Chapter

Rotational Dynamics and Equilibrium

Blinn College - Physics 1401 - Terry Honan

. - Torque and Angular Momentum

Torque about an Axis

We define torque as the rotational analog of force.  Suppose you are trying to loosen a bolt.  The axis of rotation is the center of the bolt.  If you are unable to give sufficient torque with your hand you grab a wrench.  Take (OverVector[r]) as the vector from the axis to where the force (OverVector[F]) is applied.  Clearly the important part of the force is the component of the force perpendicular to the radial vector (OverVector[r]).  Moreover the larger r is the larger the torque.  This motivates the definition of torque

τ=r (Subscript[F, ⊥])

If θ is the angle between (OverVector[r]) and (OverVector[F]) then we can write (Subscript[F, ⊥])=F sin θ.  Similarly we can  (Subscript[r, ⊥])=r sin θ as the component of (OverVector[r]) perpendicular to (OverVector[F]).  This gives us other ways of writing the torque.

τ=r (Subscript[F, ⊥])=r F sin θ= (Subscript[r, ⊥]) F

The sign of torque depends on the sign convention for kinematics.  If a force tends to make something rotate in the positive direction then the torque is positive and similarly negative torques tend to make things rotate in the negative direction.

Example . - Torques on a Disk

Solution

τ=r (Subscript[F, ⊥])=r F sin θ= (Subscript[r, ⊥]) F.

(Subscript[τ, 30])=-(Subscript[r, ⊥]) F=-(0.30(QuantityUnit))30(QuantityUnit)=-9.00(QuantityUnit) (QuantityUnit)

(Subscript[τ, 50])=-r (Subscript[F, ⊥])=-(0.40(QuantityUnit))50(QuantityUnit)=-20.00(QuantityUnit) (QuantityUnit)

(Subscript[τ, 40])=+r (Subscript[F, ⊥])=+(0.40(QuantityUnit))(40(QuantityUnit) cos 35 °)=13.11(QuantityUnit) (QuantityUnit)

(or  (Subscript[τ, 40])=+r F sin θ=+(0.40(QuantityUnit))(40(QuantityUnit)) sin 55 °=13.11(QuantityUnit) (QuantityUnit))

(Subscript[τ, 70])=r (Subscript[F, ⊥])=(0.20(QuantityUnit))70(QuantityUnit)=14.00(QuantityUnit) (QuantityUnit)

Solution

(Subscript[τ, net])=(Subscript[τ, 30])+(Subscript[τ, 50])+(Subscript[τ, 70])+(Subscript[τ, 80])=-1.89(QuantityUnit) (QuantityUnit)

Solution

Angular Momentum of a Particle and Torque

Given the vectors (OverVector[r]) and (OverVector[F]), we previously defined the torque about any axis as

τ=r (Subscript[F, ⊥])=r F sin θ= (Subscript[r, ⊥]) F

We can similarly define the angular momentum of a particle with momentum (OverVector[p]) at a position (OverVector[r]); the angular momentum about any axis is

L=r (Subscript[p, ⊥])=r p sin θ= (Subscript[r, ⊥]) p

Both of these expressions are relative to an axis; here (OverVector[r]) is the perpendicular vector from the axis to a particle and (OverVector[p]) is the particle’s momentum. In one-dimension we could write the second law for a particle as: (Subscript[F, net])=(lim_(Δ t->0))Δ p/Δ t. The net torque on a particle is the torque due to the net force: (Subscript[τ, net])=(Subscript[r, ⊥])(Subscript[F, net]). Using more calculus than is possible in this course, it can be shown that

(Subscript[τ, net])=(lim_(Δ t->0))((Δ L)/(Δ t))

(Subscript[τ, net]) is the net torque on a particle about some axis and L is the particle’s angular momentum about that axis.

As an example of this, consider a free particle. Free means that there is no force acting on it and that implies no torque. It follows that the angular momentum of that free particle must be constant. This is easy to see with the figure below. No force means that p is a constant. (Subscript[r, ⊥])is the distance from the axis to the particle’s line of motion and that is also constant. Thus L= (Subscript[r, ⊥]) p is constant.

