Chapter

Circular Motion and More Applications

Blinn College - Physics 2325 - Terry Honan

This chapter discusses the kinematics and dynamics of circular motion, and considers the problem of motion in an accelerated frame of reference.  More applications on the second law will also be presented.

. - Kinematics of Circular Motion

Uniform Circular Motion - Centripetal Acceleration

We will first consider the case of uniform circular motion, where uniform means that the speed stays constant.  Even with a constant speed there an acceleration, since the direction is changing. To find the acceleration we will first consider a finite time Δ t and the average velocity and average acceleration.  We will then let the time approach zero; the average values will then approach the instantaneous values.

Take (Subscript[(OverVector[r]), i]) and (Subscript[(OverVector[r]), f]) to be the initial and final position vectors at times (Subscript[t, i]) and (Subscript[t, f]) where Δ t=(Subscript[t, f])-(Subscript[t, i]).  The velocities at these times are (Subscript[(OverVector[v]), i]) and (Subscript[(OverVector[v]), f]); since the speed is constant these have the same magnitude.

Take the angle between the two position vectors as θ.  In circular motion the velocity is perpendicular to the position vector; it follows that θ is also the angle between the velocities.  The sides opposite the angle θ in the triangles are

Δ(OverVector[r])=(Subscript[(OverVector[r]), f])-(Subscript[(OverVector[r]), i])=(Subscript[(OverVector[v]), ave]) Δ t  and  Δ(OverVector[v])=(Subscript[(OverVector[v]), f])-(Subscript[(OverVector[v]), i])=(Subscript[(OverVector[a]), ave]) Δ t

Looking at just the magnitudes of these vectors we see a pair of isosceles triangles and since the angle is the same for both, we have similar triangles.

Comparing the ratios of similar sides of similar triangles gives

(((Subscript[a, ave]) Δ t)/((Subscript[v, ave]) Δ t))=(v/r).

This can be rewritten as

(Subscript[a, ave])=((v (Subscript[v, ave]))/r).

As Δ t->0 the average velocity and average acceleration approach their instantaneous values, so the magnitudes of the average quantities approach the instantaneous magnitudes.

(Subscript[v, ave])->v  and  (Subscript[a, ave])->a

With this we can write the value of the instantaneous acceleration as

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

As Δ t->0 it is also true that θ->0.  It follows that the velocity and acceleration vectors are perpendicular.  The direction of the acceleration is toward the center.  We will refer to this direction as centripetal and denote the centripetal acceleration as (Subscript[a, c]).

Many modern texts avoid the use of the term centripetal and have adopted the term radial instead. The radial direction could be outward or inward and the term centripetal is unambiguously toward the center. The reason for avoiding the word centripetal is because students confuse it with centrifugal. The centrifugal force is the outward false force associated with the centripetal acceleration.

Uniform Circular Motion - Period, Speed and Radius

For uniform circular motion we can relate the period,  speed and the radius.  This will also give us an alternative expression for the centripetal acceleration in terms of the period.  The period T is defined as the time needed for each full revolution.

It is a simple matter to relate the period to the radius and speed.  Since the speed  is uniform, is the same as the average speed; this is the distance traveled divided by the total time.  The distance traveled in one period is one circumference.  It follows that

v=((2 π r)/T).

Inserting this into our expression for the centripetal acceleration  (Subscript[a, c])=(v^2)/r gives

(Subscript[a, c])=(((((2 π)/T)))^2) r.

This will prove to be a useful expression.  The student should keep in mind that this only applies to uniform circular motion.

Example . - Centripetal Acceleration at the Equator

Solution

r=(Subscript[R, E])=6.38×(10^6)(QuantityUnit)  and  T=24×3600(QuantityUnit)=8640(QuantityUnit)   ⟹   (Subscript[a, c])=(((((2 π)/T)))^2) r=0.0337(QuantityUnit)/(((QuantityUnit))^2)

General Circular Motion - Tangential Acceleration

Now consider the more general case of circular motion where we allow the speed to change.  Generally, the component of the acceleration perpendicular to the motion is related to the change in direction and the component parallel to the direction of motion is related to the change in speed.

We will choose orthogonal (perpendicular) coordinates in tangential and centripetal directions.  The centripetal direction is toward the center and the tangential direction is the direction of motion.  If the origin is chosen at the center of the circle then the direction of the position vector (OverVector[r]) is the unit radial vector (OverHat[r]); this points away from the center, so the unit vector in the centripetal direction (Subscript[(OverHat[u]), c]) is opposite this:

(Subscript[(OverHat[u]), c])=-(OverHat[r]).      (unit centripetal vector)

The tangential direction is the direction of motion, so it is the direction of the velocity vector (OverVector[v]):

(Subscript[(OverHat[u]), t])=(OverHat[v]).       (unit tangential vector)

The component of the acceleration in the centripetal direction must be the same as in the uniform case

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

The velocity is purely tangential

(OverVector[v])=v (Subscript[(OverHat[u]), t]).

The component of the acceleration parallel to the direction of motion depends on the change in the speed; it is the time derivative of the speed.

(Subscript[a, t])=((dv)/(dt))

Combining these two components we get a general expression for the acceleration in circular motion.

(OverVector[a])=(Subscript[a, c]) (Subscript[(OverHat[u]), c])+(Subscript[a, t]) (Subscript[(OverHat[u]), t])=((v^2)/r) (Subscript[(OverHat[u]), c])+((dv)/(dt)) (Subscript[(OverHat[u]), t])

General Two Dimensional Motion and the Effective Radius

Now we consider the most general problem of motion in a plane.  Any trajectory may be approximated by a circle at some position.  We refer to the radius of that circle as the effective radius.  We may choose centripetal and tangential directions as before and apply the above expression for the acceleration to the general two dimensional problem.  If the direction is not changing at some instant, then the effective radius is infinite; the centripetal acceleration then becomes zero.

