Question

A thin, uniform, rectangular sign 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 75.0 cm. The sign is swinging without friction, becoming 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 540 g, traveling horizontally with a velocity of 160 cm/s to the right, strikes perpendicularly the lower edge of the sign and sticks there.

(a) Calculate the angular speed of the sign immediately before the impact.

1.916 rad/s

(b) Calculate its angular speed immediately after the impact.

? rad/s

(c) The spattered sign will swing up through what maximum angle?

?°

Answer 1

Concepts and reason

The concepts required to solve this problem are conservation of energy, conservation of angular momentum, and center of mass.

Draw the free body diagram of the sign and use conservation of energy principle to determine the angular velocity of the sign just before the collision. Next, determine the initial and final angular momentum of the sign and the snowball. Use conservation of momentum for the snowball and the sign to determine the angular velocity of the sign just after the collision. Finally, determine the distance of the center of mass from the axis of rotation. Apply conservation of energy principle to determine the maximum angular displacement of the spattered sign.

Fundamentals

Kinetic energy $\left( T \right)$ possessed by a mass $\left( m \right)$ is expressed as,

$T = \frac{1}{2}m{v^2}$

Here, $v$ is the velocity of the mass.

$T$ defined above is used for a translational motion.

Potential energy $\left( U \right)$ possessed by an elevated mass is expressed as,

$U = mgh$

Here, $g$ is the acceleration due to gravity, $h$ is the elevation (from a reference level), and $m$ is the mass.

Energy conservation

If there is no net external force acting on a system of particles, the total energy (that is, kinetic energy $\left( T \right)$ and potential energy $\left( U \right)$ ) stays the same.

${T_i} + {U_i} = {T_f} + {U_f}$

Here, the kinetic energy term contains linear energy $\left( {1/2} \right)m{v^2}$ as well as rotational energy $\left( {1/2} \right)I{\omega ^2}$ term, with moment of inertia calculated about the center of mass.

Line of zero potential (that is, datum) can be chosen as per convenience.

Free body diagram (FBD):

Free-body diagram is a diagram used to represent the magnitude and the direction of all the forces that act upon a body at given loading conditions.

Mass moment of inertia $\left( I \right)$ is expressed as,

$I = {r^2}m$

Here, $m$ is the mass of the particle, and $r$ is the shortest distance between the particle and the axis of rotation.

For an extended body, the above expression is written in terms of mass element $dm$ as,

$I = \int {{r^2}dm}$

The parallel axis theorem states that the moment of inertia ${I_A}$ of a body of mass $M$ about an axis $A$ parallel to its centroid axis is expressed as.

${I_A} = I + M{r^2}$

Here, $I$ is the moment of inertia of the object about its centroid and $r$ is the shortest distance between the centroid axis and $A$ .

Momentum conservation

If there is no net external force acting on a system of particles in a given direction, the net momentum stays the same in that direction, thus,

$\sum\limits_{}^{} {{{\left( {{m_i}{{\vec v}_i}} \right)}_0}} = \sum\limits_{}^{} {{{\left( {{m_i}{{\vec v}_i}} \right)}_1}}$

Here, the 0 and 1 subscripts refer to the initial and final states of the system, respectively, and the velocities are taken in the direction of zero net external force.

Analogous to linear momentum conservation, the angular momentum is also conserved if the net external torque is zero about a given axis,

$\sum\limits_{}^{} {{{\left( {{I_i}{\omega _i}} \right)}_1}} = \sum\limits_{}^{} {{{\left( {{I_i}{\omega _i}} \right)}_2}}$

Here, the 0 and 1 subscripts refer to the initial and final states of the system, respectively, and the rotation is considered in the direction of zero net external torque.

Write the expression of angular moment $\left( H \right)$ of the particle rotating about a point separated by a distance r from the center of rotation.

$H = mvr$

Here, $m$ is the mass of the particle, $v$ is the linear velocity at the circumference of the path followed by the particle, and $r$ is the radius of rotation.

Center of mass $\left( {{C_{CM}}} \right)$ of the system of two particles is expressed as,

${C_{CM}} = \frac{{{m_1}{c_1} + {m_2}{c_2}}}{{{m_1} + {m_2}}}$

Here, ${m_1}{\rm{ and }}{m_2}$ denote the mass of the particle 1 and 2, respectively and ${c_1}{\rm{ and }}{c_2}$ denote the separation distance from the axis of rotation of the particle 1 and 2, respectively.

