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6.3: Conservation of Angular Momentum Equation

  • Page ID
    81501
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    The recommended starting point for the application of the conservation of angular momentum principle is the rate form of the equation: \[\frac{d \mathbf{L}_{O, \text { sys }}}{d t} = \sum_{j} \mathbf{M}_{O, \ j} + \sum_{in} \left(\mathbf{r}_{i} \times \mathbf{V}_{i}\right) \dot{m}_{i} - \sum_{out} \left(\mathbf{r}_{e} \times \mathbf{V}_{e}\right) \dot{m}_{e} \nonumber \] where \(\mathbf{L}_{O, \text { sys}}\) is the system angular momentum about point \(O\), \(\mathbf{M}_{O}\) is the transport rate of angular momentum about point \(O\) with moments, and \(\dot{m}(\mathbf{r} \times \mathbf{V})\) is the transport rate of angular momentum about point \(O\) with mass flow across the boundary.

    In applying the rate form of the conservation of angular momentum equation, there are many modeling assumptions that will be used to simplify the basic equation for specific systems. As always, you should focus on understanding the physical meaning of the assumptions and how they simplify the equations for a given system. Do not just memorize the results.

    Steady-state system: If a system is operating under steady-state conditions, all intensive properties and interactions are independent of time. When this assumption is applied to the conservation of angular momentum equation, we have

    \[ \begin{align*} \underbrace{ \cancel{ \frac{d \mathbf{L}_{O, \text{ sys}}}{dt} }^{=0} }_{\begin{array}{c} \mathbf{L}_{O, \text{ sys}} \text{ is} \\ \text{constant} \end{array} } &= \sum_{j} \mathbf{M}_{O, \ j} + \sum_{in} \left( \mathbf{r}_i \times \mathbf{V}_i \right) \dot{m}_i - \sum_{out} \left( \mathbf{r}_e \times \mathbf{V}_e \right) \dot{m}_e \\ 0 &= \sum_{j} \mathbf{M}_{O, \ j} + \sum_{in} \left( \mathbf{r}_i \times \mathbf{V}_i \right) \dot{m}_i - \sum_{out} \left( \mathbf{r}_e \times \mathbf{V}_e \right) \dot{m}_e \end{align*} \nonumber \]

    Closed system: A closed system has no flow at the boundary so the conservation of angular momentum equation reduces to the following: \[ \begin{align*} \frac{d \mathbf{L}_{O, \text{ sys}}}{dt} &= \sum_{j} \mathbf{M}_{O, \ j} + \sum_{in} \left( \mathbf{r}_i \times \mathbf{V}_i \right) \cancel{ \dot{m}_i }^{=0} - \sum_{out} \left( \mathbf{r}_e \times \mathbf{V}_e \right) \cancel{ \dot{m}_e }^{=0} \\ \frac{d \mathbf{L}_{O, \text{ sys}}}{dt} &= \sum_{j} \mathbf{M}_{O, \ j} \end{align*} \nonumber \]

    To make any further simplications, it is necessary to have additional information or make assumptions about the kinematics of the moving system. If we can assume that a plane system is translating then \[\left. \begin{array}{c} \mathbf{L}_{O, \text{ sys}} = m_{\text{sys}} \left( \mathbf{r}_{G} \times \mathbf{V}_{G}\right) \\ \frac{d \mathbf{L}_{O, \text{ sys}}}{d t} = \sum_{j} \mathbf{M}_{O, \ j} \end{array} \right\} \quad \rightarrow \quad m_{\text{sys}} \frac{d \left(\mathbf{r}_{G} \times \mathbf{V}_{G} \right)}{d t} = \sum_{j} \mathbf{M}_{O, \ j} \nonumber \]

    If we can assume that a plane, rigid system is rotating about an axis (the \(z\)-axis) that passes through the system center of mass then \[\left. \begin{array}{c} \mathbf{L}_{O, \text{ sys}} = \omega_{z} I_{G} \mathbf{k} \\ \frac{d \mathbf{L}_{O, \text{ sys}}}{d t} = \sum_{j} \mathbf{M}_{O, \ j} \end{array}\right\} \quad \rightarrow \quad \frac{d\left(\omega_{z} I_{G} \mathbf{k}\right)}{d t}=\sum_{j} \mathbf{M}_{O, \ j} \nonumber \] where \(\mathbf{k}\) is the unit vector the points in the positive direction of the \(z\)-axis.

    Modeling Reactions at Supports and Connections:

    One of the significant problems in selecting a system for momentum transfer is the identification of the interactions between the system and its surroundings. What exactly are the constraints on how momentum is being transmitted across the boundary at a specific interactions? Contact forces on the boundary of a system are sometimes referred to as reactions, especially at places where the boundary cuts a support or connection. According to Beer and Johnston\(^{1}\), reactions can be classified in three broad categories:

    (a) Force with a known line of action. Many connections satisfy this condition among these are rollers, rockers, frictionless surfaces, short links and cables, collars on frictionless rods, and frictionless pins in slots. Each of these reactions is characterized by the ability to only transport linear momentum by a force in a single, clearly identifiable direction. (See Figure \(\PageIndex{1}\)).

