Note8 1

PHYS1110H, 2011 Fall. Shijie Zhong Rotation, Angular Momentum, Torque, and Conservation of Angular Momentum. Rotational Motion about a Fixed–axis (kinematics) Angular velocity ω : Consider a rotating disk with a fixed axis. An effective way to describe the motion is to introduce a coordinate or angle θ (in radian). θ increases in counter-clockwise direction, and decreases in clockwise direction. Angular velocity ω (rad/second) is the rate of change of angle θ.

€

ω = dθ dt , or

€

ω = ΔθΔt , (1) for instantaneous and average angular velocities, respectively. For counter-clockwise rotation, ω is positive (a right-hand rule), while it is negative for clockwise rotation.

It is useful to make an analogy between 1-D motion and fixed-axis rotation at this point. The coordinate x and v for 1-D motion are analogous to θ and ω here. Of course, their units are different. It’s also interesting to point out the difference in directions. Velocity v is a vector, and is in either positive x axis or negative axis. For angular velocity ω , we may also assign it a direction – for positive (i.e., counter-clockwise rotation) angular velocity, ω points to z axis direction that is perpendicular to the rotational plane, and negative ω points to negative z axis direction. Angular velocity ω is also called angular frequency that can be related to the usual definition of frequency f, the number of revolution per second. We may define period T as the time it takes for a full revolution or 2π radian. T=1/f. Therefore, we have ωΤ=2π or

Note8 2

T=2π/ω, and ω=2πf. (2) Angular acceleration α : We may further define angular acceleration α as the rate of change in ω :

€

α = dω dt = d2θdt2

, (3)

with direction in either positive z or negative z directions (but not dependent on clockwise or counter-clockwise rotation anymore). For constant angular acceleration α , the following relations hold

€

ω =ω0 +αt , (4)

€

θ = θ0 +ω0t +12αt2, (5)

€

ω2 =ω02 + 2α(θ −θ0 ) , (6)

which are similar to what we learned for 1-D motion, not surprising, given that the mathematical forms are the same. Angular acceleration α differs from the centripetal acceleration ar=v2/r for circular motion in direction and meanings, but it may be related to tangential acceleration atan, that is the rate of change in magnitude of velocity.

€

atan = dvdt = d(rω)dt = r dω dt = rα , (7) The total acceleration is a vector sum: a = atan + ar or

€

| a |= atan2 + ar

2 . (8) Polar coordinate system At this point, it is useful to discuss velocity and acceleration vectors in polar coordinates. So far, we have limited ourselves in Cartesian coordinate system. Sometimes, it is convenient to use polar coordinate system where particle’s position is described in terms of r and θ.

Note8 3

In polar coordinates, position vector r is

€

r = r e r , where er is unit vector in r direction.

The velocity vector is:

€

v = d r dt

=d(r e r )

dt=

drdt e r + r d e r

dt, (9)

where notice that both er and eθ change with time (see the middle figure above) as a particle moves from positions r(t) to r(t+Δt), and we will need to compute der/dt and deθ/dt using the right figure above.

€

d e rdt

= limΔ e rΔt

= lim e r (t + Δt) − e r (t)

Δt= limΔθ

Δt e θ =

dθdt e θ =ω

e θ , (10)

€

d e θdt

= limΔ e θΔt

= lim e θ (t + Δt) − e θ (t)

Δt= lim(−Δθ

Δt e r ) = −

dθdt e r = −ω

e r , (11)

Substituting (10) to (9), we have velocity in polar coordinates:

€

v = d r dt

=drdt e r + rω e θ . (12)

We can now derive an expression for acceleration.

€

a = d v dt

=d2rdt2 e r +

drdt

d e rdt

+d(rω)

dt e θ + (rω) d e θ

dt,

€

=d2rdt2 e r +

drdtω e θ +

drdtω e θ + r dω

dt e θ − rω2 e r .