(a)

Draw the free body diagram of the sign as shown below.

In the above diagram $\omega$ is the angular velocity of the sign when it is in vertical position, m is the mass of the sign, M is the mass of the snowball, $W$ is the weight of the sign equivalent to $mg$ that acts at the center of mass of the sign (that is, at the center of the rectangular sign), $v$ is the linear velocity of the snowball, and $h$ is the elevation of the center of mass of the sign.

Consider the lowest point of the trajectory traced by the sign as the reference level or datum (that is, when the sign is in vertical position).

The initial velocity of the sign is zero. Therefore, the initial kinetic energy is zero, thus,

${T_i} = 0$

Initial potential energy is expressed as,

${U_i} = mgh$

In the above equation $h$ is calculated as,

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

Thus, substitute $\frac{L}{2}\left( {1 - \cos \theta } \right)$ for $h$ and rewrite the equation of initial potential energy ${U_i}$ .

${U_i} = \frac{{mgL}}{2}\left( {1 - \cos \theta } \right)$

The final potential energy of the sign (at the instant when it reaches the lowest point on the trajectory) is zero, since the sign is at the datum, thus,

${U_f} = 0$

The kinetic energy possessed by the sign just before it gets hit by the snowball is rotational kinetic energy.

Thus, final kinetic energy of the sign is expressed as,

${T_f} = \frac{1}{2}{I_O}{\omega ^2}$

Here, ${I_O}$ is the mass moment of inertia of the sign about O.

The given rectangular sign swings as slender rod about the hinge point at O.

Mass moment of inertia of the slender about an axis through its center of mass is expressed as,

${I_{CM}} = \frac{{m{L^2}}}{{12}}$

Here, $m$ is the mass of the sign and $L$ is the side length of the sign.

Calculate the Mass moment of inertia of the sign about the hinge (that is, ${I_O}$ ) as,

${I_O} = {I_{CM}} + m{\left( {\frac{L}{2}} \right)^2}$

Thus,

$\begin{array}{c}\\{I_O} = \frac{{m{L^2}}}{{12}} + m{\left( {\frac{L}{2}} \right)^2}\\\\ = \frac{{m{L^2}}}{3}\\\end{array}$

Apply conservation of energy principle.

${T_i} + {U_i} = {T_f} + {U_f}$

Substitute $0$ for ${T_i}$ , $0$ for ${U_f}$ , $\frac{{mgL}}{2}\left( {1 - \cos \theta } \right)$ for ${U_i}$ , and $\frac{1}{2}{I_O}{\omega ^2}$ for ${T_f}$ .

$0 + \frac{{MgL}}{2}\left( {1 - \cos \theta } \right) = \frac{1}{2}{I_O}{\omega ^2} + 0$

Substitute $\frac{{m{L^2}}}{3}$ for ${I_O}$ and solve for $\omega$ .

$\begin{array}{l}\\\frac{{mgL}}{2}\left( {1 - \cos \theta } \right) = \frac{1}{2}\left( {\frac{{m{L^2}}}{3}} \right){\omega ^2}\\\\\frac{{3mgL}}{{m{L^2}}}\left( {1 - \cos \theta } \right) = {\omega ^2}\\\\\omega = \sqrt {\frac{{3g}}{L}\left( {1 - \cos \theta } \right)} \\\end{array}$

Substitute $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ , $25^\circ$ for $\theta$ , and $75{\rm{ cm}}$ for $L$ .

$\begin{array}{c}\\\omega = \sqrt {\frac{{3\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{75{\rm{ cm}}\left( {1{\rm{ m}}/{{10}^2}{\rm{ cm}}} \right)}}\left( {1 - \cos \left( {25^\circ } \right)} \right)} \\\\ = 1.917{\rm{ rad/s}}\\\end{array}$

(b)

Consider the velocity in the right direction as negative and the velocity in the left direction as positive.

Use the sign convention given above and calculate the initial angular momentum $\left( {{H_{i,{\rm{ snowball}}}}} \right)$ of the snowball.