    (b) Force of unknown direction and magnitude. Hardware or connections that provide this type of reaction include frictionless pins in fitted holes, frictionless ball-and-socket joints, hinges, and rough surfaces (Although frictionless surfaces only transmit a force normal to the surface, friction introduces shear forces at the surface.)

    (c) Force of unknown direction and magnitude and a couple. This last type of reaction is usually referred to as a fixed support. The reaction of the ground on a telephone pole stuck vertically in the ground is an excellent example. When you push of the telephone pole in any direction, the ground exerts three forces on the pole—a horizontal shear force, a vertical normal force, and a distributed force that is a force couple. The couple, as shown earlier, only transmits angular momentum.

    Rollers, rockers, and frictionless surfaces experience a single reaction force normal to the surface which that support is in contact with. A short cable or link experiences a single support reaction along the line of action of the cable or link. A frictionless slider on a bar experiences a single reaction force normal to the bar. A frictionless pin and a rough surface can experience support forces in any direction. A fixed support can experience reaction forces and a reaction couple/moment in any direction.

    Figure \(\PageIndex{1}\): Surface force reactions.

    As one might expect, these reactions can be seen in everyday life if one is observant. Look around you right now and see if you can find a physical connection that falls in each of these categories. For example, look at the caster or wheel on your desk chair or the hinge on the door. Next time you drive under an overpass on the highway, look at how the bridge roadbed is supported on the columns. You very likely will see a roller.

    \(^{1}\) F. P. Beer and E. R. Johnston, Jr., Vector Mechanics for Engineers: Statics, 6th ed., McGraw-Hill, New York, 1996.

    Example — What a reaction!

    Determine the reactions at \(A\) and \(C\). Assume the weight of the mechanism is negligible and that all joints are frictionless.

    An L-shaped body has a long horizontal arm of length L=800 mm, whose left endpoint A has a roller support resting on a 30-degree incline. The shorter, 400-mm arm of the L extends upwards from the right end of the horizontal arm. The upper endpoint of this arm, C, is pinned to a support. Two rightward horizontal forces F, each of magnitude 300 N, are applied to the vertical arm, one at its base and one halfway up its height.

    Figure \(\PageIndex{2}\): An L-shaped bar is supported at two ends and has point loads applied at two points.

    Solution

    Known: A specific device with given loads.

    Find: The reactions (magnitude and direction) at points \(A\) and \(C\).

    Given: See figure drawn above. (Notice how all physical quantities have been given symbols.)

    Analysis:

    Strategy \(\rightarrow\) Since this problem concerns forces, at a minimum it will require linear momentum and possibly angular momentum. (The last part is a giveaway since this problem is in the angular momentum chapter. Unfortunately, problems in real life never come with the chapter number.)

    System \(\rightarrow\) Closed, non-deforming system including the pinned joint \(C\), the roller at \(A\) and the connecting bar.
    Property to count \(\rightarrow\) Linear and angular momentum
    Time interval \(\rightarrow\) Try rate form. (Momentum is flowing through this system.)

    The system drawn below shows the reactions at \(A\) and \(C\) and all pertinent dimensions on the figure. Assuming that this is a closed and steady-state system, the linear and angular momentum equations become the following: \[0=\sum_{j} \mathbf{F}_{j} \quad \text { and } \quad 0=\sum_{j} \mathbf{M}_{0, \ j} \nonumber \] (Could you explicitly show how these two equations were obtained given the assumptions above?)

    Free-body diagram of the L-shaped bar. Reaction force R_A at point A points upwards and to the right, perpendicular to the incline. At point C, reaction forces R_Cx and R_Cy are point towards the right and upwards directions, respectively. Positive x-direction is to the right, positive y-direction is upwards, and positive direction for angular momentum is counterclockwise.

    Figure \(\PageIndex{3}\): Free-body diagram.

    Before writing the component equations in scalar form we must clearly indicate a coordinate system and what we are assuming as positive directions. This is done with the small coordinate system in the figure. Note that we are assuming counterclockwise (CCW) as the positive direction for angular momentum.

    Now looking at the \(x\)-momentum:

    \[ \boxed{ \rightarrow + } \quad\quad 0 = R_{A}(\sin \theta) + 2 F + R_{C x} \quad \rightarrow \quad -2 F = R_{A}(\sin \theta)+R_{C x} \quad \rightarrow \quad -(600 \mathrm{~N}) = R_{A}\left(\sin 30^{\circ}\right)+R_{C x} \nonumber \]

    For the \(y\)-momentum:

    \[ \boxed{ \uparrow +} \quad\quad 0 = R_A (\cos \theta) + R_{C y} \quad \rightarrow \quad R_{C y} = - R_A \left( \cos 30 ^{\circ} \right) \nonumber \]

    This gives two equations for three unknowns — \(R_{A}\), \(R_{C x}\), and \(R_{C y}\). To get the remaining equation we must apply conservation of angular momentum.

