Angular Momentum of a Particle

Angular Momentum (l) is the vector cross product of translational momentum and the displacement of that momentum relative to a fixed point :

x

y

z

mvr

l = r x p

l

Since angular momentum is a vector cross product its direction follows the right hand rule!

In order to produce a change in linear momentum, a force must act : ∑F = m dv

d tIn order to produce a change in the angular momentum, a net torque must act : ∑ = d l

d t

A particle of mass m is held a horizontal distance x from point P and dropped. A) Find the torque acting at any time t with respect to P:

A) P

x

mg

ø

At any given time:

= rFsinørsinø = x and F = mg

therefore torque is constant:

= mgx

By using the right hand rule the direction would be perpendicular to the plane of the page and into the page (negative)!

into

out of

r

B) Find the angular momentum of m at any given time t with respect to P:

l = rpsinø rsinø = x and p = mv = m(gt)

l = mgxt Perpendicular into the plane

C) Show that ∑ = dl / dt = dl

dt

mgx = d(mgxt)

dt mgx = mgx

Systems of Particles

The sum of the net forces acting on an object will equal the timed rate change of linear momentum:

∑F = m∆v = dp

dt dt

The sum of the net torque acting on a system of particles is equal to the time rate change of total angular momentum:

∑ = dL

dt

For a rigid body that is rotating with axial symmetry (the body is symmetric to the axis of rotation), total angular momentum can be found:

L = I units: kg•m2/s

A sanding disk with a rotational inertia of 1.22 X 10-3 kg-m2 is attached to an electric drill whose motor delivers a torque of 15.8 N-m. Find A) the angular momentum and B) the angular speed of the disk 33.0 ms after the motor is turned on.

I = 1.22 x 10-3 kg•m2

= 15.8 N•mt = .033 s

= I = / I

= 13,000 rad/s2

= o + t = 427 rad/s

L = I = .521 kg•m2/s

A wheel of radius 24.7 cm, moving initially at 43.3 m/s, rolls to a stop in 225 m. Calculate its linear and angular acceleration. If the moment of inertia of the wheel is .155 kg-m2, what torque does friction exert on the wheel?

r = .247 mvo = 43.3 m/s

v = 0s = 225 m

a = v2 - vo2

2s= - 4.17 m/s2

= a / r = - 16.9 rad/s2

I = .155 kg•m2 = I = - 2.62 N•m

A uniform stick with a mass of 4.42 kg and a length of 1.23 m is initially lying flat and motionless on a frictionless surface. It is then struck perpendicularly by a puck imparting an impulse of 12.8 N-s at a distance of 46.4 cm from the center. Determine the subsequent motion of the stick. 2.90 m/s & 10.7 rad/s

The angular momentum of a flywheel having a rotational inertia of .142 kgm2 decreases from 3.07 to .788 kgm2/s in 1.53 s. A)Find the average torque acting on the flywheel during this time. B) Assuming a uniform angular acceleration, through what angle will the flywheel have turned during this time? C) How much work was done on the wheel? D) How much average power was supplied by the flywheel?

11.27.2 A 0.16 meter long, 0.15 kg thin rigid rod has a small 0.22 kg mass stuck on one of its ends and a small 0.080 kg mass stuck on the other end. The rod rotates at 1.7 rad/s through its physical center without friction. What is the magnitude of the angular momentum of the system taking the center of the rod as the origin? Treat the masses on the ends as point masses.

11.27.3 A solid uniform cylinder rolls down a ramp, starting from a stationary position at height 1.4 m. At the bottom of the ramp, the cylinder has no potential energy. If the mass of the cylinder is 2.9 kg and its radius is 0.076 m, what is the magnitude of angular momentum of the cylinder at the bottom of the ramp with respect to the cylinder's center of mass?

11.31.3 A string is wound around the edge of a solid 1.60 kg disk with a 0.130 m radius. The disk is initially at rest when the string is pulled, applying a force of 6.50 N in the plane of the disk and tangent to its edge. If the force is applied for 1.90 seconds, what is the magnitude of its final angular velocity?

Conservation of Angular Momentum

If not net external torque acts upon a system, the total angular momentum of the system will remain constant.

• a spinning skater will spin faster when pulling in her arms because she reduces her rotational inertia in doing so.

Li = Lf Iii = If f

A disk of 125 g and radius 7.2 cm is spinning with and angular speed of .84 rev/s about a vertical axis through its center and perpendicular to the plane of the disk. Two additional identical disks are then dropped onto the first disk and friction between the disks cause them to all end up rotating at the same angular speed. What is that angular speed?A man stands on a frictionless platform that is rotating at 1.22 rev/s. His arms are outstretched and in each hand he holds a weight. In this position, the man, the platform and the weights have a total moment of inertia of 6.13 kg-m2. By moving his hands in, the man reduces the rotational inertia by 4.16 kg-m2. A) What is the angular speed of the platform now?

1 = 1.22 rev/s

I1 = 6.13 kg•m2

I2 = 1.97 kg•m2

A) 2 = I1 1

I2

Li = Lf

= 3.80 rev/s

B) K1 = .5I1 12

K2 .5I2 22

= 3.21

B) What is the ratio of total kinetic energy of the man with outstretched arms to the man with the arms pulled in?

A wheel with a rotational inertia of 1.27 kg-m2 is rotating with an angular speed of 824 rpm when a second wheel with a rotational inertia of 4.85 kg-m2 is suddenly clamped onto it. What is the angular speed of both wheels and what is the percent change of kinetic energy of the system?

I1 = 1.27 kg•m2

1 =824 rev/min

I2 = 4.85 kg•m2

Li = Lf

I1 1 = (I1 + I2) 2

2 = I1 1

I1 + I2

= 171 rpm

B) %∆K = ?

Ki = .5I1 12 = 4.31 x 105

Kf = .5(I1 + I2) 22 = 8.95 x 104

∆K = Kf - Ki = -3.42 x 105

% = ∆K

Ki

= .792 = 79.2%

A)

A 45.0 kg cube is resting 0.750 m from the center of a solid 125 kg disk that is rotating at 2.80 rad/s. The radius of the disk is 2.00 m. The cube slides radially to a location 1.60 m from the center of the disk, changing the moment of inertia of the system. What is the new angular velocity of the system? Treat the cube as a point mass in your calculation.

Gyroscopic Precession

mg

N F

O

L

There is no net torque here, so the axis of rotation remains constantly fixed!

Precession- the movement of

the axis of rotation

No Precession!

mg

N

O L

When the end of the axle is released:

r

The weight will produce an

unbalanced torque directed into the

page.

∆L

The torque changes the direction of the axis of rotation and the shaft moves horizontally in that direction.

The Spinning Top

The spinning top is a common example of when the axis of rotation will move in its own circle of rotation. The torque is provided by the weight of the top itself.

The speed of precession can be found in terms of the weight, the radius, and the angular momentum:

p = Mgr

L