System of Particles and the Conservation of Angular Momentum

In Chapter 9 we saw that for a system of particles that

(Subscript[(OverVector[F]), net]^ext)=(lim_(Δ t->0))((Δ(Subscript[(OverVector[p]), tot]))/(Δ t))    or    (Subscript[(OverVector[F]), net,ave]^ext)=((Δ(Subscript[(OverVector[p]), tot]))/(Δ t))

A similar but a bit more complicated derivation in the case of torques and angular momenta about an axis we have a similar result.

(Subscript[τ, net]^ext)=(lim_(Δ t->0))((Δ (Subscript[L, tot]))/(Δ t))    or    (Subscript[τ, net,ave]^ext)=((Δ (Subscript[L, tot]))/(Δ t))

This result is true for all axes.

The  conservation of angular momentum follows from the expression above.  If there are no external torques on a system then the total angular momentum of the system is conserved.

(Subscript[τ, net]^ext)=0 ⟹  ((Δ (Subscript[L, tot]))/(Δ t))=0 ⟹  Δ (Subscript[L, tot])=0

This derivation mirrors the conservation of linear momentum.

This is a very fundamental result.  It has deep implications on the very large scale; in astrophysics it is crucial in the dynamics of planets, stars, solar systems and galaxies.  It is also important on the very small scale; in particle accelerators where elementary particles are collided and created, angular momentum is always conserved.

. - More on Rigid Bodies

Angular Momentum of a Rigid Body

As before, we view our rigid body as a collection of point masses where the perpendicular distance form the axis to (Subscript[m, i]) is (Subscript[r, i]). Since all the (Subscript[r, i]) are fixed we get the momentum related to the tangential velocity, which is then related to the angular velocity.

(Subscript[p, i,⊥])=(Subscript[m, i]) (Subscript[v, i,t])=(Subscript[m, i]) (Subscript[r, i]) ω

The angular momentum of the (i^th) mass becomes

(Subscript[L, i])=(Subscript[r, i]) (Subscript[p, i,⊥])=(Subscript[r, i]) (Subscript[m, i]) (Subscript[v, i,t])=(Subscript[m, i]) (Subscript[r, i]^2) ω

The total angular momentum is the sum over all these terms  L=(Subscript[L, 1])+(Subscript[L, 2])+….  Using  I=(Subscript[m, 1]) (Subscript[r, 1]^2)+(Subscript[m, 1]) (Subscript[r, 1]^2)+… we get the angular momentum of a rotating rigid body

L=(Subscript[L, 1])+(Subscript[L, 2])+…=(Subscript[m, i]) (Subscript[r, i]^2) ω+(Subscript[m, i]) (Subscript[r, i]^2) ω+…=((Subscript[m, i]) (Subscript[r, i]^2) ω+(Subscript[m, i]) (Subscript[r, i]^2)) ω.

This gives the result we had in our table from last chapter that related rotations about a fixed axis to one dimensional linear motion. It is the rotational analog of p=m v.

L=I ω

Example . - Angular Momentum of the Earth

(Subscript[M, E])=5.97×(10^24)(QuantityUnit) ,  (Subscript[R, E])=6.38×(10^6)(QuantityUnit)  and  (Subscript[R, ES])=1.50×(10^11)(QuantityUnit).

Solution

I=(2/5)(Subscript[M, E]) (Subscript[R, E]^2)=9.72×(10^37)(QuantityUnit) (((QuantityUnit))^2)

(Subscript[ω, rot])=((2 π)/T)=((2π)/(1 day))=((2π)/(24×3600(QuantityUnit)))=7.27×(10^(-5))(((QuantityUnit))^(-1))

L=I (Subscript[ω, rot])=7.07×(10^33)(QuantityUnit) (((QuantityUnit))^2)/(QuantityUnit)

Solution

Solution

(Subscript[ω, orbit])=((2 π)/T)=((2π)/(1 (QuantityUnit)))=((2π)/(365.24 24×3600(QuantityUnit)))=1.992×(10^(-7))(((QuantityUnit))^(-1)).

v=(Subscript[v, t])=r ω  ⟹  L=r (Subscript[p, ⊥])=r p=r m v=m (r^2)ω

L=(Subscript[M, E])(Subscript[R, ES]^2)(Subscript[ω, orbit])=2.67×(10^40)(QuantityUnit) (((QuantityUnit))^2)/(QuantityUnit)

