Chapter

Rotational Dynamics and Equilibrium

Blinn College - Physics 2325 - Terry Honan

. - The Vector Nature of Rotational Quantities

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 Velocity and Torque as Vectors

A rotation has a magnitude, the angle of rotation, and a direction, along the direction of the axis.  Rotations are not vectors, though.  Vector addition is commutative but rotations are not.  It turns out that infinitesimal rotations do commute and are vectors.  We can write an infinitesimal rotation as d(OverVector[θ]).  Since angular velocity about an axis requires only an infinitesimal rotation, ω=dθ/dt, we can define the angular velocity vector

(OverVector[ω])=((d(OverVector[θ]))/(dt)) .

There are two possible directions along an axis.  We decide which direction by using the right hand rule.  Wrap the fingers of your right hand in the direction of rotation.  The thumb points in the direction of the vector.  If angular velocity is a vector then we can also make angular acceleration a vector.

(OverVector[α])=(d/(dt)) (OverVector[ω])

Wit these considerations we can now make a vector out of the torque.  We can assign its direction to the sense of rotation due to that torque.  (OverVector[r]) and (OverVector[F]) are vectors; we will define the cross product so that  the cross product of  (OverVector[r]) and (OverVector[F]) is the torque (OverVector[τ]).

(OverVector[τ])=(OverVector[r])×(OverVector[F])

The Cross or Vector Product

The dot product, or scalar product, is a way of multiplying of two vectors that gives a scalar.  The cross product, also known as the vector product, is a multiplication that gives a vector.

(OverVector[A])×(OverVector[B])  is a vector.

The magnitude of this vector is  A B sin θ.  We will specify the direction with a unit vector (OverHat[u]).  The two vectors (OverVector[A]) and (OverVector[B]) define a plane; their cross product is perpendicular to that plane.  There are two unit vectors perpendicular to any plane; we use the right hand rule to find the correct one.  Put your right thumb in the direction of the first entry (OverVector[A]) and your fingers in the direction of the second entry (OverVector[B]).  The palm of your hand is in the direction (OverHat[u]), giving the direction of the cross product.

(OverVector[A])×(OverVector[B])=A B sin θ  (OverHat[u])  ((OverHat[u]) by right hand rule)

Properties of the Cross Product

(OverVector[A])×(OverVector[B])=-(OverVector[B])×(OverVector[A])  (antisymmetry - not commutative)

((OverVector[A])×(OverVector[B]))×(OverVector[C])≠(OverVector[A])×((OverVector[B])×(OverVector[C]))  (not associative)

(c(OverVector[A]))×(OverVector[B])=c ((OverVector[A])×(OverVector[B]))=(OverVector[A])×(c (OverVector[B]))  (associative w.r.t. scalar mult.)

((OverVector[A])+(OverVector[B]))×(OverVector[C])=(OverVector[A])×(OverVector[C])+(OverVector[B])× (OverVector[C])  and            
(OverVector[A])×((OverVector[B])+(OverVector[C]))=(OverVector[A])×(OverVector[B])+(OverVector[A])× (OverVector[C])  (distributive)

The Cross Product and Components

With the dot product we were able to write it in terms of components.  We can do the same for the cross product.  As before we have to find the products of the basis unit vectors.  Because of the antisymmetry property we get (OverVector[A])×(OverVector[A])=(OverVector[0]).  It follows then that

(OverVector[0])=(OverHat[x])×(OverHat[x])=(OverHat[y])×(OverHat[y])=(OverHat[z])×(OverHat[z]).

Using our definition of the cross product we can see that the cross product of two perpendicular unit vectors is a third unit vector perpendicular to the two.

(OverVector[A])×(OverVector[B])=A B sin θ  (OverHat[u])  ⟹  (OverHat[v])×(OverHat[w])=1·1·1  (OverHat[u])

The cross product of the unit vectors (OverHat[x]) and (OverHat[y]) is thus a unit vector perpendicular to the xy plane.  This is either (OverHat[z]) or -(OverHat[z]).  We insist that our coordinate system is right-handed; this means that

(OverHat[x])×(OverHat[y])=(OverHat[z]).

