Chapter

Rotational Kinematics and Energy

Blinn College - Physics 1401 - Terry Honan

. - Kinematics of Rotations about a Fixed Axis

Rigid Bodies and Rotations in General

The distance between any two positions in a rigid body is fixed.  A book can be viewed as a rigid body as long as it is kept closed; when it is opened then the distance between a point on the back cover and a point on the front cover varies and it is not rigid.

A rotation is described by an axis and an angle.  An axis is a line.  The axis of rotation of a door is its hinge.  The axis of a tire is its axle.  Often in a planar diagram we will draw an axis as a point.  The axis is then the line perpendicular to that plane through the point.  A rotation is about some axis and by some angle.  Note that when a rigid body rotates different points move different distances.  The distance a point moves s is proportional to the (perpendicular) distance from the axis r, but the ratio s/r is the same for any two points.  This ratio is just the angle of rotation in radians.

Kinematical Variables

To understand rotational kinematics it is essential to appreciate the analogy to one dimensional kinematics.  Recall that a one dimensional vector is a real number and that its direction is given by its sign.  The rotational analog of the position is the angle of rotation.  The other kinematical variables follow:

({{ , One Dimensional
Linear Motion, Rotations about
a Fixed Axes}, {Position, x   (m), θ  (angle in rad)}, {Velocity, v   (m/s), ω   (angular  velocity in rad/s)}, {Average, (Subscript[v, ave])=((Δ x)/(Δ t)), (Subscript[ω, ave])=((Δ θ)/(Δ t))}, {Instantaneous, v=(lim_(Δ t->0)) ((Δ x)/(Δ t)), ω=(lim_(Δ t->0)) ((Δ θ)/(Δ t))}, {Acceleration, a   (m/(s^2)), α   (angular acceleration in rad/(s^2))}, {Average, (Subscript[a, ave])=((Δ v)/(Δ t)), (Subscript[α, ave])=((Δ ω)/(Δ t))}, {Instantaneous, a=(lim_(Δ t->0)) ((Δ v)/(Δ t)), α=(lim_(Δ t->0)) ((Δ ω)/(Δ t))}})

Constant  Angular Acceleration

Since the rotational variables θ, ω and α are interrelated the same as x, v and a, we can find expressions for constant angular acceleration.

({{One Dimensional
Linear Motion, Rotations about
a Fixed Axes}, { v=(Subscript[v, 0])+a t
Δ x=(1/2)((Subscript[v, 0])+v)  t
Δ x=(Subscript[v, 0]) t+(1/2)a (t^2)
(v^2)=(Subscript[v, 0]^2)+2 a  Δ x,  ω=(Subscript[ω, 0])+α t
Δ θ=(1/2)((Subscript[ω, 0])+ω)  t
Δ θ=(Subscript[ω, 0]) t+(1/2)α (t^2)
(ω^2)=(Subscript[ω, 0]^2)+2 α  Δ θ}})

Example . - Decelerating Ceiling Fan

Solution

(Subscript[ω, 0])=20 (rev/((QuantityUnit)))×((2π (QuantityUnit))/rev)×((1(QuantityUnit))/(60(QuantityUnit)))=2.0944(((QuantityUnit))/((QuantityUnit)))

ω=8 (rev/((QuantityUnit)))×((2π (QuantityUnit))/rev)×((1(QuantityUnit))/(60(QuantityUnit)))=0.83776(((QuantityUnit))/((QuantityUnit)))

ω=(Subscript[ω, 0])+α t   ⟹   α=((ω-(Subscript[ω, 0]))/t)=-0.157(((QuantityUnit))/(((QuantityUnit))^2))

Solution

# of rotations = ((Δθ)/(2π))

Δ θ=(1/2)((Subscript[ω, 0])+ω)  t=11.729(QuantityUnit)   ⟹   ((Δθ)/(2π))=1.87

Relation Between Linear and Rotational Variables

In chapter 6 we described circular motion in terms of centripetal and tangential directions, where the centripetal direction is toward the center of the circle and the tangential direction is in the direction of motion, meaning the direction of the velocity. A point on a rigid body a distance r from the center moves in a circle of radius r, so that discussion is relevant here.