Example . - The Rotating Figure Skater

Solution

(Subscript[L, out])=(Subscript[L, in])  ⟹  (Subscript[I, out])(Subscript[ω, out])=(Subscript[I, in])(Subscript[ω, in])  ⟹  (Subscript[ω, in])=((Subscript[I, out])/(Subscript[I, in]))(Subscript[ω, out])

Solution

K=(1/2)I (ω^2)  and  L=I ω  ⟹  K=((L^2)/(2 I))

(Subscript[I, in])<(Subscript[I, out])  ⟹  (Subscript[K, in])=((L^2)/(2 (Subscript[I, in])))>((L^2)/(2 (Subscript[I, out])))=(Subscript[K, out])

Example . - Bullet Shot in Door

Solution

(Subscript[L, i])= (Subscript[r, ⊥]) p=d m v.

(Subscript[I, door])=(1/3)M (L^2)=(1/3)M (w^2)

(Subscript[I, bullet])=(∑_i) (Subscript[m, i]) (Subscript[r, i]^2)=m (d^2)

(Subscript[L, i])=(Subscript[L, f])  ⟹  d m v=((1/3)M (w^2)+m (d^2)) (Subscript[ω, f])  ⟹  (Subscript[ω, f])=((d m v)/((1/3)M (w^2)+m (d^2)))

The Rotational Second Law

We can now, finally, derive the rotational equivalent of the second law (Subscript[τ, net])=I α.  Start with the the expression for a system of particles.

(Subscript[τ, net]^ext)=(lim_(Δ t->0))((Δ (Subscript[L, tot]))/(Δ t))

When the system is the rigid body then the net external torque on the rigid body is just the net torque on it. Similarly, the total angular momentum is just I ω the angular momentum of the body.  We get

(Subscript[τ, net])=(lim_(Δ t->0))((Δ L)/(Δ t)).

Using  L=I ω and α=(lim_(Δ t->0))Δ ω/Δ t we get our result.

(Subscript[τ, net])=I α

Example . - Torques on a Disk (Continued)

Solution

I=(1/2)m (R^2)=0.28(QuantityUnit)(((QuantityUnit))^2)

(Subscript[τ, net])=I α   ⟹   α=((Subscript[τ, net])/I)=6.75(((QuantityUnit))/((QuantityUnit)))

The Torque Due to Gravity

We saw in the previous chapter that to calculate the potential energy due to gravity we treat the object as if all the mass is at the center of mass.  The same is true for finding the torque due to gravity.

(Subscript[τ, grav])=(Subscript[r, cm,⊥])M g

It is straightforward to verify this. Write the torque as the sum over the torques on all the point masses in the body. Then use the definition of center of mass to get the result.

(Subscript[τ, grav])=(Subscript[r, 1,⊥])(Subscript[m, 1]) g+(Subscript[r, 2,⊥])(Subscript[m, 2]) g…=((Subscript[m, 1])(Subscript[r, 1,⊥])+(Subscript[m, 2])(Subscript[r, 2,⊥]) …)g=(M (Subscript[r, cm,⊥]) )g

Note that since gravity is vertical the perpendicular component of the (OverVector[r]) vectors will always be the horizontal component.

This is why the term center of gravity is commonly used for the center of mass. To calculate  both the gravitational potential energy and the torque due to gravity, you treat a system as if all of the mass is located at the center of mass.

Example . - Swinging Disk

Solution

(Subscript[τ, net])=(Subscript[τ, grav])=r F sin θ=R m g sin θ

I=(3/2)m (R^2)

(Subscript[τ, net])=I α  ⟹  R m g sin θ=(3/2)m (R^2) α  ⟹  α=((2  g)/(3 R)) sin θ

Example . - Atwood’s Machine with a Massive Pulley

Solution

Δ (Subscript[x, 1])=Δ (Subscript[x, 2])=Δ x=R Δ θ

(Subscript[a, 1])=(Subscript[a, 2])=a=R α

(Subscript[F, net,1])=(Subscript[m, 1]) (Subscript[a, 1])  ⟹  (Subscript[T, 1])-(Subscript[m, 1]) g=(Subscript[m, 1]) a