For the other combinations of unit vectors there is a simple rule to keep track of their cross products.  Arrange x, y and z around a circle.

({{x}, {z      y}})

If the order of the three coordinates has the same sense of rotation as  x, y, z  it gains a positive sign.  If opposite it gets a minus sign.

(OverHat[y])×(OverHat[z])=(OverHat[x]),    (OverHat[z])×(OverHat[x])=(OverHat[y]) ,
(OverHat[y])×(OverHat[x])=-(OverHat[z]),    (OverHat[x])×(OverHat[z])=-(OverHat[y]) and  (OverHat[z])×(OverHat[y])=-(OverHat[x])

We can put all this together and get the cross product in terms of components.

(OverVector[A])×(OverVector[B])=((Subscript[A, x]) (OverHat[x])+(Subscript[A, y]) (OverHat[y])+(Subscript[A, z]) (OverHat[z]))×((Subscript[B, x]) (OverHat[x])+(Subscript[B, y]) (OverHat[y])+(Subscript[B, z]) (OverHat[z]))        
           =(OverHat[x]) ((Subscript[A, y]) (Subscript[B, z])-(Subscript[A, z]) (Subscript[B, y]))+(OverHat[y]) ((Subscript[A, z]) (Subscript[B, x])-(Subscript[A, x]) (Subscript[B, z]))+(OverHat[z]) ((Subscript[A, x]) (Subscript[B, y])-(Subscript[A, y]) (Subscript[B, x]))

The determinant method is a common way to write this.  A determinant is a mathematical operation that completely antisymmetrizes a square matrix.  On a 2×2 matrix we have:

|({{a, b}, {c, d}})|=a d-b c .

On a higher order matrix we write it as an alternating sum of determinants of lower order.  We will consider only the case of the cross product.

(OverVector[A])×(OverVector[B])=|({{(OverHat[x]), (OverHat[y]), (OverHat[z])}, {(Subscript[A, x]), (Subscript[A, y]), (Subscript[A, z])}, {(Subscript[B, x]), (Subscript[B, y]), (Subscript[B, z])}})|=(OverHat[x]) |({{(Subscript[A, y]), (Subscript[A, z])}, {(Subscript[B, y]), (Subscript[B, z])}})|-(OverHat[y]) |({{(Subscript[A, x]), (Subscript[A, z])}, {(Subscript[B, x]), (Subscript[B, z])}})|+(OverHat[z]) |({{(Subscript[A, x]), (Subscript[A, y])}, {(Subscript[B, x]), (Subscript[B, y])}})| 
           =(OverHat[x]) ((Subscript[A, y]) (Subscript[B, z])-(Subscript[A, z]) (Subscript[B, y]))-(OverHat[y]) ((Subscript[A, x]) (Subscript[B, z])-(Subscript[A, z]) (Subscript[B, x]))+(OverHat[z]) ((Subscript[A, x]) (Subscript[B, y])-(Subscript[A, y]) (Subscript[B, x]))

Note that there is not a discrepancy between the differing signs of the middle term for the y-component, since the factor multiplying  (OverHat[y]) has terms in reverse order.