The velocity is purely tangential, where (Subscript[v, t]) is just the speed, which is just (Subscript[v, t])=(lim_(Δ t->0))((Δ s)/(Δ t)) where Δ s is the small arc length traveled in the small time Δ t. An angle in radians is defined as the arc length per unit radius, θ=s/r. It follows that the small arc length Δ s is related to small angle Δ θ by  Δ s=r Δ θ.  Since ω=(lim_(Δ t->0)) ((Δ θ)/(Δ t)), we get

(Subscript[v, t])=(lim_(Δ t->0))((Δ s)/(Δ t))=(lim_(Δ t->0))((r Δ θ)/(Δ t))=r ((lim_(Δ t->0))((Δ θ)/(Δ t)))=r ω

In chapter 6 we saw the acceleration had a centripetal component (Subscript[a, c]) and a tangential component that depended on the change in the speed. The centripetal component (Subscript[a, c]) can be written in terms of the radius r and the angular velocity ω.

(Subscript[a, c])=((v^2)/r)=((((r ω))^2)/r)=(ω^2)r

The tangential component (Subscript[a, t]) can similarly be written in terms of the radius r and the angular acceleration α.

(Subscript[a, t])=(lim_(Δ t->0))((Δ (Subscript[v, t]))/(Δ t))=(lim_(Δ t->0))((Δ (r ω))/(Δ t))=r (lim_(Δ t->0))((Δ ω)/(Δ t)) =r α.

The tangential and centripetal directions are perpendicular. It follows that the magnitude of the acceleration can be found by applying the Pythagorean theorem to the perpendicular components (Subscript[a, c]) and (Subscript[a, t]).

a=(Sqrt[(Subscript[a, c]^2)+(Subscript[a, t]^2)])

Example . - Decelerating Ceiling Fan (continued)

Solution

ω=0.85(QuantityUnit)/(QuantityUnit) ,  r=0.90(QuantityUnit)  and  α=-0.157(((QuantityUnit))/(((QuantityUnit))^2))  (from part (a))

(Subscript[v, t])=r ω=0.765(QuantityUnit)/(QuantityUnit)

(Subscript[a, c])=(ω^2)r=0.65025(((QuantityUnit))/(((QuantityUnit))^2))  and  (Subscript[a, t])=r α=-0.14137(((QuantityUnit))/(((QuantityUnit))^2))   ⟹   a=(Sqrt[(Subscript[a, c]^2)+(Subscript[a, t]^2)])=0.665(((QuantityUnit))/(((QuantityUnit))^2))

. - Dynamics of Rigid Bodies Rotating about an Axis

Summary and Analogy with One Dimensional Motion

({{ , One Dimensional
Linear Motion, Rotations about
a Fixed Axes}, {Kinematics, x, v, a, θ, ω, α}, {Force, F, τ   (torque)}, {Inertia, m, I   (moment of inertia)}, {Momentum, p=m v, L=I ω   (angular momentum)}, {Second
Law, (Subscript[F, net])=m a
(Subscript[F, net])=(lim_(Δ t->0))((Δ (OverVector[p]))/(Δ t)), (Subscript[τ, net])=I α
(Subscript[τ, net])=(lim_(Δ t->0))((Δ L)/(Δ t))}, {Conservation
of Momentum, (Subscript[F, net]^ext)=0
⟹ Δ (Subscript[p, tot])=0, (Subscript[τ, net]^ext)=0
⟹ Δ (Subscript[L, tot])=0}, {Kinetic Energy, K=(1/2)m (v^2), K=(1/2)I (ω^2)}, {Work, W= F Δ x, W= τ Δ θ}, {Work–Energy Theorem, (Subscript[W, net])=Δ K, (Subscript[W, net])=Δ K}, {Power, (Subscript[𝒫, ave])=(W/(Δ t))=F v, (Subscript[𝒫, ave])=(W/(Δ t))=τ ω}})

This table is an extension of the preceding tables for kinematics.  Now we consider dynamics.  Dynamical quantities are things like force and mass.  The rotational analog of force is called torque and the rotational analog of mass is the moment of inertia.  These two quantities are undefined in the table; their definitions follow.  For all the other quantities, the above table serves as the definitions of the variables.

Kinetic Energy and the Definition of the Moment of Inertia

Consider a rigid body consisting of point masses (Subscript[m, i]).  The perpendicular distance from the axis to the (i^th) mass is (Subscript[r, i]).  If the rigid  body  rotates with angular velocity ω then the speed of the (i^th) mass is

(Subscript[v, i])=(Subscript[r, i]) ω.