(OverVector[a])=(Subscript[a, c]) (Subscript[(OverHat[u]), c])+(Subscript[a, t]) (Subscript[(OverHat[u]), t])=((v^2)/r) (Subscript[(OverHat[u]), c])+((dv)/(dt)) (Subscript[(OverHat[u]), t])

. - Dynamics of Circular Motion

Newton’s Second Law (Subscript[(OverVector[F]), net])= m (OverVector[a])  is the foundation of classical mechanics and, of course, it must still apply when we have circular motion.  All that is new here is we now have new formulas for the acceleration.  Now we will consider examples of dynamics with circular motion.

Examples with Uniform Circular Motion

Example . - Block on a Rotating Disk

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  (Subscript[f, s])=m (Subscript[a, c])

(Subscript[F, net,ver])=m (Subscript[a, ver])=0  ⟹  N=m g

(Subscript[a, c])=(((((2 π)/T)))^2)r

(Subscript[f, s])≤(Subscript[μ, s]) N  ⟹  m (Subscript[a, c])≤(Subscript[μ, s]) m g  ⟹  (Subscript[a, c])≤(Subscript[μ, s]) g  ⟹  (((((2 π)/T)))^2) r≤(Subscript[μ, s]) g

(((((2 π)/T)))^2) (Subscript[r, max])=(Subscript[μ, s]) g  ⟹  (Subscript[r, max])=(((Subscript[μ, s]) g)/(((2 π/T))^2))=0.434(QuantityUnit)

Example . - Rotating Vertical Cylinder in an Amusement Park

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  N=m (Subscript[a, c])

(Subscript[F, net,ver])=m (Subscript[a, ver])=0  ⟹  (Subscript[f, s])=m g

(Subscript[f, s])≤(Subscript[μ, s]) N  ⟹  m g ≤(Subscript[μ, s]) m (Subscript[a, c])  ⟹  g≤(Subscript[μ, s]) (Subscript[a, c])  ⟹  g≤(Subscript[μ, s]) ((v^2)/R)

g=(Subscript[μ, s]) ((Subscript[v, min]^2)/R)   ⟹  (Subscript[v, min])=(Sqrt[((R g)/(Subscript[μ, s]))])

Nonuniform Examples

Example . - Car at the Bottom of a Trough

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  N-m g=m ((v^2)/R)  ⟹  N=m g+m ((v^2)/R)

Example . - Car at the Top of a Hill

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  m g-N=m ((v^2)/R)  ⟹  N=m g-m ((v^2)/R)

N≥0  ⟹  v≤(Sqrt[R g])  ⟹  (Subscript[v, max])=(Sqrt[R g])

Example . - Roller-Coaster Loop-to-Loop

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  m g+N=m ((v^2)/R)  ⟹  N=m ((v^2)/R)-m g

N≥0  ⟹  v≥(Sqrt[R g])  ⟹  (Subscript[v, min])=(Sqrt[R g])

Example . - Ball Swung in Vertical Circle

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  m g+T=m ((v^2)/L)  ⟹  T=m ((v^2)/L)-m g

T≥0  ⟹  v≥(Sqrt[L g])  ⟹  (Subscript[v, min])=(Sqrt[L g])

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  m g cos θ+T=m ((v^2)/L)  ⟹  T=m ((v^2)/L)-m g cos θ

. - Accelerated Frames and False Forces

False Force and Acceleration

Newton's second law applies to inertial (non-accelerated) frames.  In accelerated frames there are false forces.  In any case of acceleration there is a false force opposite the acceleration.  We will refer to as an artificial gravity.

(Subscript[(OverVector[g]), art])=-(OverVector[a])

The false force is  m (Subscript[(OverVector[g]), art]).  In a braking car the acceleration is backward so the false force is forward.  In the case of circular motion the acceleration is centripetal (toward the center) so the false force is outward.  The false force associated with a centripetal acceleration is called a centrifugal force.

The Principle of Equivalence

The principle of equivalence states that a constant acceleration is equivalent to a uniform gravitational field in the direction opposite the acceleration.

(Subscript[(OverVector[g]), art])=-(OverVector[a])

The equivalence principle states that this artificial gravity for a uniformly accelerated frame is indistinguishable from a real uniform gravitational field in the opposite direction.  A rocket in space accelerating upward at  9.80 m/(s^2) is indistinguishable from standing in a uniform gravity of the same magnitude.

Effective Gravity

Where there is acceleration and gravity we can define an effective gravity.

(Subscript[(OverVector[g]), eff])=(OverVector[g])+(Subscript[(OverVector[g]), art])

The earth is a rotating frame and there is a centrifugal force.  When we drop a plum-bob it does not hang in the direction of gravity but in the direction of the effective gravity.  Clearly the centrifugal force is significantly smaller than actual gravity.

Example . - Rotating Space Station

Solution

(Subscript[g, art])=(Subscript[a, c])=(((((2π)/T)))^2)r

(Subscript[g, art])=g=9.80(((QuantityUnit))/(((QuantityUnit))^2))  and  r=R=150(QuantityUnit)  ⟹   T=2π(Sqrt[(R/g)])=24.6(QuantityUnit)

Nonuniform Acceleration and Velocity Dependent Forces

Note that in an accelerated frame with a nonuniform acceleration there are additional velocity-dependent false forces other than the artificial gravity discussed here.  For instance, in the case of a rotating frame the artificial gravity force is the outward centrifugal force but there is also the velocity dependent force called the Coriolis force.  The Coriolis force is responsible for weather patterns having different senses of rotation in different hemispheres.