    Applying conservation of angular momentum about Point \(A\) we eliminate any moment due to one of the applied forces \(F\) and the reaction \(R_{A}\) : \[ \boxed{ \text {CCW }+ } \quad\quad 0=\left[ +L R_{C y} \right] + \left[-2 h R_{C x}\right] + [-h F] \quad \rightarrow \quad F=\left(\frac{L}{h}\right) R_{C y}-2 R_{C x} \quad \rightarrow \quad (300 \mathrm{~N}) = (4) R_{C y} - 2 R_{C x} \nonumber \]

    Solving these three equations simultaneously, we get the following: \[\left. \begin{array}{rlrl} -(600 \mathrm{~N}) & =R_{A} \sin 30^{\circ}-2 R_{C x} \\ R_{C y} &= -R_{A} \cos 30^{\circ} \\ (300 \mathrm{~N}) &= R_{A} \sin 30^{\circ}+R_{C x} \end{array} \right\} \quad \rightarrow \quad \begin{aligned} R_{A} &= 365.2 \mathrm{~N} \\ R_{C x} &= -782.6 \mathrm{~N} \\ R_{C y} &= -316.3 \mathrm{~N} \end{aligned} \nonumber \]

    Notice that the reaction at \(C\) occurs in a direction opposite to that assumed on the drawing.

    Example — Going to the wall for a reaction

    A water jet steadily impacts a blade that is attached to a cantilever beam as shown in the figure. The cantilever beam is fixed to the wall at \(A\) and holds the blade stationary. The water hits the blade at \(B\) and leaves the blade at \(C\).

    The mass of the blade/beam combination is \(100 \ \text{lbm}\) with a known center of mass at point \(G\) located at a distance \(L/2\) from the wall. The water has a mass flow rate of \(150 \text{lbm} / \text{s}\), and the speed of the water is the same entering and leaving the blade, \(V_1 = V_2 = 20 \ \text{ft} / \text{s}\). The water leaves the blade at an angle of \(30^{\circ}\) below the horizontal.

    Other information is as follows: \(L=5 \ \mathrm{ft}\), \(d=7 \ \mathrm{ft}\), and \(h=2 \ \mathrm{ft}\).

    Determine the reactions on the cantilever beam at the wall \((A)\).

    A fixed horizontal beam, with left endpoint A, extends out from a vertical wall at the left of the figure. Point G is located within this section of beam. A blade curves downwards a distance h and to the right from the right end of the beam. A vertical stream of water falls at velocity V_1 onto the blade at point B, a distance L from the wall. Water runs off the right end of the blade at point C, a distance d from the wall, at velocity V_2 and angle theta below the horizontal.

    Figure \(\PageIndex{4}\): Water falls onto a curved blade attached to a cantilever beam and runs off the blade's lower end.

    Solution

    Known: Water is deflected by a blade supported by a cantilever beam.

    Find: The reactions on the cantilever beam at the wall.

    Given: See the figure above. Again, note how all physical quantities have symbols.

    Analysis:

    Strategy \(\rightarrow\) Since we are asked for reactions (forces and moments) we should try using conservation of linear and angular momentum.

    System \(\rightarrow\) Open, non-deforming system that includes the blade-beam-water with water flowing in and out as shown below.
    Property to count \(\rightarrow\) Angular and linear momentum (and possibly mass).
    Time interval \(\rightarrow\) Probably an infinitesimal time interval, i.e. the rate form.

    Free-body diagram of the system consisting of the beam, blade, and water in contact with the blade. Wall is replaced with two perpendicular reaction forces R_x and R_y, plus a counterclockwise moment M_A. Weight W acts downwards at point G, the product of mass flow rate 1 and velocity 1 acts downwards at point B, and the product of mass flow rate 2 and velocity 2 acts at point C at an angle theta below the horizontal.

    Figure \(\PageIndex{5}\): Free-body diagram of the beam, blade, and water system.

    The figure above shows the free-body (or system interaction) diagram. Notice that the connection at the wall is replaced by two forces and a moment because the wall-beam connection is a fixed support. Also note that the mass transports of linear momentum are shown on the figure along with the weight acting at point \(G\).

    As a starting point, let's write the conservation of mass and see what it tells us: \[\underbrace{ \cancel{ \frac{d m_{\text{sys}}}{d t} }^{=0} }_{\text {steady state}} = \dot{m}_{1} - \dot{m}_{2} \quad \rightarrow \quad \dot{m}_{1} = \dot{m}_{2}=\dot{m} \nonumber \] This may seem like a trivial result, but if you didn't see it immediately it isn't trivial. Furthermore, this just confirms your intuition.