${H_{i,{\rm{ snowball}}}} = - mvL$

Calculate the initial angular momentum $\left( {{H_{i,{\rm{ sign}}}}} \right)$ of the sign.

${H_{i,{\rm{ sign}}}} = {I_O}{\omega _i}$

Here, ${\omega _i}$ is the initial angular velocity of the sign.

Calculate the final angular momentum $\left( {{H_{f,{\rm{system}}}}} \right)$ of the system of combined mass snowball and the sign.

${H_{f,{\rm{system}}}} = {I_{O,system}}{\omega _f}$

Here, ${\omega _f}$ is the final angular velocity of the sign and ${I_{O,system}}$ is the mass moment of inertia of the system of combined snowball and the sign.

Calculate moment of inertia of the system, that is, ${I_{O,system}}$ as,

${I_{O,system}} = \frac{{m{L^2}}}{3} + M{L^2}$

Here, $m{L^2}/3$ gives the moment of inertia of the sign about the hinge, that is, point O and $M{L^2}$ gives the moment of inertia of the snowball about the hinge.

Apply conservation of angular momentum principle.

${H_{i,{\rm{snowball}}}} + {H_{i,s{\rm{ign}}}} = {H_{f,{\rm{snowball}}}} + {H_{f,{\rm{sign}}}}$

Or,

${H_{i,{\rm{snowball}}}} + {H_{i,s{\rm{ign}}}} = {H_{f,{\rm{system}}}}$

Substitute ${I_{O,system}}{\omega _f}$ for ${H_{f,{\rm{sign}}}}$ , $- mvL$ for ${H_{i,{\rm{ snowball}}}}$ , and ${I_O}{\omega _i}$ for ${H_{i,{\rm{sign}}}}$ .

$- mvL + {I_O}{\omega _i} = {I_{O,system}}{\omega _f}$

Substitute $\frac{{m{L^2}}}{3} + M{L^2}$ for ${I_{O,system}}$ and $\frac{{m{L^2}}}{3}$ for ${I_O}$ .

$- MvL + \left( {\frac{{m{L^2}}}{3}} \right){\omega _i} = {\omega _f}\left( {\frac{{m{L^2}}}{3} + M{L^2}} \right)$

Substitute $1.917{\rm{ rad/s}}$ for ${\omega _i}$ , $2.40{\rm{ kg}}$ for $m$ , $540{\rm{ g}}$ for $M$ , $75{\rm{ cm}}$ for $L$ and solve for ${\omega _f}$ .

$\begin{array}{l}\\ - \left( {0.540{\rm{ kg}}} \right)\left( {160{\rm{ }}\frac{{{\rm{cm}}}}{{\rm{s}}}\left( {\frac{{1{\rm{ m}}}}{{{{10}^2}{\rm{ cm}}}}} \right)} \right)\left( {0.75{\rm{ m}}} \right) + \left( {\frac{{\left( {2.40{\rm{ kg}}} \right){{\left( {75{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{{{10}^2}{\rm{ cm}}}}} \right)} \right)}^2}}}{3}} \right)\left( {1.917{\rm{ rad/s}}} \right)\\\\ = {\omega _f}\left( {\frac{{2.40{\rm{ kg}}{{\left( {0.75{\rm{ m}}} \right)}^2}}}{3} + \left( {540{\rm{ g}}} \right)\left( {\frac{{1{\rm{ kg}}}}{{{{10}^3}{\rm{ g}}}}} \right){{\left( {75{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{{{10}^2}{\rm{ cm}}}}} \right)} \right)}^2}} \right)\\\end{array}$

Or,

$\begin{array}{c}\\{\omega _f} = \frac{{0.215{\rm{ kg}} \cdot {{\rm{m}}^2}{\rm{/s}}}}{{0.754{\rm{ kg}} \cdot {{\rm{m}}^2}}}\\\\ = 0.285{\rm{ rad/s}}\\\end{array}$

(c)

Consider, the axis of rotation (at the hinge) is at the origin and the distance of the center of mass of the combined system of snowball and the sign is at a distance ${h_{CM}}$ from the axis of rotation.

Calculate the distance ${h_{CM}}$ of the center of mass of system.