The total kinetic energy is the sum of  the kinetic energies of all the masses.  Using the above expression for the speed we get

({{K=(1/2)( (Subscript[m, 1]) (Subscript[v, 1]^2)+(Subscript[m, 2]) (Subscript[v, 2]^2)+…)=(1/2) ((Subscript[m, 1]) (Subscript[r, 1]^2)(ω^2)+(Subscript[m, 2]) (Subscript[r, 2]^2)(ω^2)+…)=(1/2)((Subscript[m, 1]) (Subscript[r, 1]^2)+(Subscript[m, 2]) (Subscript[r, 2]^2)+…)(ω^2).}})

Using our desired expression for the kinetic energy  K=(1/2)I (ω^2) we get the expression for I moment of inertia for a rigid body about some axis.

I=(Subscript[m, 1]) (Subscript[r, 1]^2)+(Subscript[m, 2]) (Subscript[r, 2]^2)+…=∑m (r^2)

This expression is for a discrete distribution; this means that the distribution is a collection of point masses.

Moment of Inertia

The moment of inertia is a property of a rigid body and an axis. The further the mass is from the axis, the larger the moment.

I=(Subscript[m, 1]) (Subscript[r, 1]^2)+(Subscript[m, 2]) (Subscript[r, 2]^2)+…=∑m (r^2)

Example . - A Discrete Distribution

Solution

I=(Subscript[m, 1])(Subscript[x, 1]^2)+(Subscript[m, 2])(Subscript[x, 2]^2)+(Subscript[m, 3])(Subscript[x, 3]^2)=3(QuantityUnit) (((-2(QuantityUnit)))^2)+5(QuantityUnit) (0^2)+2(QuantityUnit) (((4(QuantityUnit)))^2)=44(QuantityUnit) (((QuantityUnit))^2)

Solution

I=(Subscript[m, 1])(Subscript[y, 1]^2)+(Subscript[m, 2])(Subscript[y, 2]^2)+(Subscript[m, 3])(Subscript[y, 3]^2)=3(QuantityUnit) (((-4(QuantityUnit)))^2)+5(QuantityUnit) (((-3(QuantityUnit)))^2)+2(QuantityUnit) (0^2)=93(QuantityUnit) (((QuantityUnit))^2)

Solution

({{I=(Subscript[m, 1]) ((Subscript[x, 1]^2)+(Subscript[y, 1]^2))+(Subscript[m, 2])((Subscript[x, 2]^2)+(Subscript[y, 2]^2))+(Subscript[m, 3])((Subscript[x, 2]^2)+(Subscript[y, 2]^2))}, {=3(QuantityUnit) ((((-2(QuantityUnit)))^2)+(((-4(QuantityUnit)))^2))+5(QuantityUnit) ((0^2)+(((-3(QuantityUnit)))^2))+2(QuantityUnit) ((((4(QuantityUnit)))^2)+(0^2))}, {=137(QuantityUnit) (((QuantityUnit))^2)}})

K=(1/2)I (ω^2)=616.5(QuantityUnit)

Example . - Moments for Different Axes

Solution

(Subscript[I, 3])<(Subscript[I, 1])<(Subscript[I, 2])

Moments of Inertia for Uniform Bodies

Hoop or Thin-shelled Hollow Cylinder about Perpendicular Axis through Center

First consider a hoop of mass M and radius R rotating about a perpendicular axis through the center.  r is the distance from the axis to the infinitesimal mass dm.  All the mass is at the same distance

r=R=constant.

It is possible to find I without actually performing an integral.

I=∑m (r^2)=∑m  (R^2)=(R^2)∑m

Since M=∑m we get

I=M (R^2).

Now consider a thin-shelled hollow cylinder about the central axis.  It is still true that all the mass is the same perpendicular distance of  r=R from the axis and the above formula still applies.

Disk or Solid Cylinder about Perpendicular Axis through the Center

It should now be clear that the moment for a disk should be the same as a solid cylinder.  We can break up a disk into concentric thin rings of radius r with thickness dr.  The limits of integration become

0≤r≤R.

If all the mass were at r=0, then I=0. If all were at r=R, then it would be the same as a hoop and I=M (R^2). It then must satisfy: 0<I<M (R^2). It can be shown using calculus that

I=(1/2)M (R^2).

Table of Moments of Inertia for Uniform Distributions with Different Geometries and Axes

By a uniform distribution of mass we mean the density is constant throughout the body.

. - Energy and Rigid Bodies

Gravitational Potential Energy

It is a straightforward matter to find the potential energy of a rigid body.

