign is hinged to a stationary horizontal rod along its top edge. The mass of the sign is 2.40 kg, and its vertical dimension is 50.0 cm. The sign is swinging without friction, so it is a tempting target for children armed with snowballs. The maximum angular displacement of the sign is 25.0° on both sides of the vertical. At a moment when the sign is vertical and moving to the left, a snowball of mass 400 g, traveling horizontally with a velocity of 160 cm/s to the right, strikes perpendicularly at the lower edge of the sign and sticks there. (a) Calculate the angular speed of the sign immediately before the impact. (b) Calculate its angular speed immediately after the impact. (c) The spattered sign will swing up through what maximum angle?

To determine

The angular speed of the sign immediate before the collision.

Answer to Problem 11.38P

The angular speed of the sign immediate before the collision is $2.35\text{rad}/\text{s}$.

Explanation of Solution

The mass of sign is $2.40\text{ kg}$, the vertical length of sign is $\text{50 cm}$, the maximum angular displacement is $25°$, the mass snowball is $400\text{ g}$ and the velocity of snowball is $160\text{cm}/\text{s}$.

Consider figure given below.

Figure (I)

The height of the sign is calculated as,

    $\begin{array}{c}h=\frac{1}{2}L-\frac{1}{2}L\cos \theta \\ =\frac{1}{2}L\left(1-\cos \theta \right)\end{array}$

Here, $L$ is the vertical length of the sign and $\theta$ is the angular displacement.

The formula for the conservation of the energy is,

    $\begin{array}{c}Mgh=\frac{I{\omega }^2}{2}\\ {\omega }^2=\frac{2Mgh}{I}\end{array}$        (I)

Here, $M$ is the mass of sign, $g$ is the acceleration due to gravity, $I$ is the moment of inertia and ${\omega }_{\text{i}}$ is the angular speed before the collision.

The formula to calculate initial moment of inertia is,

    $\begin{array}{c}{I}_{\text{i}}=\frac{M{L}^2}{12}+M{\left(\frac{L}{2}\right)}^2\\ =\frac{M{L}^2}{12}+\frac{M{L}^2}{4}\\ =\frac{M{L}^2}{3}\end{array}$

Substitute $\frac{1}{2}L\left(1-\cos \theta \right)$ for $h$ and $\frac{M{L}^2}{3}$ for $I_{\text{i}}$ in equation (I) to find ${\omega }_{\text{i}}$.

  $\begin{array}{c}{\omega }_{\text{i}}^2=\frac{2Mg\frac{1}{2}L\left(1-\cos \theta \right)}{\frac{M{L}^2}{3}}\\ =\frac{3g\left(1-\cos \theta \right)}{L}\\ {\omega }_{\text{i}}=\sqrt{\frac{3g\left(1-\cos \theta \right)}{L}}\end{array}$

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, $25°$ for $\theta$ and $\text{50 cm}$ for $L$ in above equation to find ${\omega }_{\text{i}}$.

    $\begin{array}{c}{\omega }_{\text{i}}=\sqrt{\frac{3\left(9.8\text{m}/{\text{s}}^2\right)\left(1-\cos \left(25°\right)\right)}{\text{50 cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}}\\ =2.35\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the angular speed of the sign immediate before the collision is $2.35\text{rad}/\text{s}$.

To determine

The angular speed of the sign immediate after the collision.

Answer to Problem 11.38P

The angular speed of the sign immediate after the collision is $0.498\text{rad}/\text{s}$.

Explanation of Solution

The formula for the conservation of the angular momentum is,

    $I_{\text{f}}{\omega }_{\text{f}}=I_{\text{i}}{\omega }_{\text{i}}-mvL$        (II)

Here, $m$ is the mass of snowball, $v$ is the velocity of snowball and ${\omega }_{\text{f}}$ is the angular speed after the collision.

The formula to calculate final moment of inertia is,

    $I_{\text{f}}=\left(\frac{M{L}^2}{3}+m{L}^2\right)$

Substitute $\left(\frac{M{L}^2}{3}+m{L}^2\right)$ for $I_{\text{f}}$ and $\frac{M{L}^2}{3}$ for $I_{\text{i}}$ in equation (II) to find ${\omega }_{\text{f}}$.

  $\begin{array}{c}\left(\frac{M{L}^2}{3}+m{L}^2\right){\omega }_{\text{f}}=\left(\frac{M{L}^2}{3}\right){\omega }_{\text{i}}-mvL\\ {\omega }_{\text{f}}=\frac{\left(\frac{M{L}^2}{3}\right){\omega }_{\text{i}}-mvL}{\left(\frac{M{L}^2}{3}+m{L}^2\right)}\\ =\frac{\left(\frac{M{L}^2}{3}\right){\omega }_{\text{i}}-mvL}{L^2\left(\frac{M}{3}+m\right)}\end{array}$

Substitute $2.40\text{ kg}$ for $M$, $400\text{ g}$ for $m$, $2.35\text{rad}/\text{s}$ for ${\omega }_{\text{i}}$, $160\text{cm}/\text{s}$ for $v$ and $\text{50 cm}$ for $L$ in above equation to find $\omega$.