    Now let's write the conservation of linear momentum: \[ \cancel{ \frac{d \mathbf{P}_{\text{sys}}}{d t} }^{=0} = \mathbf{Rx} + \mathbf{Ry} + \mathbf{W} \ + \ \dot{m}_{1} \mathbf{V}_{1} - \dot{m} \mathbf{V}_{2} \quad \rightarrow \quad 0=\mathbf{Rx} + \mathbf{Ry} + \mathbf{W} \ + \ \dot{m}_{1} \mathbf{V}_{1} - \dot{m} \mathbf{V}_{2} \nonumber \]

    Now writing the scalar form of this equation, assume positive \(x\) is to the right and positive \(y\) is up.

    \[\begin{aligned} \boxed{\rightarrow +} \quad\quad 0 &= Rx - \dot{m}\left( V_{2} \cos \theta \right) \\ Rx &= \dot{m}\left( V_{2} \cos \theta \right) \quad \rightarrow \quad Rx = \left( 150 \ \dfrac{\mathrm{lbm}}{\mathrm{s}}\right) \left(20 \ \frac{\mathrm{ft}}{\mathrm{s}} \right) \left( \cos 30^{\circ} \right) \left(\frac{1 \ \mathrm{lbf}}{32.174 \ \dfrac{\mathrm{lbm} \cdot \mathrm{ft}}{\mathrm{s}^{2}}} \right) = 80.8 \ \mathrm{lbf} \\ \\ \boxed{\uparrow +} \quad\quad 0 &= Ry - W + \dot{m}\left(-V_{1}\right) - \dot{m}\left(-V_{2} \sin 30^{\circ}\right) \\[4pt] Ry &= W + \dot{m}\left(V_{1}-V_{2} \sin 30^{\circ}\right) \\[4pt] &= W + \dot{m} V \left(1-\sin 30^{\circ}\right) \\[4pt] Ry &= \left[(100 \ \mathrm{lbm}) \left(\frac{1 \ \mathrm{slug}}{32.174 \ \mathrm{lbm}}\right) \left(32.174 \ \dfrac{\mathrm{ft}}{\mathrm{s}^{2}}\right) \right] \\[4pt] &=[100+46.6] \ \mathrm{lbf} \\[4pt] &=146.6 \ \mathrm{lbf} \end{aligned} \nonumber \]

    Be very careful about the signs on the terms, especially the mass flow rate terms. Note that the \(+ \ / \ -\) in front of the mass flow rates depend upon whether the flow is into or out of the system; the signs inside of the parenthesis on the velocity (specific linear momentum) terms are the result of translating the velocity vector to this specific coordinate system.

    With one remaining unknown \(M_A\), we need to apply conservation of angular momentum. Although we can sum moments about any point, let's use point \(C\). This has the benefit of eliminating the angular momentum carried out in the water jet at \(C\).

    \[\begin{aligned} \frac{d \mathbf{L}_{C, \text{ sys}}}{d t} &= M_{A} + \left(\mathbf{r}_{A} \times \mathbf{Rx} \right) + \left( \mathbf{r}_{A} \times \mathbf{R y}\right) + \left(\mathbf{r}_{G} \times \mathbf{W}\right) + \dot{m} \left(\mathbf{r}_{B} \times \mathbf{V}_{1}\right) \\ \boxed{\mathrm{CW} +} \quad\quad 0 &= \left(-M_{A}\right) + (+h \cdot Rx) + (d \cdot Ry) + \left[ -\left(d-\dfrac{L}{2}\right) W \right] + \dot{m} \left[-(d-L) V_{1}\right] \\ M_{A} &= (h \cdot Rx) + (d \cdot Ry) + \left( -\left(d-\frac{L}{2}\right) W \right) + \dot{m}\left[-(d-L) V_{1}\right] \\ M_{A} &= [(2 \ \mathrm{ft})(80.8 \ \mathrm{lbf})] + [(7 \ \mathrm{ft})(146.6 \ \mathrm{lbf})] + \left[ -\left(7-\frac{5}{2}\right)(\mathrm{ft})(100 \ \mathrm{lbf}) \right] + \left(150 \ \frac{\mathrm{lbm}}{\mathrm{s}}\right) \left[-(2 \ \mathrm{ft})\left(20 \ \frac{\mathrm{ft}}{\mathrm{s}}\right) \right] \\ &=(161.6 \ \mathrm{ft} \cdot \mathrm{lbf}) + (1026.2 \ \mathrm{ft} \cdot \mathrm{lbf}) + (-450 \ \mathrm{ft} \cdot \mathrm{lbf}) + \underbrace{ \left(-6000 \ \frac{\mathrm{lbm} \cdot \mathrm{ft}}{\mathrm{s}^2} \right) \left( \frac{1 \ \mathrm{lbf}}{32.174 \ \dfrac{\mathrm{lbm} \cdot \mathrm{ft}}{\mathrm{s}^2}} \right) }_{= -186.5 \ \mathrm{ft} \cdot \mathrm{lbm}} \\ &=[161.6+1026.2-450-186.5] \ \mathrm{ft} \cdot \mathrm{lbf} \\ \\ M_{A} &= 551.3 \ \mathrm{ft} \cdot \mathrm{lbf} \end{aligned} \nonumber \]

    Similar comments about the signs apply to angular momentum as given for linear momentum.

