At a particular instant the location of an object relative to location \(A\) is given by the vector \({\overrightarrow r _A} = \left\langle {6,6,0} \right\rangle {\rm{m}}\). At this instant the momentum of the object is \(\overrightarrow p = \left\langle { - 11,13,0} \right\rangle {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}.\) What is the angular momentum of the object about location \(A\)?

Short Answer

Expert verified

The angular momentum of the object about location \(A\) is \(\left( {0,1,144} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\)

Step by step solution

01

Definition of Angular Momentum

The rotating inertia of an object or system of objects in motion about an axis that may or may not pass through the object or system is described by angular momentum.

02

Given data

Position of an object to point \(A\) is \({\overrightarrow r _A} = \left\langle {6,6,0} \right\rangle m\)

Here\(x - \) component of the position of the object, \({r_x} = 6\)

\(y - \)component of the position of the object, \({r_y} = 6\)

\(z - \)component of the position of the object,\({r_z} = 0\)

Momentum of the object relative to the point \(A\),\({\overrightarrow p _A} = \left\langle { - 11,13,0} \right\rangle kg \cdot m/s\)

Here\(x - \) component of the momentum of the object, \({p_x} = - 11\)

\(y - \)component of the momentum of the object, \({p_y} = 13\)

\(z - \)component of the momentum of the object, \({p_z} = 0\)

03

Find the Angular Momentum of the object at point A 

Angular momentum of the object relative to point \(A\)is

\({\overrightarrow L _A} = {\overrightarrow r _A} \times \overrightarrow p \)

\(\begin{aligned}{}{\overrightarrow L _A} &= \left( {{r_y}{p_z} - {r_z}{p_x} - {r_x}{p_z},{r_x}{p_y} - {r_y}{p_x}} \right)\\ &= \left( {0 - 0,0 - 0,\left( 6 \right)\left( {13} \right) - \left( 6 \right)\left( { - 11} \right)} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\\ &= \left( {0,078 + 66} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\\ &= \left( {0,1,144} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\end{aligned}\)

Hence, the Angular momentum of the object about point \(A\)is\(\left( {0,1,144} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\).

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Most popular questions from this chapter

A rotating uniform-density disk of radius 0.6mis mounted in the vertical plane, as shown in Figure 11.88.The axle is held up by supports that are not shown, and the disk is free to rotate on the nearly frictionless axle. The disk has mass 5kg.A lump of clay with mass0.4kgfalls and sticks to the outer edge of the wheel at the location -0.36,0.480,0m,relative to an origin at the centre of the axle. Just before the impact the clay has speed 8m/s,and the disk is rotating clockwise with angular speed0.51radians/s.

(a) Just before the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the centerC?(As usual,xis to the right,yis up, andzis out of the screen, toward you.) (b) Just after the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the centerC?(c) Just after the impact, what is the angular velocity (magnitude and direction) of the wheel? (d) Qualitatively, what happens to the linear momentum of the combined system? Why? (A) There is no change because linear momentum is always conserved. (B) Some of the linear momentum is changed into angular momentum. (C) Some of the linear momentum is changed into energy. (D) The downward linear momentum decreases because the axle exerts an upward force.

A uniform-density wheel of mass 6 kg and radius 0.3 m rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of15Nfor0.6s(a) what is the final angular speed? (b) what is the average angular speed? (c) Through how big an angle did the wheel turn? (d) How much string come off the wheel?

A device consists of eight balls, each of massattached to the ends of low-mass spokes of length L so the radius of rotation of ball is L/2. The device is mounted in the vertical plane, as shown in Figure 11.73. The axle is help up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. A lump of clay with massm falls and sticks to one of the balls at the location shown, when the spoke attached to that ball is 45°to the horizontal. Just before the impact the clay has a speed v, and the wheel is rotating counter clock wise with angular speedω .

(a.) Which of the following statements are true about the device and the clay, for angular momentum relative to the axle of the device? (1) the angular momentum of the device + clay just after the collision is equal to the angular momentum of the device +clay just before the collision. (2) The angular momentum of the falling clay is zero because the clay is moving in a straight line. (3) Just before the collision, the angular momentum of the wheel is 0. (4) The angular momentum of the device is the sum of the angular momenta of all eight balls. (5) The angular momentum of the device is the same before and after the collision. (b) Just before the impact, what is the (vector) angular momentum of the combined system of device plus clay about the center C? (As usual, xis to the right, yis up, and zis out of the screen, toward you) (c) Just after the impact, what is the angular momentum of the combined system of device plus clay about the center C? (d) Just after the impact, what is the (vector) angular velocity of the device? (e) Qualitatively. What happens to the total linear momentum is changed system? Why? (1) some of the linear momentum is changed into energy. (2) some of the linear momentum is changed into angular momentum. (3) There is no change because linear momentum is always conserved. (4) The downward linear momentum decreases because the axle exerts an upwards force. (f) qualitatively, what happens to the total kinetic energy of the combined system? Why? (1) some of the kinetic energy is changed into linear momentum. (2) some of the kinetic energy is changed into angular momentum. (3) The total kinetic energy decreases because there is an increase of internal energy in this inelastic collision. (4) There is no change because kinetic energy is always conserved.

A stationary bicycle wheel of radius 0.9mis mounted in the vertical plane (figure). The axle is held up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. The wheel has mass all 4.8kgconcentrated in the rim (the spokes have negligible mass). A lump of clay with mass 0.5kgfalls and sticks to the outer edge of the wheel at the location shown. Just before the impact the clay has speed 5m/sand the wheel is rotating clockwise with angular speed0.33rad/s.

(a) Just before the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the center C? (As usual, is to the right, is up, and is out of the screen, towards you) (b) Just after the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the center C? (c) Just after the impact, what is the angular velocity (magnitude and direction) of the wheel? (d) Qualitatively, what happens to the linear momentum of the combined system? Why? (1) The downward linear momentum decreases because the axle exerts into angular momentum. (3) some of the linear momentum is changed into energy. (4) There is no change because linear momentum is always conserved.

In Figure 11.26, if rA=3m, and θ=30°, what is the magnitude of the torque about locationAincluding units? If the force in Figure 11.26 were perpendicular to rA but gave the same torque as before, what would be its magnitude?

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