    $\begin{array}{c}{\omega }_{\text{f}}=\frac{\left(\frac{\left(2.40\text{ kg}\right){\left(\text{50 cm}\right)}^2}{3}\right)\left(2.35\text{rad}/\text{s}\right)-\left(400\text{ g}\right)\left(160\text{cm}/\text{s}\right)\left(\text{50 cm}\right)}{{\left(\text{50 cm}\right)}^2\left(\frac{\left(2.40\text{ kg}\right)}{3}+\left(400\text{ g}\right)\right)}\\ =\frac{\left(\text{2000 kg}\cdot {\text{cm}}^2\right)\left(2.35\text{rad}/\text{s}\right)-\left(400\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\right)\left(\text{160 }\text{cm}/\text{s}\right)\left(\text{50 cm}\right)}{{\left(\text{50 cm}\right)}^2\left(\frac{\left(2.40\text{ kg}\right)}{3}+\left(400\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\right)\right)}\\ =\frac{\left(\text{2000 kg}\cdot {\text{cm}}^2\left(\frac{{10}^{-4}{\text{ m}}^2}{1{\text{ cm}}^2}\right)\right)\left(2.35\text{rad}/\text{s}\right)-\left(\text{3200 }\text{kg}\cdot {\text{cm}}^2\left(\frac{{10}^{-4}{\text{ m}}^2}{1{\text{ cm}}^2}\right)/\text{s}\right)}{{\left(\text{50 cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2\left(1.2\text{ kg}\right)}\\ =0.498\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the angular speed of the sign immediate after the collision is $0.498\text{rad}/\text{s}$.

To determine

The maximum angle.

Answer to Problem 11.38P

The maximum angle is $5.58°$.

Explanation of Solution

The formula distance of the centre of mass from the axis of rotation is,

    $h_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

Substitute $2.40\text{ kg}$ for $m_1$, $400\text{ g}$ for $m_2$, $\frac{\text{50 cm}}{2}$ for $y_1$ and $\text{50 cm}$ for $y_2$ in above equation.

    $\begin{array}{c}{h}_{\text{CM}}=\frac{\left(2.40\text{ kg}\right)\left(\left(\frac{\text{50 cm}}{2}\right)\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)+\left(400\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\right)\left(\text{50 cm}\right)\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}{\left(2.40\text{ kg}\right)+\left(400\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\right)}\\ =\frac{\left(2.40\text{ kg}\right)\left(0.250\text{ m}\right)+\left(0.400\text{ kg}\right)\left(\text{0}\text{.500 m}\right)}{\left(2.40\text{ kg}\right)+\left(0.400\text{ kg}\right)}\\ =0.28\text{ m}\end{array}$

The conservation of mechanical energy is,

    $\begin{array}{c}\left(M+m\right)g{h}_{\text{CM}}\left(1-\cos \theta \right)=\frac{1}{2}\left(\frac{M{L}^2}{3}+m{L}^2\right){\omega }_{\text{f}}^2\\ \left(1-\cos \theta \right)=\frac{\frac{1}{2}\left(\frac{M{L}^2}{3}+m{L}^2\right){\omega }_{\text{f}}^2}{\left(M+m\right)g{h}_{\text{CM}}}\end{array}$

Rearrange the above expression for $\cos \theta$.

    $\cos \theta =\left[1-\frac{\left(\frac{M}{3}+m\right)L^2{\omega }_{\text{f}}^2}{2\left(M+m\right)g{h}_{\text{CM}}}\right]$

Substitute $2.40\text{ kg}$ for $M$, $400\text{ g}$ for $m$, $0.498\text{rad}/\text{s}$ for ${\omega }_{\text{f}}$, $\text{50 cm}$ for $L$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $0.28\text{ m}$ for $h_{\text{CM}}$ in above equation.

    $\begin{array}{c}\cos \theta =\left[1-\frac{\left(\frac{\left(2.40\text{ kg}\right)}{3}+\left(400\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\right)\right){\left(\text{50 cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2{\left(0.498\text{rad}/\text{s}\right)}^2}{2\left(\left(2.40\text{ kg}\right)+\left(400\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\right)\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(0.28\text{ m}\right)}\right]\\ \theta ={\cos }^{-1}\left[1-\frac{\left(0.80\text{ kg}+\left(0.400\text{ kg}\right)\right){\left(\text{0}\text{.500 m}\right)}^2{\left(0.498\text{rad}/\text{s}\right)}^2}{2\left(\left(2.40\text{ kg}\right)+\left(0.400\text{ kg}\right)\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(0.28\text{ m}\right)}\right]\\ =5.58°\end{array}$

Conclusion:

Therefore, the maximum angle is $5.58°$.