    Thus the reactions at \(A\) are \(Rx = 80.8 \ \mathrm{lbf}\), \(Ry =146.6 \ \mathrm{lbf}\), and \(M_{A} = 551.3 \ \mathrm{ft} \cdot \mathrm{lbf}\) in the directions as indicated on the diagram.

    Example — Tipping a box

    You have been asked to help move a filing cabinet by pushing it on the floor. Luckily for you the floor has very low friction coefficients (\(\mu_{\text {static}} = 0.3\) and \(\mu_{\text {kinetic}}=0.1\)). The filing cabinet is \(0.3 \mathrm{~m}\) wide and \(1.6 \mathrm{~m}\) high and has a mass of \(100 \mathrm{~kg}\). You will push it at a position above the floor equal to \(3 / 4\) of its height.

    (a) Determine the force \(F\) required to get the cabinet moving. Will it begin to tip before it slips?

    (b) If it tips before it slips, what force \(F\) is required to begin tipping the stationary cabinet?

    (c) If it slips before it tips, what is the maximum force \(F\) that can be applied before the slipping cabinet tips?

    Solution

    Known: A filing cabinet is being pushed along the floor.

    Find:

    (a) The force required to slip the cabinet

    (b) If it tips before it slips, the force required to begin tipping

    (c) If it slips before it tips, the maximum force that can be applied before the slipping cabinet tips

    Given: \[\begin{aligned} &h=1.6 \mathrm{~m} \quad m=100 \mathrm{~kg} \\ &L=0.3 \mathrm{~m} \\ &d=1.2 \mathrm{~m} \\ &G \text { in center of rectangle } \\ \\ &\text {Positive directions on coordinate system } \end{aligned} \nonumber \]

    A cabinet is represented by a tall, thin rectangle of uniform mass distribution with width L and height h that rests on a horizontal surface. The lower left corner is labeled A and the lower right corner is labeled B. A rightwards force F is applied at the left side of the cabinet, a distance d above the ground. Positive x-direction is to the right, positive y-direction is upwards, and positive moment is clockwise.

    Figure \(\PageIndex{6}\): Labeled diagram showing all given information.

    Analysis

    Strategy \(\rightarrow\) Try conservation of linear and angular momentum since we are considering friction forces and motion and tipping.

    System \(\rightarrow\) Closed, non-deforming system that includes the cabinet only.
    Property to count \(\rightarrow\) Linear momentum and angular momentum.
    Time interval \(\rightarrow\) Probably the rate form since we are interested in forces.

    First we need to draw the free-body diagram. Notice that the direction of the friction force along the bottom of the cabinet opposes the motion of the cabinet. You should also notice that the normal force \(F_{\mathrm{N}}\) on the bottom of the cabinet can act anywhere between \(A\) and \(B\) depending on the value of the applied force \(F\). Clearly when \(F\) is zero the normal force is applied immediately under the center of mass.

    First, let's determine what force will cause the cabinet to slip. To do this we apply conservation of linear momentum to a stationary, closed system.

    \[\underbrace{ \cancel{ \frac{d \mathbf{P}_{\text{sys}}}{dt} }^{=0} }_{ \begin{array}{c} \mathbf{P}=0 \\ \text{a constant} \end{array} } = \sum_{j} \mathbf{F}_{j} + \underbrace{ \cancel{ \sum_{\text {in}} \dot{m}_{i} \mathbf{V}_{i} - \sum_{\text {out}} \dot{m}_{e} \mathbf{V}_{e} }^{=0} }_{\text {closed system, no mass flow}} \quad \rightarrow \quad 0 = \mathbf{F} + \mathbf{F}_{f} + \mathbf{F}_{\mathrm{N}} +m \mathbf{g} \nonumber \]

    Where the friction force must equal the maximum static friction force possible \[\begin{array}{lll} & \boxed{\rightarrow +} \quad & 0 =-F_{f}+F & \rightarrow \quad F=F_{f} \leq \mu_{\text {static}} F_{N} \\ & \boxed{\uparrow +} & 0=-m g+F_{N} & \rightarrow \quad F{N}=mg \\ & F=F_{f} \leq \mu_{\text {static}} m g = (0.3)\left[(100 \mathrm{~kg})\left(9.81 \ \frac{\mathrm{m}}{\mathrm{s}^{2}}\right)\right] = 294.3 \mathrm{~N} \end{array} \nonumber \]

    So the minimum force that must be applied to get the cabinet to slide is \(294.3 \mathrm{~N}\).