${h_{CM}} = \frac{{m{c_1} + M{c_2}}}{{m + M}}$

The distance ${c_1}$ is equivalent to $L/2$ and distance ${c_2}$ is equivalent to $L$ .

Substitute $2.40{\rm{ kg}}$ for $m$ , $0.540{\rm{ kg}}$ for $M$ , and $0.75{\rm{ m}}$ for $L$ .

$\begin{array}{c}\\{h_{CM}} = \frac{{2.40{\rm{ kg}}\left( {0.75{\rm{ m}}/2} \right) + 0.540{\rm{ kg}}\left( {0.75{\rm{ m}}} \right)}}{{2.40{\rm{ kg}} + 0.540{\rm{ kg}}}}\\\\ = 0.444{\rm{ m}}\\\end{array}$
‎

The initial potential energy of the system (at the instant when the sign is at the lowest point on the trajectory) is zero, since the sign is at the datum, thus,

${U_i} = 0$

The initial angular velocity of the system is ${\omega _f}$ . Therefore, the initial kinetic energy is calculated as, thus,

${T_i} = \frac{1}{2}{I_{O,system}}\omega _f^2$

Or,

${T_i} = \frac{1}{2}\left( {\frac{{m{L^2}}}{3} + M{L^2}} \right)\omega _f^2$

Final potential energy of the system is expressed as,

${U_f} = \left( {m + M} \right)g{h_{CM}}\left( {1 - \cos \theta } \right)$

In the above equation ${h_{CM}}\left( {1 - \cos \theta } \right)$ denotes the vertical projection of ${h_{CM}}$ .

The final kinetic energy of the system (at the instant when it reaches the highest point on the trajectory traced by the spattered sign) is zero, since the system momentarily comes to rest at highest point, thus,

${T_f} = 0$

Apply conservation of energy principle.

${T_i} + {U_i} = {T_f} + {U_f}$

Substitute $0$ for ${T_f}$ , $0$ for ${U_i}$ , $\left( {m + M} \right)g{h_{CM}}\left( {1 - \cos \theta } \right)$ for ${U_f}$ , and $\frac{1}{2}\left( {\frac{{m{L^2}}}{3} + M{L^2}} \right)\omega _f^2$ for ${T_i}$ .

$\frac{1}{2}\left( {\frac{{m{L^2}}}{3} + M{L^2}} \right)\omega _f^2 = \left( {m + M} \right)g{h_{CM}}\left( {1 - \cos \theta } \right)$

Substitute $0.444{\rm{ m}}$ for ${h_{CM}}$ , $0.285{\rm{ rad/s}}$ for ${\omega _f}$ , $2.40{\rm{ kg}}$ for $m$ , $0.540{\rm{ kg}}$ for $M$ , $0.75{\rm{ m}}$ for $L$ , $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ and solve for $\theta$ .

$\begin{array}{l}\\\frac{1}{2}\left( {\frac{{2.40{\rm{ kg}}{{\left( {0.75{\rm{ m}}} \right)}^2}}}{3} + 0.540{\rm{ kg}}{{\left( {0.75{\rm{ m}}} \right)}^2}} \right){\left( {0.285{\rm{ rad/s}}} \right)^2}\\\\ = \left( {2.40{\rm{ kg}} + 0.540{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {0.444{\rm{ m}}} \right)\left( {1 - \cos \theta } \right)\\\end{array}$

Or,

$\begin{array}{l}\\0.031{\rm{ kg}} \cdot \frac{{{{\rm{m}}^2}}}{{{{\rm{s}}^2}}} = 12.806{\rm{ kg}} \cdot \frac{{{{\rm{m}}^2}}}{{{{\rm{s}}^2}}} - \left( {12.806{\rm{ kg}} \cdot \frac{{{{\rm{m}}^2}}}{{{{\rm{s}}^2}}}\left( {\cos \theta } \right)} \right)\\\\\cos \theta = \frac{{12.806{\rm{ kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2} - 0.031{\rm{ kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2}}}{{12.806{\rm{ kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2}}}\\\\\theta = {\cos ^{ - 1}}\left( {0.99758} \right)\\\\\theta = 3.98^\circ \\\end{array}$

Ans: Part a

Angular velocity of the sign just before the collision is $1.917{\rm{ rad/s}}$ .