Example . -  Dynamics of a Particle - Net Torque

(OverVector[r])(t)=⟨a (t^2),b t, -c (t^3)⟩ ,

Solution

(OverVector[v])(t)=((d(OverVector[r])(t))/(dt))=⟨2a t,b,-3c (t^2)⟩

(OverVector[a])(t)=((d(OverVector[v])(t))/(dt))=⟨2a,0,-6c t⟩

(Subscript[(OverVector[F]), net])(t)=m (OverVector[a])(t)=m ⟨2a,0,-6c t⟩

(Subscript[(OverVector[τ]), net])(t)=(OverVector[r])(t)×(Subscript[(OverVector[F]), net])(t)=m ⟨a (t^2),b t, -c (t^3)⟩ ×⟨2a,0,-6c t⟩

({{(Subscript[(OverVector[τ]), net])=m|({{(OverHat[x]), (OverHat[y]), (OverHat[z])}, {a (t^2), b t, -c (t^3)}, {2a, 0, -6c t}})|=m⟨-6b c (t^2)-0,-(-6a c (t^3)+2 a c (t^3)),0-2a b t⟩}, {=m⟨-6b c (t^2),4a c (t^3),-2a b t⟩}})

Angular Momentum of a Particle and Torque

We previously defined the torque as

(OverVector[τ])=(OverVector[r])×(OverVector[F]).

We can similarly define the angular momentum of a particle as

(OverVector[L])=(OverVector[r])×(OverVector[p]).

Both of  these expressions are relative to an origin; (OverVector[r]) is the position vector.  It is from the origin to the position to where the force is applied in the case of torque.  It is from the origin to the position of the particle for the angular momentum.

The net torque on a particle (a point) is the torque due to the net force.  If the particle is at position (OverVector[r]) then the net torque is (Subscript[(OverVector[τ]), net])=(OverVector[r])×(Subscript[(OverVector[F]), net]).  It is straightforward to verify that the cross product satisfies the usually product rule for differentiation.  Using this we can differentiate the angular momentum for a particle.  This gives:

(d/(dt))(OverVector[L])=((d/(dt))(OverVector[r]))×(OverVector[p])+(OverVector[r])×(d/(dt))(OverVector[p])=(OverVector[v])×m (OverVector[v])+(OverVector[r])×(Subscript[(OverVector[F]), net])=(OverVector[0])+(Subscript[(OverVector[τ]), net]).

This gives the analog of the momentum form of the second law  (Subscript[(OverVector[F]), net])=d(OverVector[p])/dt.  This is

(Subscript[(OverVector[τ]), net])=(d/(dt))(OverVector[L]).

Example . - Dynamics of a Particle (continued) - Angular Momentum

(OverVector[r])(t)=⟨a (t^2),b t, -c (t^3)⟩ ,

Solution

(OverVector[v])(t)=((d(OverVector[r])(t))/(dt))=⟨2a t,b,-3c (t^2)⟩

(OverVector[L])(t)=(OverVector[r])(t)×(OverVector[p])(t)=(OverVector[r])(t)× m (OverVector[v])(t)

({{(OverVector[L])(t)=m|({{(OverHat[x]), (OverHat[y]), (OverHat[z])}, {a (t^2), b t, -c (t^3)}, {2a t, b, -3c (t^2)}})|=m⟨-3b c (t^3)+b c (t^3),-(-3a c (t^4)+2 a c (t^4)),a b (t^2)-2a b (t^2)⟩}, {=m⟨-2b c (t^3),a c (t^4),-a b (t^2)⟩}})

Solution

((d(OverVector[L])(t))/(dt))=m⟨-6b c (t^2),4a c (t^3),-2a b t⟩

A System of Particles

In the preceding chapter we considered a three particle system with masses (Subscript[m, 1]),  (Subscript[m, 2])and (Subscript[m, 3]) at positions (Subscript[(OverVector[r]), 1]), (Subscript[(OverVector[r]), 2]) and (Subscript[(OverVector[r]), 3]).  As before, we write the forces on (Subscript[m, 1]) as a sum of internal forces (Subscript[(OverVector[F]), 2,1]) and (Subscript[(OverVector[F]), 3,1]) and external forces (Subscript[(OverVector[F]), 1]^ext).  The cross products of this with (Subscript[(OverVector[r]), 1]) gives the net torque.  The torques for (Subscript[m, 2]) and (Subscript[m, 3]) break up similarly.