    Now to determine if it tips, apply the conservation of angular momentum to the same system. To do this we must recognize that just at it starts to tip (tipping impending) the normal force is acting at point \(B\) (the right side of the cabinet). So let's calculate angular momentum about point \(B\) and solve for the force that must be applied to tip the cabinet assuming it is stationary, \(F_{\text {tip}}\).

    \[ \begin{array}{lll} & \underbrace{ \cancel{ \dfrac{d \mathbf{L}_{B, \text{ sys}}}{dt} }^{=0} }_{\begin{array}{c} \mathbf{L}=0 \text{ since} \\ \text{stationary} \end{array}} = \sum_j \mathbf{M}_{B, \ j} \ + \underbrace{ \cancel{ \sum - \sum}^{=0} }_{\begin{array}{c} \text{Closed system} \\ \text{No mass flow terms} \end{array}} \quad & \rightarrow \quad & 0 = \left( \mathbf{r}_F \times \mathbf{F}_{\text{tip}} \right) + \left( \mathbf{r}_G \times m \mathbf{g} \right) \\ \\ & \begin{align*} \boxed{\text{CW } +} \quad\quad 0 &= \left( d \ F_{\text{tip}} \right) + \left( \frac{L}{2} mg \right) \\ F_{\text{tip}} &= \frac{1}{2} \frac{L}{d} mg \end{align*} & \rightarrow & F_{\text{tip}} = \dfrac{1}{2} \left( \dfrac{0.3 \mathrm{~m}}{1.2 \mathrm{~m}} \right) (981.0 \mathrm{~N}) = 122.6 \mathrm{~N} \end{array} \nonumber \]

    Comparing the force to slip against the force to tip, we find that the force to tip is less than the force to slip the cabinet—the cabinet tips before it slips. The maximum force that can be applied to the cabinet before it tips is \(122.6 \mathrm{~N}\) and the cabinet will never slip.

    The answer to Part (c) is academic since the cabinet never slips under these conditions.

    Comment:

    But wait, we need to move the cabinet! Suppose we place the cabinet on some very small roller bearings so it essentially glides over the floor with no static or kinetic friction. How much force can I apply now before it tips?

    The \(y\)-momentum results will be identical \(\rightarrow \ F_{\mathrm{N}} = mg = 981 \mathrm{~N}\).

    The \(x\)-momentum results must consider motion in the \(x\)-direction so we have \(\rightarrow m \frac{d V_{G}}{d t}=F\).

    The angular momentum equation is trickier. First we must recognize that the cabinet is translating, and that when tipping is impending the normal force \(F_{\mathrm{N}}\) will act at the right corner:

    \[ \dfrac{d \mathbf{L}_{B}}{dt} = \left(\mathbf{r}_{F} \times \mathbf{F}\right) + \left(\mathbf{r}_{G} \times m \mathbf{g}\right) \quad \text { and } \quad \mathbf{L}_{B} = m \left(\mathbf{r}_{G} \times \mathbf{V}_{G}\right) \quad\quad\quad\quad\quad\quad\quad \nonumber \]

    \[ \begin{align*} \boxed{ \text{CW }+ } \quad\quad \dfrac{d}{dt} \underbrace{ \left[ m \left( \dfrac{h}{2} V_G \right) \right] }_{\begin{array}{c} \text{Angular momentum} \\ \text{of system about} \\ \text{point }B \end{array}} &= \underbrace{ \left( d \ F \right) }_{\begin{array}{c} \text{Moment of} \\ \text{force } F \text{ about} \\ \text{point }B \end{array}} + \underbrace{ \left( - \dfrac{L}{2} mg \right) }_{\begin{array}{c} \text{Moment of} \\ \text{the weight about} \\ \text{point }B \end{array}} \\[4pt] m \dfrac{d}{dt} \left( \dfrac{h}{2} V_G \right) &= dF - \dfrac{L}{2} mg \\[4pt] m \dfrac{h}{2} \dfrac{d V_G}{dt} &= dF - \dfrac{L}{2} mg \quad\quad\quad \rightarrow \quad\quad\quad \begin{array}{l} \\{ } \\ F = \dfrac{m}{2} \dfrac{h}{d} \dfrac{d V_G}{dt} + \dfrac{m}{2} \dfrac{L}{2} g \\[4pt] F = \dfrac{m}{2} \left( \dfrac{h}{d} \right) \left[ \dfrac{d V_G}{dt} + g \left( \dfrac{L}{h} \right) \right] \end{array} \end{align*} \nonumber \]