U=(∑_i) (Subscript[m, i]) g (Subscript[y, i])=g (∑_i) (Subscript[m, i])(Subscript[y, i])=g M (Subscript[y, cm])

Here M is the total mass and (Subscript[y, cm]) is the height of the center of mass.  It follows that the total potential energy of a rigid body is

U=M g (Subscript[y, cm]).

This is easy to interpret.  When calculating the potential energy of a rigid body we treat the body as if all its mass is at its center of mass.

Example . - Swinging Rod

Solution

E=(1/2)I (ω^2)+m g (Subscript[y, cm])

(Subscript[E, i])=(Subscript[E, f])  ⟹   0+m g (L/2)=(1/2)I (Subscript[ω, f]^2)+0

m g (L/2)=(1/2)I (Subscript[ω, f]^2)=(1/2)((1/3)m (L^2))(Subscript[ω, f]^2)  ⟹  (Subscript[ω, f])=(Sqrt[((3 g)/L)])

v=r ω=L (Subscript[ω, f])=L (Sqrt[((3 g)/L)])=(Sqrt[3 g L])

Rotation with Translation

If a body is rotating and translating then there is kinetic energy in the rotation and in the translation. The total kinetic energy (Subscript[K, tot]) is the sum of (Subscript[K, tran]), the translational kinetic energy and (Subscript[K, rot]), the rotational kinetic energy. If v is the speed of the center of mass and I is the moment of inertia through the center of mass then

(Subscript[K, tot])=(Subscript[K, tran])+(Subscript[K, rot])=(1/2)M (v^2)+(1/2)I (ω^2)

Rolling Motion

In the case of a rolling body we have a rolling constraint that a body rolls without slipping.  This is: the arc length along the rolling radius of the body is the same as the distance Δ x it moves along the surface it rolls on.  If it rotates by an angle Δ θ then the arc length is R Δ θ.  The constraint becomes

R Δ θ=Δ x.

Since the velocity is v=(lim_(Δ t->0))Δ x/Δ t and the angular velocity is ω=(lim_(Δ t->0))Δ θ/Δ t the rolling constraint becomes

R ω=v

and similarly, with the acceleration and angular acceleration and we get

R α=a.

(Subscript[K, tot])=(Subscript[K, tran])+(Subscript[K, rot])=(1/2)M (v^2)+(1/2)I (ω^2)=(1/2)(M+(I/(R^2))) (v^2)

The last expression above follows from v=R ω and replacing the angular velocity ω with v/R. The total kinetic energy depends on the total mass M and the speed of the center of mass. The rotational kinetic energy depends on the moment of inertia about the axis through the center of mass.

Example . - Rigid Body Races

Solution

({{Object    I=κ M (R^2), κ}, {Solid Sphere, 2/5}, {Solid Cylinder, 1/2}, {Hollow Spherical Shell, 2/3}, {Hollow Cylndrical Shell, 1}})

U=M g (Subscript[y, cm])

(Subscript[K, tot])=(Subscript[K, trans])+(Subscript[K, rot])=(1/2)M (v^2)+(1/2)I (ω^2)

(Subscript[K, tot])=(1/2)M (v^2)+(1/2)I (ω^2)=(1/2)M (v^2)+(1/2)(κ M (R^2)) ((((v/R)))^2)=(1/2)(1+κ)M (v^2).

E=(Subscript[K, tot])+U=(1/2)(1+κ)M (v^2)+M g (Subscript[y, cm]).

(Subscript[E, i])=(Subscript[E, f])  ⟹  0+M g h=(1/2)(1+κ)M (v^2)+0  ⟹  v=(Sqrt[((2 g h)/(1+κ))]).

Previously, when considering pulleys we used ideal pulleys, which were frictionless and light, meaning that the mass of the pulley was negligible compared to the other masses in the problem. Now we consider an example with a pulley with mass that is not small.

Example . - A Pulley with Mass

Solution

E=(1/2)m (v^2)+(1/2)I (ω^2)+m g y

R Δ θ=Δ x

R ω=v

I=(1/2)M (R^2)

E=(1/2)m (v^2)+(1/2)×(1/2)M (R^2) ((((v/R)))^2)+m g y=(1/2)(m+(1/2)M) (v^2)+m g y

(Subscript[E, i])=(Subscript[E, f])  ⟹  0+m g h=(1/2)(m+(1/2)M) (Subscript[v, f]^2)+0  ⟹  (Subscript[v, f])=(Sqrt[((2 m g h)/(m+M/2))])