€

a = (d2r

dt2− rω2 ) e r + (2 dr

dtω + rα) e θ . (13)

For circular motion with r as a constant, (12) and (13) become

€

v = rω e θ , and

€

a = −rω2 e r + rα e θ .

Note8 4

There are two components in acceleration -- the radial component is our usual centripetal acceleration, while the θ-component is tangential acceleration. They can be related back to (7) and (8). For circular motion at a constant rate, tangential component vanishes, and

€

a = −rω2 e r is our usual centripetal acceleration. Torque. For a particle with position vector r in a given coordinate system with force F, its torque τ is given as cross product of r and F,

€

τ = r × F . (14)

The unit of τ is kg m2/s2. Recall all the properties we discussed about the cross product. If r and F are in xy plane, τ is parallel to z axis,

€

τ = r × F = rF sinθ e z . (15)

where θ is the angle between vectors r and F when they share the same origin (remember that we can move a vector with no rotation). However, vector r often starts from the origin of coordinates or some reference point (e.g., hinge of rotation) and ends at the location of a particle, while vector F starts at the particle location. τ points to either positive or negative z direction that would bring in either counter-clockwise (positive z) or clockwise (negative z) rotation of the particle around the reference point. If vectors r and F are given in 3-D as

€

r = (x,y,z) and

€

F = (Fx ,Fy ,Fz ) , we

can use the general form -- determinant form for cross product.

€

τ =

e x e y

e zx y z

Fx Fy Fz

. (16)

Note8 5

Notice that torque τ is always perpendicular to vectors r and F.

If several forces acting on the same object (e.g., a wheel or a bin), the total or net torque is

€

τ net =

τ j∑ =

r j∑ × F j , (17)

where j is index for each torque or position and force vector pair. Net force and net torque are not closely related to each other. The following 4 scenarios are all possible. 1)

€

τ net =

F net = 0 (in equilibrium), 2)

€

τ net = 0,

F net ≠ 0, 3)

€

τ net ≠ 0,

F net = 0, and 4) they are both nonzero.

Moment of Inertia about a rotation axis. For a system of particles with mass mj and distance to a rotation axis rj, the moment of inertia of the particles about the rotation axis:

€

I = m jrj2

j∑ , (18)

For an object (e.g., a ring or sphere) with density ρ, (18) can be generalized:

€

I = r2dm∫ = ρr2dV∫ , (19) where r is the distance of the element of volume to the rotation axis. If the rotation axis is along z axis, r2=x2+y2.

Note8 6

As we will see below, the torque is analogous to force in linear motion, and moment of inertia is analogous to mass. Rotational Dynamics about a Fixed-axis. We drew comparisons between 1-D linear motion kinematics and fixed-axis rotational motion. The analogy can be further extended in dynamics as well with torque, moment of inertia, and angular acceleration in rotational motion analogous to a force, mass, and linear acceleration in 1-D linear motion. Apply force F to a block of mass m that is attached to a rod with the other end fixed at a free-to-rotate axis (see figure). The force F is always perpendicular to the rod and causes the mass to accelerate in tangential direction, according to the Newton’s 2nd law:

€

F = matan = mrα or . (20)

is the moment of inertia.

(20) becomes

. (21) (21) is analogous to the Newton’s 2nd law F=ma. Example 1. I for a uniform ring around axis passing the center. For a ring with mass M and radius R, the perimeter of the ring is 2πR, and the linear density λ=M/(2πR).

€

I = λR2dl∫ = λR2Rdθ0

2π∫ = 2πλR3 = MR2.

Example 2. I for a uniform disk around axis passing the center. For a disk with mass M and radius R, the area of the disk is πR2, and the surface density λ=M/(πR2). We may divide the disk into concentric rings,

!

rF = mr2"

!

I = mr2

!

" = I#

Note8 7

and for a given ring with radius r and thickness dr, the surface area of this ring is 2πrdr. The moment of inertia I for the disk is

€

I = λr2dS∫ = r2λ2πrdr0

R∫ = 2πλ 1

4R4 =

12MR2 .