    Now we need to use our results from the \(x\)-momentum to eliminate the acceleration term \[\begin{aligned} F = \frac{m}{2} \left(\frac{h}{d}\right) \left[\frac{dV_{G}}{dt} + g \left(\frac{L}{h}\right) \right] &= \frac{m}{2}\left(\frac{h}{d}\right) \underbrace{\left[\left(\frac{F}{m}\right)+g\left(\frac{L}{h}\right)\right]}_{\begin{array}{c} \text { Here's where we } \\ \text { made the substitution } \end{array}} \\[4pt] \begin{array}{c} F - \dfrac{m}{2} \left(\dfrac{h}{d}\right) \left(\dfrac{F}{m}\right) = \dfrac{m}{2} \left(\dfrac{h}{d}\right) \left(\dfrac{L}{h}\right) g \\[4pt] F \left[1-\dfrac{1}{2}\left(\dfrac{h}{d}\right)\right] = \dfrac{m}{2} \left(\dfrac{L}{d}\right) g \\ { } \\ { } \end{array} \quad\quad\quad &\rightarrow \quad\quad\quad F = \dfrac{\dfrac{m}{2} \left(\dfrac{L}{d}\right) g}{\left[1-\dfrac{1}{2}\left(\dfrac{h}{d}\right)\right]} = (m g) \dfrac{\dfrac{1}{2} \left(\dfrac{L}{d}\right)}{\left[1-\dfrac{1}{2} \left(\dfrac{h}{d}\right)\right]} \end{aligned} \nonumber \]

    Now substituting the numerical values we have \[F = (mg) \frac{\dfrac{1}{2}\left(\dfrac{L}{d}\right)}{\left[1 - \dfrac{1}{2} \left(\dfrac{h}{d}\right)\right]} =(981 \mathrm{~N}) \frac{\dfrac{1}{2}\left(\dfrac{0.3}{1.2}\right)}{\left[1 - \dfrac{1}{2} \left(\dfrac{1.6}{1.2}\right)\right]} = 367.9 \mathrm{~N} \nonumber \] Notice that the value of \(F\) depends on the weight of the object, the ratio \(h / d\), and the ratio \(L / d\). What happens when \(d=h / 2\) ? What does this mean physically? What about when \(L=0\) ? What does this mean?

    Example — It's Turning!

    Two metal cylinders \(A\) and \(B\) are suspended from a frictionless pulley. Cylinder \(A\) has a mass of \(30 \mathrm{~kg}\) and cylinder \(B\) has a mass of \(60 \mathrm{~kg}\). The pulley is essentially a flat disk with a diameter of \(0.5 \mathrm{~m}\) and a mass of \(10 \mathrm{~kg}\). The cables have negligible mass.

    A pulley is attached to a ceiling, and a cord passes over it. Cylinder A hangs from the left end of the cord and cylinder B hangs from the right end.

    Figure \(\PageIndex{7}\): Two cylinders suspended from a frictionless pulley.

    Initially the cylinders are stationary. If they are suddenly released, calculate the acceleration of the cylinders.

    Solution

    Known: Two masses suspended from a pulley suddenly start moving.

    Find: The acceleration of the cylinders after they are released.

    Given: See the figure above.

    \[ \begin{align*} & \text{Diameter of the pulley} = D = 0.5 \mathrm{~m}\\ & \text{Mass of the pulley} = m_{\text {Pulley}} = 10 \mathrm{~kg} \\ & \text{Mass of cylinder } A = m_{\mathrm{A}} = 30 \mathrm{~kg} \\ & \text{Mass of cylinder } B = m_{\mathrm{B}} = 60 \mathrm{~kg} \end{align*} \nonumber \]

    Analysis:

    Strategy \(\rightarrow\) Try angular momentum since pulley is turning and we have not been told to ignore the mass of the pulley.
    System \(\rightarrow\) Closed system containing the cylinders, the cable, and the pulley.
    Property to count \(\rightarrow\) Angular momentum
    Time interval \(\rightarrow\) Start with rate equation

    The free-body diagram is shown below, indicating the known forces on the system. Since this is a closed system there are no mass transfers of momentum. See the arrows on the diagram to indicate the positive directions for \(x_{A}, x_{B}\), and \(\omega\).

    Downwards weight of the pulley and upwards reaction force R act on the pulley center. Pulley rotates clockwise at rate omega. Block A experiences downwards weight force W_A and moves upward at velocity V_A, with the positive x_A direction being upwards. Block B experiences downwards weight force W_B and moves downward at velocity V_B, with the positive x_B direction being downwards.

    Figure \(\PageIndex{8}\): Free-body diagram of the system consisting of the pulley and its attached blocks.

    Writing angular momentum about the pulley axis of rotation for the system and assuming clockwise rotation is positive, we have the following result:

    \[\begin{aligned} \frac{d \mathbf{L}_{0, \text{ sys}}}{d t} &= \left(\mathbf{r}_{A} \times \mathbf{W}_{A}\right) + \left(\mathbf{r}_{B} \times \mathbf{W}_{B}\right) \\ \text{where} \quad\quad\quad \mathbf{L}_{0, \text{ sys}} &= m_{A} \left(\mathbf{r}_{A} \times \mathbf{V}_{A}\right) + m_{B} \left(\mathbf{r}_{B} \times \mathbf{V}_{B}\right) + \mathbf{\omega} I_{G} \\[4pt] \\ \boxed{\text{CW } +} \quad\quad \frac{d L_{0, \ \mathrm{sys}}}{d t} &= \left(\dfrac{D}{2} W_{B}\right) + \left(-\dfrac{D}{2} W_{A}\right) \\[4pt] \text{where} \quad\quad\quad L_{0, \mathrm{sys}} &= \underbrace{ \left(\dfrac{D}{2} m_{A} V_{A}\right) }_{\begin{array}{c} \text{Angular momentum of} \\ \text{Cylinder A about axle with} \\ \text{positive velocity } V_A \text{ upward} \end{array}} + \underbrace{ \left(\dfrac{D}{2} m_{A} V_{A}\right) }_{\begin{array}{c} \text{Angular momentum of} \\ \text{Cylinder B about axle with} \\ \text{positive velocity } V_B \text{ downward} \end{array}} + \underbrace{ \omega I_{G} }_{\begin{array}{c} \text{Angular momentum of} \\ \text{the pulley assuming positive} \\ \omega \text{ is in clockwise direction} \end{array}} \end{aligned} \nonumber \]