(Subscript[(OverVector[τ]), net,1])=(Subscript[(OverVector[r]), 1])×(Subscript[(OverVector[F]), 1]^ext)+ (Subscript[(OverVector[r]), 1])×(Subscript[(OverVector[F]), 2,1]) + (Subscript[(OverVector[r]), 1])×(Subscript[(OverVector[F]), 3,1])=(d/(dt))(Subscript[(OverVector[L]), 1])

(Subscript[(OverVector[τ]), net,2])=(Subscript[(OverVector[r]), 2])×(Subscript[(OverVector[F]), 2]^ext)+ (Subscript[(OverVector[r]), 2])×(Subscript[(OverVector[F]), 1,2]) + (Subscript[(OverVector[r]), 2])×(Subscript[(OverVector[F]), 3,2])=(d/(dt))(Subscript[(OverVector[L]), 2])

(Subscript[(OverVector[τ]), net,3])=(Subscript[(OverVector[r]), 3])×(Subscript[(OverVector[F]), 3]^ext)+ (Subscript[(OverVector[r]), 3])×(Subscript[(OverVector[F]), 1,3]) + (Subscript[(OverVector[r]), 3])×(Subscript[(OverVector[F]), 2,3])=(d/(dt))(Subscript[(OverVector[L]), 3])

To concentrate on the bulk motion of our system we sum over these expressions.  In the previous case with forces the internal forces canceled due to Newton's third law.  Here we need to make an additional assumption that the forces are central forces; this is that (Subscript[(OverVector[F]), 1,2]), the force of mass 2 on mass 1, is directed parallel (or antiparallel) to the line between the masses.

(Subscript[(OverVector[F]), 2,1]) ∥  ((Subscript[(OverVector[r]), 1])-(Subscript[(OverVector[r]), 2]))   ⟺   ((Subscript[(OverVector[r]), 1])-(Subscript[(OverVector[r]), 2]))×(Subscript[(OverVector[F]), 2,1])=(OverVector[0])

Now when we sum the net torques we get a cancellation of the internal torques.  The internal torques cancel for all pairs of masses.  The cancellation between (Subscript[m, 1]) and (Subscript[m, 2])follows from

(Subscript[(OverVector[r]), 1])×(Subscript[(OverVector[F]), 2,1]) +(Subscript[(OverVector[r]), 2])×(Subscript[(OverVector[F]), 1,2])=(Subscript[(OverVector[r]), 1])×(Subscript[(OverVector[F]), 2,1]) +(Subscript[(OverVector[r]), 2])×(-(Subscript[(OverVector[F]), 2,1])) =((Subscript[(OverVector[r]), 1])-(Subscript[(OverVector[r]), 2]))×(Subscript[(OverVector[F]), 2,1])=(OverVector[0]).

The other internal forces cancel similarly.  We end up with

(Subscript[(OverVector[τ]), 1]^ext)+ (Subscript[(OverVector[τ]), 2]^ext) + (Subscript[(OverVector[τ]), 3]^ext)=(d/(dt))((Subscript[(OverVector[L]), 1])+(Subscript[(OverVector[L]), 2])+(Subscript[(OverVector[L]), 3])) .

It  should be clear how this could be generalized to four, or an arbitrary number, of particles.  This gives the very fundamental result that for a system of particles

(Subscript[(OverVector[τ]), net]^ext)=(d/(dt))(Subscript[(OverVector[L]), tot]).

Conservation of Angular Momentum

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

Axes and Origins

We began with a discussion of rigid bodies rotating about a fixed axis.  Then we considered quantities like angular velocity, angular acceleration, torque and angular momentum as vectors.  How are the two points of view related?  Torque and angular momentum vectors are relative to an origin, where the position vector (OverVector[r]) is based at the origin.  If the origin is chosen as some point on the axis then the vector relative to the axis is just the component in the direction of the axis.  The torque about some origin is the vector

(OverVector[τ])=(OverVector[r])×(OverVector[F]).