Notice from examples 1 and 2 that for a given mass M, the more the mass is near the center or rotational axis, the smaller the moment of inertia.

For a uniform rod of mass M and length L,

€

I =112

ML2 , around the axis

passing the center of mass. For a uniform sphere of mass M and radius R

around the axis passing the center of mass,

€

I =25MR2 .

Example 3. I for a uniform disk around an axis at a distance l to the center – parallel axis theorem. Without loss of generality, let the rotation axis to be on positive x axis at a distance of l from the center of the disk which is also used as the origin of the coordinates (see figure below). Consider a small element with surface area dS and mass λdS, with λ=M/(πR2) as in example 2. The distance between this element and the rotation axis is r, and its distance to the origin is r’. For the given triangle in the figure below, the law of cosine is

€

r2 = l2 + r'2 −2r'l cosθ . The moment of inertia with respect to the rotation axis:

€

I = λr2dS∫ = λ (l2 + r'2 −2lr'cosθ)dS∫

€

= λ l2dS∫ + λr'2 dS∫ − 2lλ r'cosθdS∫

€

= πR2λl2 + ICM = Ml2 + ICM , where ICM is the moment of inertia relative to the axis passing through the center of mass (i.e., 1/2MR2 from example 2), and the integral with cosθ is zero, because of the definition of CM R=0.

€

λr'cosθdS∫ = λ r '⋅ l dS∫ = λ

r 'dS∫ ⋅ l = R ⋅ l = 0 .

€

I = Ml2 + ICM , (22)

Note8 8

(22) is called parallel axis theorem. Although we showed this for a special case of a disk, this theorem is true for all the cases. Example 4. Atwood’s machine with a massive pulley. Two masses M1 and M2 hang on two sides of a pulley with mass Mp and radius R. Find acceleration for the masses. We solved similar problems before with massless pulley, but we now need to consider mass of pulley. While the acceleration for the two masses remains the same, the tensional force on the string T1 and T2 differ now. Newton’s 2nd law for the two masses:

€

M1g −T1 = M1a , (i)

€

T2 −M2g = M2a , (ii) For the pulley, we may use (14) that relates net torque to tangential acceleration and also notice

€

atan = a = Rα ,

€

T1R −T2R = Iα , or

€

T1R −T2R =12M pR

2 aR

,

€

T1 −T2 =12M pa. (iii)

Equations (i), (ii), and (iii) can be solved for a, T1 and T2.

€

a =(M1 −M2 )g

M1 + M2 + M p /2.

Rotational kinetic energy. If we place a rotating disk with zero translational velocity on a ramp, the disk may rotate and climb the ramp. This indicates that a rotating object has rotational kinetic energy. For a pure rotational motion, the kinetic energy can be considered as sum for small elements of masses that are all rotated around a fixed axis at constant angular velocity ω.

€

KE =12m j∑ v j

2 =12m j∑ rj

2ω2 =12ω2 m j∑ rj

2 =12Iω2 , (23)

Note8 9

For an object that rotates around an axis passing the center of mass and that the center of mass has a translational motion vCM (e.g., a rolling wheel), the total kinetic energy for this object is

€

KEtotal = KEtrans + KErot =12mvCM

2 +12Iω2. (24)

Example 5. Down a ramp, which goes faster? a sphere, a hoop and a cylinder, each with mass M and radius R and starts at rest at a height h.

Conservation of energy:

€

Mgh =12Mv f

2 +12Iω2 =

12Mv f

2 +12I(v fR)2 .

€

v f = 2Mgh /(M + I /R2 ) . Substituting I for a sphere, a hoop, and a cylinder, we found that vf is

€

v f = gh ,

€

4gh /3 , and

€

10gh /7 for a hoop, cylinder, and sphere, respectively, in the order from slower to faster. The sphere has a smaller moment of inertia, thus taking less rotational kinetic energy, and hence moving faster. One can easily show that for a block sliding down the ramp with no rotation and no friction,

€

v f = 2gh is the largest among them all.