    Now substituting the system angular momentum into the momentum balance we have

    \[ \boxed{\mathrm{CW} \ +} \quad\quad \frac{d}{dt} \left[ \left(\frac{D}{2} m_{A} V_{A}\right) + \left(\frac{D}{2} m_{A} V_{A}\right) + \omega I_{G}\right] = \left(\frac{D}{2} W_{B}\right) + \left(-\frac{D}{2} W_{A}\right) \nonumber \]

    This equation can be further simplified by recognizing that only the velocities are constants, so bringing the constants outside the derivative and dividing through by \(D / 2\) gives the following \[\boxed{\mathrm{CW} \ +} \quad\quad m_{A} \frac{d V_{A}}{d t} + m_{B} \frac{d V_{B}}{d t} + \left(\frac{2}{D}\right) I_{G} \frac{d \omega}{d t} = -W_{A}+W_{B} \nonumber \]

    To go further, we must relate the translational and the rotational velocities. Assuming that a positive rotational direction is clockwise which corresponds with the velocity arrows on the diagram, we have the following: \[V_{A} = V_{B} = \omega (D / 2) \nonumber \] Using this result to replace \(V_{\text {В }}\) and \(\omega\) in the angular momentum balance gives the following:

    \[\begin{aligned} \boxed{\mathrm{CW} \ +} \quad\quad m_{A} \frac{d V_{A}}{dt} + m_{B} \frac{d V_{B}}{dt} + \left(\frac{2}{D}\right) I_{G} \frac{d}{dt} \left(\frac{2}{D} V_{A}\right) &= W_{B}-W_{A} \\[4pt] m_{A} \frac{d V_{A}}{dt} + m_{B} \frac{d V_{A}}{dt} + \left(\frac{2}{D}\right)^{2} I_{G} \frac{d V_{A}}{d t} &= W_{B}-W_{A} \end{aligned} \nonumber \]

    To finish, we must write the weights and the mass moment of inertia of the pulley in terms of their mass as follows:

    \[\begin{aligned} m_{A} \frac{d V_{A}}{dt} + m_{B} \frac{d V_{A}}{dt} + \left(\frac{2}{D}\right)^{2} I_{G} \frac{d V_{A}}{dt} &= W_{B}-W_{A} \\[4pt] m_{A} \frac{d V_{A}}{dt} + m_{B} \frac{d V_{B}}{dt} + \left(\frac{2}{D}\right)^2 \underbrace{ \left[\frac{1}{2} m_{\text {Pulley }} \left(\frac{D}{2}\right)^{2}\right] }_{\begin{array}{c} I_G \text{ for the pulley treated} \\ \text{as a disk} \end{array}} \frac{d V_A}{dt} &= m_B \ g - m_A \ g \\ m_A \frac{d V_A}{dt} + m_B \frac{d V_A}{dt} + \left( \frac{m_{\text{Pulley}}}{2} \right) \left[\frac{d V_A}{dt}\right] &= \left(m_B - m_A\right) g \\[4pt] \frac{d V_A}{dt} &= \frac{ \left(m_B - m_A\right) }{\left(m_A + m_B + \dfrac{m_{\text{Pulley}}}{2}\right)} g \end{aligned} \nonumber \]

    Substituting in the numbers gives \[\frac{d V_{A}}{dt} = \frac{\left(m_{B}-m_{A}\right)}{\left(m_{A} + m_{B} + \frac{m_{\text {Pulley}}}{2}\right)} g = \frac{(60-30)}{\left(30+60+\dfrac{10}{2}\right)} g = \left(\frac{30}{95}\right) g = 3.098 \ \frac{\mathrm{m}}{\mathrm{s}^{2}} \nonumber \]

    Thus cylinder \(A\) accelerates up and cylinder \(B\) accelerates down at the rate of \(3.098 \mathrm{~m} / \mathrm{s}^{2}\).

    Comment:

    Determine the error if we had neglected the mass of the pulley: i.e., what would the acceleration be if we had neglected the pulley mass?


    This page titled 6.3: Conservation of Angular Momentum Equation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Donald E. Richards (Rose-Hulman Scholar) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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