The torque about the z axis is just the z component of this  (Subscript[τ, z])=τ  where

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

Example . - Axis and Origin

(OverVector[F])=-F (OverHat[x])

Solution

Solution

({{(OverVector[τ])=(OverVector[r])×(OverVector[F])=(y (OverHat[y])+z (OverHat[z]))×(-F (OverHat[x]))=-y F(OverHat[y])× (OverHat[x])-z F (OverHat[z])× (OverHat[x])}, {=y F(OverHat[z])-z F (OverHat[y])=⟨0,-z, y⟩F}})

Solution

Similarly, the angular momentum of a particle relative to an origin

(OverVector[L])=(OverVector[r])×(OverVector[p])

can be written relative to an axis.  If the axis is the z axis then L about the axis is just the z component of L about the origin.  (Subscript[L, z])=L where

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

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=(∑_( i)) (Subscript[L, i]).  Using  I=(∑_( i)) (Subscript[m, i]) (Subscript[r, i]^2) we get the angular momentum of a rotating rigid body

L=I ω.

This is the result we had in our table relating rotations about a fixed axis to one dimensional linear motion.

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 Second Law

We can now, finally, derive the rotational equivalent of the second law  (Subscript[τ, net])=I α.  Start with the momentum form of the second law.

(Subscript[(OverVector[τ]), net]^ext)=(d/(dt))(Subscript[(OverVector[L]), tot])

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

(Subscript[τ, net])=(d/(dt))L.

Using  L=I ω and α=dω/dt we get our result.

(Subscript[τ, net])=I α

The Torque Due to Gravity

We saw earlier 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[(OverVector[τ]), grav])=(Subscript[(OverVector[r]), cm])×M (OverVector[g])

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

(Subscript[(OverVector[τ]), grav])=(∑_i) (Subscript[(OverVector[r]), i])×(Subscript[m, i]) (OverVector[g])=((∑_i) (Subscript[m, i]) (Subscript[(OverVector[r]), i]))×(OverVector[g])=(M(Subscript[(OverVector[r]), cm]))×(OverVector[g])=(Subscript[(OverVector[r]), cm])×M (OverVector[g])

Example . - Swinging Disk

Solution

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

I=(Subscript[I, cm])+m (d^2)=(1/2)m (R^2)+m (R^2)=(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

. - 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[(OverVector[τ]), net])= (OverVector[0])

For the examples we will consider all possible rotation will be in a plane and thus we only need to consider torques relative to an axis.

The Origin (or Axis) is Arbitrary

When considering an equilibrium problem sometimes the choice of axis is clear.  Often in isn't clear, though, and there isn't a 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 origin and the forces balance then the torques balance about any origin.  If the vector from one origin to another is (Subscript[(OverVector[r]), 0]).  If  (Subscript[(OverVector[r]), i]^′) is the vector from the new origin at (Subscript[(OverVector[r]), 0])  to the mass (Subscript[m, i]) and (Subscript[(OverVector[r]), i])  is from the first origin to the mass then

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

Take the net torques about these axes to be (Subscript[(OverVector[τ]), net]) and (Subscript[(OverVector[τ]), net]^′).  If  (Subscript[(OverVector[τ]), net])=(OverVector[0]) and (Subscript[(OverVector[F]), net])=(OverVector[0]) then (Subscript[(OverVector[τ]), net]^′)=(OverVector[0]).

(OverVector[0])=(Subscript[(OverVector[τ]), net])=(∑_i) (Subscript[(OverVector[r]), i])× (Subscript[(OverVector[F]), i])=(∑_i) (Subscript[(OverVector[r]), i]^′)× (Subscript[(OverVector[F]), i])+ (Subscript[(OverVector[r]), 0])×(∑_i) (Subscript[(OverVector[F]), i])=(Subscript[(OverVector[τ]), net]^′)+(OverVector[0])

This proves our result.

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))