Note8 10

Angular momentum of a particle. For a particle with position vector r in a given coordinate system with momentum p, its angular momentum L is given as cross product of r and p,

€

L = r × p . (25) The unit of L is kg m2/s. If r and p are in xy plane, L is parallel to z axis, or

€

L = r × p = rpsinφ e z, (26) where φ is the angle between vectors r and p when they share the same origin. However, often vector r starts from the origin of coordinates and ends at the location of a particle, while vector p starts at the particle location and points to the direction of motion or v. L points to either positive or negative z direction that represents counter-clockwise (positive z) or clockwise (negative z) rotation of the particle. Suppose that

€

r = (x,y) and

€

p = (px ,py ), we can use the determinant form for cross product.

€

L = r × p =

e x e y

e zx y 0

px py 0= (xpy − ypx )

e z. (27)

Angular momentum depends on the reference point where vector r starts, as in the following example. Torque can be related to the rate of change of angular momentum.

€

d L

dt=

ddt( r × p ) =

d r dt× p + r × d p

dt= v × p + r × d p

dt= r × d p

dt

€

= r × F = τ ,

where we considered that v x p = 0 as they are parallel to each other.

€

τ =

d L

dt. (28)

Note8 11

Example 6. Angular momentum of a sliding block. A block with mass m slides along x axis at a velocity v.

First, angular momentum relative to the origin or point A,

€

L A =

r × p = r e x ×mv e x = 0. We did not specify any force on the block, but we can infer based on motion along x axis, if there is a net force, the force must be also along x axis. Therefore, the torque with respect to point A would also be zero:

€

τ A =

r × F = r e x ×F e x = 0 .

The torque and angular momentum satisfy (28). Now let’s compute the angular momentum relative to point B along y axis.

€

L B =

r × p =

e x e y e z

x −l 0mv 0 0

= mlv e z .

We can compute the torque due to a net force F along x direction, given the motion of the block is along x direction.

€

τ B =

r × F =

e x e y e z

x −l 0F 0 0

= Fl e z .

Again, the net torque and angular momentum satisfy (28), as demonstrated below:

€

d L Bdt

=d(mlv e z )

dt= lm dv

dt e z = lF e z =

τ B.

Note8 12

Example 7. Angular momentum of a conical pendulum (the particle mass M, velocity v, and length of the string l).

point A caused by the net force. The net force acting on the pendulum is the sum of the tension along the string and gravity, and should point to A or parallel to vector r. Therefore, the net torque relative to point A is ZERO. Again, this is consistent with (28), because vector LA is a constant, and its time derivative is also zero. Now, let’s compute angular momentum relative to the hinge point at B, and its magnitude is given as below,

€

| L B |=|

r '× p |= Mvl, where vector r’ starts from B and ends at the particle with length of l, and is perpendicular to vector v or p. The direction of vector LB, as shown, changes with time (around z axis). To show (28) is more tedious for this case, and we skip it. Angular momentum for a fixed rotation axis Equation (28) is equivalent to Newton’s second law, and can be reduced to (21) for a fixed axis rotation. Without loss of generality, we may consider an object rotating around z axis with angular velocity ω also in z direction, and the object may be divided into many small elements

First, let’s compute angular momentum relative to the center of the circle on the plane of motion, or point A. Since r and p are perpendicular to each other in this case, and the motion is counter-clockwise,

€

L A =

r × p = Mvr e z , which is along z direction. We can also compute the torque with respect to

Note8 13

of volumes. The total angular momentum for the rotating object

€

L = r j ×

p j∑ = r j ×m j

v j∑ . Let vector r’j be position vector to j-th element from the axis, we have

€

v j = r' j ω e vj where

€

e vj is the unit vector in xy plane in the direction of velocity (see figure above). This is because the velocity must be perpendicular to z axis and in xy plane. The z-component of the angular momentum,

€

Lz = [( r j × p j ) ⋅ e z ]∑ = [( r j ×m jr' j ω

e vj ) ⋅ e z ]∑ = m jr' j

2ω∑ = Iω .

where we considered

€

( r j × e vj ) ⋅

e z = r' j , and

€

m jr' j2∑ = I .

