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Angular momentum of a rigid body

We have investigated the angular momentum of a single particle, which we generalized to a system of particles. Now we can use the principles discussed in the previous section to develop the concept of the angular momentum of a rigid body. Celestial objects such as planets have angular momentum due to their spin and orbits around stars. In engineering, anything that rotates about an axis carries angular momentum, such as flywheels, propellers, and rotating parts in engines. Knowledge of the angular momenta of these objects is crucial to the design of the system in which they are a part.

To develop the angular momentum of a rigid body, we model a rigid body as being made up of small mass segments, Δ m i . In [link] , a rigid body is constrained to rotate about the z -axis with angular velocity ω . All mass segments that make up the rigid body undergo circular motion about the z -axis with the same angular velocity. Part (a) of the figure shows mass segment Δ m i with position vector r i from the origin and radius R i to the z -axis. The magnitude of its tangential velocity is v i = R i ω . Because the vectors v i and r i are perpendicular to each other, the magnitude of the angular momentum of this mass segment is

l i = r i ( Δ m v i ) sin 90 ° .
Figure a shows a door-knob shaped object and an x y z coordinate system. The object is arranged vertically and centered on the z axis, with the wide knob at the top. The object is rotating about the z axis, counterclockwise as viewed from above, with angular velocity omega. A small part of the object is highlighted. This mass segment, labeled Delta m sub i, is located at vector r sub i, moves with velocity vector v sub i, and traces a counterclockwise circle of radius R sub i. The vector r sub i extends from the origin to the mass segment and makes an angle of theta sub i with the z axis. The vector v sub i is tangent to the circle traced by the mass segment. Figure b shows the x y coordinate system and the mass segment. Vectors r sub i and v sub i are shown again, as is the angle theta sub i between the vector r sub i and the z axis. The angular momentum vector of the mass segment, vector l sub i, is also shown. The vector l sub i is perpendicular to both r and v, as given by the right hand rule, and has a z component upward, shown on the diagram and labeled l sub i z. The remaining side of the right triangle whose hypotenuse is l sub i and vertical side is l sub i z is shown as a dashed line. The angle adjacent to this side, and opposite the vertical side l sub i z, is theta sub i.
(a) A rigid body is constrained to rotate around the z -axis. The rigid body is symmetrical about the z -axis. A mass segment Δ m i is located at position r i , which makes angle θ i with respect to the z -axis. The circular motion of an infinitesimal mass segment is shown. (b) l i is the angular momentum of the mass segment and has a component along the z -axis ( l i ) z .

Using the right-hand rule, the angular momentum vector points in the direction shown in part (b). The sum of the angular momenta of all the mass segments contains components both along and perpendicular to the axis of rotation. Every mass segment has a perpendicular component of the angular momentum that will be cancelled by the perpendicular component of an identical mass segment on the opposite side of the rigid body. Thus, the component along the axis of rotation is the only component that gives a nonzero value when summed over all the mass segments. From part (b), the component of l i along the axis of rotation is

( l i ) z = l i sin θ i = ( r i Δ m i v i ) sin θ i , = ( r i sin θ i ) ( Δ m i v i ) = R i Δ m i v i .

The net angular momentum of the rigid body along the axis of rotation is

L = i ( l i ) z = i R i Δ m i v i = i R i Δ m i ( R i ω ) = ω i Δ m i ( R i ) 2 .

The summation i Δ m i ( R i ) 2 is simply the moment of inertia I of the rigid body about the axis of rotation. For a thin hoop rotating about an axis perpendicular to the plane of the hoop, all of the R i ’s are equal to R so the summation reduces to R 2 i Δ m i = m R 2 , which is the moment of inertia for a thin hoop found in [link] . Thus, the magnitude of the angular momentum along the axis of rotation of a rigid body rotating with angular velocity ω about the axis is

L = I ω .

This equation is analogous to the magnitude of the linear momentum p = m v . The direction of the angular momentum vector is directed along the axis of rotation given by the right-hand rule.

Questions & Answers

a particle projected from origin moving on x-y plane passes through P & Q having consituents (9,7) , (18,4) respectively.find eq. of trajectry.
rahul Reply
definition of inertia
philip Reply
the reluctance of a body to start moving when it is at rest and to stop moving when it is in motion
An inherent property by virtue of which the body remains in its pure state or initial state
why current is not a vector quantity , whereas it have magnitude as well as direction.
Aniket Reply
the flow of current is not current
bcoz it doesn't satisfy the algabric laws of vectors
The Electric current can be defined as the dot product of the current density and the differential cross-sectional area vector : ... So the electric current is a scalar quantity . Scalars are related to tensors by the fact that a scalar is a tensor of order or rank zero .
what is binomial theorem
Tollum Reply
hello are you ready to ask aquestion?
Saadaq Reply
what is binary operations
What is the formula to calculat parallel forces that acts in opposite direction?
Martan Reply
position, velocity and acceleration of vector
Manuel Reply
*a plane flies with a velocity of 1000km/hr in a direction North60degree east.find it effective velocity in the easterly and northerly direction.*
hello Lydia.
What is momentum
A rail way truck of mass 2400kg is hung onto a stationary trunk on a level track and collides with it at 4.7m|s. After collision the two trunk move together with a common speed of 1.2m|s. Calculate the mass of the stationary trunk
Ekuri Reply
I need the solving for this question
is the eye the same like the camera
I can't understand
same here please
I think the question is that ,,, the working principal of eye and camera same or not?
yes i think is same as the camera
what are the dimensions of surface tension
why is the "_" sign used for a wave to the right instead of to the left?
why classical mechanics is necessary for graduate students?
khyam Reply
classical mechanics?
principle of superposition?
Naveen Reply
principle of superposition allows us to find the electric field on a charge by finding the x and y components
Two Masses,m and 2m,approach each along a path at right angles to each other .After collision,they stick together and move off at 2m/s at angle 37° to the original direction of the mass m. What where the initial speeds of the two particles
2m & m initial velocity 1.8m/s & 4.8m/s respectively,apply conservation of linear momentum in two perpendicular directions.
A body on circular orbit makes an angular displacement given by teta(t)=2(t)+5(t)+5.if time t is in seconds calculate the angular velocity at t=2s
2+5+0=7sec differentiate above equation w.r.t time, as angular velocity is rate of change of angular displacement.
Ok i got a question I'm not asking how gravity works. I would like to know why gravity works. like why is gravity the way it is. What is the true nature of gravity?
Daniel Reply
gravity pulls towards a mass...like every object is pulled towards earth
An automobile traveling with an initial velocity of 25m/s is accelerated to 35m/s in 6s,the wheel of the automobile is 80cm in diameter. find * The angular acceleration
Goodness Reply
(10/6) ÷0.4=4.167 per sec
what is the formula for pressure?
Goodness Reply
force is newtom
and area is meter squared
so in SI units pressure is N/m^2
In customary United States units pressure is lb/in^2. pound per square inch
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