(Subscript[F, net,2])=(Subscript[m, 2]) (Subscript[a, 2])  ⟹  (Subscript[m, 2]) g-(Subscript[T, 2])=(Subscript[m, 2]) a

(Subscript[τ, net])=I α  ⟹  R (Subscript[T, 2])-R (Subscript[T, 1])=I α=((1/2)M (R^2))α

R (Subscript[T, 2])-R (Subscript[T, 1])=I α=((1/2)M (R^2)) (a/R)  ⟹  (Subscript[T, 2])-(Subscript[T, 1])=(M/2)a

(Subscript[m, 2]) g-(Subscript[m, 1]) g=((Subscript[m, 1])+(Subscript[m, 2])+M/2) a  ⟹  a=(((Subscript[m, 2])-(Subscript[m, 1]))/((Subscript[m, 1])+(Subscript[m, 2])+M/2))g

. - Static Equilibrium

The Conditions for Equilibrium

If a body is in equilibrium then there is no acceleration and there is no angular acceleration.  This implies that the net force and the net torque must vanish.

(Subscript[(OverVector[F]), net])= (OverVector[0])   and   (Subscript[τ, net])= 0

The Choice of Axis is Arbitrary

When considering an equilibrium problem sometimes the choice of axis is clear.  Often, though, it is not clear; there is no natural choice.  The key point is that the choice of origin or axis is arbitrary. When something is arbitrary then we have the luxury of making a choice that simplifies the problem. The basic result is this: If the torques balance about one axis and the forces balance, then the torques balance about any axis parallel to the first. Suppose (Subscript[(OverVector[F]), i]) is one of the forces acting on a body. Take the perpendicular vector from one axis to another is (Subscript[(OverVector[r]), 0]). Also take (Subscript[(OverVector[r]), i]) as the perpendicular vector from the original axis to the where the force (Subscript[(OverVector[F]), i])  acts and  (Subscript[(OverVector[r]), i]^′) is from the new axis to the same point.

(Subscript[(OverVector[r]), i])=(Subscript[(OverVector[r]), i]^′)+(Subscript[(OverVector[r]), 0]).

The torques due to (Subscript[(OverVector[F]), i]) can then be written as

(Subscript[τ, i])=(Subscript[r, i,⊥])(Subscript[F, i])=(Subscript[r, i,⊥]^′)(Subscript[F, i])+(Subscript[r, 0,⊥])(Subscript[F, i])=(Subscript[τ, i]^′)+(Subscript[r, 0])(Subscript[F, i.⊥])

Summing over all the forces (Subscript[(OverVector[F]), i]) gives

(Subscript[τ, net])=(Subscript[τ, net]^′)+(Subscript[r, 0])(Subscript[F, net.⊥])

It follows that if the net torque is zero about one axis and the net force is zero ((Subscript[(OverVector[F]), net])=(OverVector[0]) ), then the net torque is zero about any axis parallel to the first.

0=(Subscript[τ, net])  and  (Subscript[(OverVector[F]), net])=(OverVector[0])   ⟹   (Subscript[τ, net]^′)=0

Example . - Hanging Meter Stick

Solution

(Subscript[F, net])=0  ⟹  (Subscript[T, L])+(Subscript[T, R])=W

0=(Subscript[τ, net,A])=0+(0.30(QuantityUnit)) W-(0.40(QuantityUnit)) (Subscript[T, R])  ⟹  (Subscript[T, R])=(3/4)W

(Subscript[T, L])+(Subscript[T, R])=W  ⟹  (Subscript[T, L])=W -(Subscript[T, R])=(1/4)W

Example . - Leaning Ladder

Solution

({{(Subscript[F, net,hor])=0  ⟹  N=H
(Subscript[F, net,ver])=0  ⟹  V=W}})

(Subscript[τ, W])=(Subscript[r, ⊥])F=+((L/2)cos θ) W

(Subscript[τ, N])=-(Subscript[r, ⊥])F=-(L sin θ) N

0=(Subscript[τ, net])=(Subscript[τ, W])+(Subscript[τ, N])+(Subscript[τ, H])+(Subscript[τ, V])=+((L/2)cos θ) W-(L sin θ) N+0+0

H=N=(W/(2 tan θ))