€

Lz = Iω . (29) Substituting (29) into (28) leads to (21). We should point out that equation (29) is the angular momentum of a rotating object with a fixed axis in the direction of angular velocity. Motion with both translation and rotation A rolling disk on a plane or a spinning planet orbiting a star has both translation and rotation motion. Here we only consider the rotation motion with rotation axis parallel to z axis (i.e., having a fixed direction in space).

Note8 14

First, let’s consider the z component angular momentum and state that

€

Lz = I0ω + ( R ×M

V cm )z, (30)

where I0 is the moment of inertia of the body relative to a rotation axis passing the center of mass CM, ω is the angular velocity of the rotation relative to the same axis, M is the mass of the body, R is the position vector for CM and Vcm is CM’s velocity, both relative to an inertial reference frame. Equation (30) can be proved as following. Let’s divide the object into many small elements, and for j-th element with position vector rj, in an inertial reference frame, the angular momentum is

€

L = r j ×m j

v j∑ = r j ×m j

˙ r j∑ . (31) CM’s position vector is

€

R = m j

r j∑ /M . Suppose that r’j is the position vector for j-th element relative to the CM, we have

€

r j = R + r ' j ,

and substituting it to (31), we have

€

L = (

R + r ' j ) ×m j (

˙ R + ˙ r ' j )∑

€

= R × m j

˙ R ∑ + m j r ' j × R ∑ + R × m j

˙ r ' j∑ + m j r ' j × ˙ r ' j∑ . (32)

The 2nd and 3rd terms in (32) are zero, because

€

m j r ' j∑ = m j (

r j − R )∑ = m j

r j − M R = 0∑ .

Therefore, we have

€

L = R × m j

˙ R ∑ + m j r ' j × ˙ r ' j∑ =

R ×M

˙ R + m j r ' j × ˙ r ' j∑ .

€

Lz = ( R ×M

V cm )z + ( m j

r ' j × ˙ r ' j∑ )z . (33)

The 2nd term in (33) is for the rotational motion about the axis that is parallel to z axis and passes the CM, and it can be written as (29) or I0ω, according to derivation there. Therefore, we proved equation (30). Similarly, we may write total torque acting on the body in the same inertial frame as

€

τ =

r i × f i∑ = ( r 'i +

R ) × f i∑ = ( r 'i ×

f i )∑ +

R ×

f i∑ . (34)

Let

€

F =

f i∑ be the total force on the body. The z component of the total

torque is

€

τ z = ( r 'i × f i )∑ z + (

R × F )z = τ0 + (

R × F )z, (35)

Note8 15

where τ0 is the z component of the torque relative to the CM. Clearly, we should have

€

τ z = dLz /dt , with z-component of torque and angular momentum defined in (35) and (30), respectively. More specifically, we have

€

τ0 = d(I0ω) /dt = I0α , (36) where α is angular acceleration. Equation (36) states that the rotation about the axis passing the CM only depends on the torque relative to the CM. Recall that equation (24) for total kinetic energy with both rotational and translational motions. Example 8. A uniform disk with radius R and mass M rolls with no slipping down a ramp with angle θ. Find acceleration.

