Chapter 11, Problem 33AP

(a)

To determine

The angular speed of the sign immediate before the collision.

Answer to Problem 33AP

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

Explanation of Solution

Given information:

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,

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

• $L$ is the vertical length of the sign.
• $\theta$ 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)

• $M$ is the mass of sign.
• $g$ is the acceleration due to gravity.
• $I$ is the moment of inertia.
• ${\omega }_{\text{i}}$ 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 $2.35\text{rad}/\text{s}$.

(b)

To determine

The angular speed of the sign immediate after the collision.

Answer to Problem 33AP

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

Explanation of Solution

Given information:

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 $160\text{cm}/\text{s}$.

The formula for the conservation of the angular momentum is,

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

• $m$ is the mass of snowball.
• $v$ is the velocity of snowball.
• ${\omega }_{\text{f}}$ 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 $0.498\text{rad}/\text{s}$.

(c)

To determine

The maximum angle.

Answer to Problem 33AP

maximum angle is $5.58°$.

Explanation of Solution

Section 1:

To determine: The distance of the centre of mass from the axis of rotation.

Answer: The distance of the centre of mass from the axis of rotation is $0.28\text{ m}$.

Given information:

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 $160\text{cm}/\text{s}$.

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.

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

Section 2:

To determine: The maximum angle.

Answer: The maximum angle is $5.58°$.

Given information:

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 $160\text{cm}/\text{s}$.

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 $5.58°$.