Method 1: See the left diagram, using Newton’s 2nd law and also (36),

€

W sinθ − f = Ma , (W=Mg)

€

fR = I0α , (Only frictional force contributes to torque relative to the CM)

€

a = Rα . (because of no slipping) Considering I0=1/2MR2, solve these three equations for three unknowns, f, a, and α, and we have

€

a = 2/3gsinθ . Method 2: See the right diagram, and use a coordinate system with origin at the contact point. Use (30) to compute the angular momentum:

€

Lz = I0ω + ( R ×M

V cm )z =

12

MR2ω + MRVcm =32

MRVcm .

where we consider no slipping condition or Vcm=Rω. With the contact point as the origin, only gravity contributes to the torque,

€

τ z = ( r i × F i

i=1

3∑ )z = MgRsinθ .

€

τ z = dLz /dt , or

€

MgRsinθ = 3/2MRdVcm /dt , or

€

a = dVcm /dt = 2/3gsinθ .

Note8 16

Conservation of Angular momentum. From (28), if the net torque on a particle is zero, then its angular momentum L is conserved or constant with time. Conservation of angular momentum is an important conservation law, similar to conservation laws of momentum and energy. Example 9. Proof of Kepler’s 2nd law – equal area. The gravitational force between a planet and its star always points to the star, or F(r)=f(r)r, therefore with the gravity as the only force on the planet, the torque on the planet is zero --

€

τ = r × ( f r ) = 0 . Hence, the angular

momentum of a planet is conserved or constant with time.

Consider the planet at times t and t+Δt, at coordinates (r,θ) and (r+Δr,θ+Δθ). The area swept out is

€

ΔA ≈ 0.5(r + Δr)(rΔθ) = 0.5r2Δθ + 0.5rΔrΔθ . The rate of sweeping the area is

€

dAdt

= limΔt→0ΔAΔt

= 0.5r2 dθdt

+ 0 =12r2 dθ

dt.

The velocity of the planet can be written as

€

v = d r dt

=d(r e r )

dt=

drdt e r + r d e r

dt=

drdt e r + r dθ

dt e θ ,

which is the velocity vector expressed in a polar coordinate system, and in deriving it we considered

€

d e rdt

=dθdt e θ ,

in the polar coordinate system, as in (10). Then the angular momentum of the planet is

€

L = r ×m v = mr e r × (

drdt e r + r dθ

dt e θ ) = mr2 dθ

dt e z.

Since L is constant with time, dA/dt must be a constant.

Note8 17

Example 10. Capture cross section of a planet. Suppose that we send a spacecraft to a planet with radius R. If gravity of the planet plays no role, then the spacecraft has to aim at a cross section area πR2 to land the spacecraft to the planet. However, when the planet’s gravity field is considered, then the cross section area to capture the spacecraft would be larger. Here we determine this cross section area.

Suppose the distance between where the spacecraft is launched and the planet is much larger than R. The distance between the initial trajectory of spacecraft and the line aa is b, and is called impact parameter of the trajectory. We need to determine the largest b (let’s call it b’) for which the spacecraft can still be captured by the planet. The capture cross section area is πb’2. Because the position vector and gravitational force on the spacecraft are parallel, the net torque on the spacecraft is zero, and the angular momentum of the spacecraft is conserved (is the linear momentum of the spacecraft conserved?). The total energy of the spacecraft is E=K+U and is

€

E =12mv2 − GmM

r.

The angular momentum of the spacecraft about the center of the planet is

€

L = −mvr sinφ . Initially, at very large distance r, the spacecraft moves at a speed of v0 and is

a trajectory with impact parameter b’.

€

E =12mv0

2 and L= -mv0b’.

The point of collision occurs at a distance of closest approach of the orbit, that is r=R. And at this point, vectors r and v are perpendicular to each other, i.e., L= -mvR. Therefore, conservation of angular momentum and energy lead to:

€

12mv0

2 =12mv2 − GmM

R,

Note8 18

€

−mv0b'= −mvR. Two unknowns are v and b’ that can be solved from these two equations.

€

b'2 = R2 (1+GmM /Rmv0

2 /2), or

€

πb'2 = πR2 (1+GmM /Rmv0

2 /2) = πR2 (1−U (R)

E) .