example of reference frames in newtonian physics

Inertial Reference Frames in Newtonian Physics: Examples and Complete Worked Solutions

This companion develops the ideas of inertial and non-inertial reference frames through concrete Newtonian examples. The emphasis is on the Galilean relations

r = R + r′,

(1)

v = V + v′,

(2)

and

a = A + a′.

(3)

If the moving frame has constant velocity, A = 0 and therefore

a ′ = a.

(4)

This acceleration invariance is the central reason Newton’s equations have the same form in frames related by an ordinary Galilean transformation.

Figure 1. Two frames with parallel axes. When S′ translates at constant velocity V relative to inertial frame S, the positions satisfy r = R + r′.

Part I: Exercises

Exercise 1: Is the observation consistent with an inertial frame?

A puck is observed to move according to

r(t) = (2 + 3t)ex + (1 − 2t)ey m.

(5)

Find its velocity and acceleration. If the puck is known to be free of net real force, is this observation consistent with the observer being in an inertial frame? Does this one trajectory prove that the frame is inertial?

Exercise 2: Galilean position and velocity transformation

Frame S′ moves in the +x direction at a constant speed of 5 m∕s relative to inertial frame S. The origins coincide at t = 0. A particle has

x (t) = 20 + 8t m

(6)

in S.

Find x′(t) and the velocity measured in S′. Find x and x′ at t = 4 s.

Exercise 3: Acceleration invariance

In inertial frame S, a particle moves along the x axis according to

x (t) = 4 + 2t + t2 m.

(7)

Frame S′ moves at constant velocity 7 m∕s in the +x direction relative to S, with coincident origins at t = 0.

Find x′(t), v, v′, a, and a′. If the particle has mass 3 kg, find the net force in either inertial frame.

Exercise 4: Correct sign for an accelerating observer

A free particle moves in inertial frame S with constant velocity

v = 12 m ∕s.

(8)

The origin of frame S′ moves according to

R (t) = t2 m.

(9)

Thus the observer velocity is u(t) = 2t.

Find the particle’s relative velocity v′ and relative acceleration a′. If the particle mass is 4 kg, what apparent force must be introduced in S′ to write the equation in Newton-like form?

Exercise 5: Apparent force in an accelerating elevator

A 5 kg mass rests on the floor of an elevator accelerating upward at

2 A = 3.0 m ∕s .

(10)

Take g = 9.81 m∕s².

Find the normal force using (a) an inertial ground frame and (b) the accelerating elevator frame with an inertial force.

Figure 2. In the inertial frame the elevator and mass accelerate upward. In the elevator frame the mass is at rest, but the translational inertial force −mA must be included.

Exercise 6: Projectile inside a uniformly moving train

A train travels at constant velocity 20 m∕s relative to the ground. A passenger throws a ball vertically upward at 10 m∕s relative to the train. Neglect air resistance.

Write the motion in the train frame and in the ground frame. Find the total flight time and the horizontal distance traveled relative to the ground before the ball returns to its launch height.

Figure 3. A vertical throw in the train frame is a parabola in the ground frame. Both frames are inertial because their relative velocity is constant.

Exercise 7: Vector velocity addition

Frame S′ moves with velocity

V = 12e m ∕s x

(11)

relative to S. An object has velocity

v ′ = 5e m ∕s y

(12)

in S′.

Find its velocity and speed in S.

Exercise 8: Newton’s second law in two inertial frames

A 2 kg particle experiences the uniform real force

F = 6ex − 4ey N.

(13)

Frame S′ moves with constant velocity relative to inertial frame S.

Find the acceleration in S and S′. Explain why no inertial force is required in S′.

Exercise 9: Prove that uniformly translating frames form a family of inertial frames

Let S be inertial. Let the origin of S′ satisfy

R (t) = R0 + Vt,

(14)

where V is constant, and suppose the axes of S′ do not rotate relative to S.

Starting from r = R + r′, prove that a free particle that has constant velocity in S also has constant velocity in S′.

Exercise 10: A rotating frame is not inertial

A disk rotates at constant angular speed

ω = 2.0 rad∕s.

(15)

A 0.50 kg mass is fixed to the disk at radius 1.0 m.

Find its acceleration and required real radial force in an inertial frame. In the rotating disk frame the mass is stationary. What apparent force must be included to describe equilibrium there?

Figure 4. Constant angular speed does not make a rotating frame inertial. Points fixed in the rotating frame have centripetal acceleration relative to an inertial observer.

Exercise 11: How non-inertial is the Earth-fixed frame?

At Earth’s equator, use

−5 Ω = 7.292115 × 10 rad ∕s

(16)

and

R = 6.371 × 106 m.

(17)

Calculate the rotational acceleration Ω²R. Express it as a fraction of g = 9.81 m∕s². Explain why the Earth-fixed frame is not exactly inertial even though it is often treated as approximately inertial in elementary mechanics.

Exercise 12: Center-of-mass frame of an isolated system

Two particles move along one dimension. Their masses and velocities in an inertial laboratory frame are

m1 = 2 kg, v1 = 4 m∕s,

(18)

m2 = 3 kg, v2 = − 1 m ∕s.

(19)

Find the center-of-mass velocity and the two velocities in the center-of-mass frame. Verify that the total momentum in that frame is zero. If the system is isolated, explain why the center-of-mass frame is itself inertial relative to the laboratory frame.

Exercise 13: Momentum conservation under a Galilean transformation

A system of total mass M = 5 kg has constant total momentum

p = 10 kg m∕s

(20)

in frame S. Frame S′ moves at the constant speed

V = 3 m ∕s

(21)

in the same direction.

Find the total momentum p′ in S′. Show generally that if p is conserved and V is constant, then p′ is also conserved.

Exercise 14: An elastic collision in the center-of-mass frame

Two equal masses collide elastically in one dimension. In the laboratory frame the initial velocities are

u1 = 6 m ∕s, u2 = 0.

(22)

Find the center-of-mass velocity. Transform to the center-of-mass frame, solve the elastic collision there, and transform back to the laboratory frame.

Exercise 15: A freely falling frame in a uniform gravitational field

Near Earth’s surface, approximate the ground frame as inertial and take upward as positive. A freely falling particle has

a = − g.

(23)

Now choose a frame whose origin also accelerates downward with

A = − g.

(24)

Find the particle acceleration a′ in the falling frame. For a particle of mass m, identify the translational inertial force and show how it combines with the real gravitational force.

Exercise 16: What Galilean invariance does not mean

A particle moves at constant speed

v = 10 m ∕s

(25)

in inertial frame S. Frame S′ moves in the same direction at

V = 4 m ∕s.

(26)

Over a time interval

Δt = 3 s,

(27)

find the displacement measured in each frame. Are the displacements equal? What quantities relevant to Newton’s second law are invariant between the two frames?

Part II: Complete Worked Solutions

Solution 1: Is the observation consistent with an inertial frame?

Differentiate the position:

v = 3e − 2e m∕s. x y

(28)

The velocity is constant, so

a = 0.

(29)

If the puck is known to be free of net real force, this behavior is exactly what Newton’s first law predicts in an inertial frame. Thus the observation is consistent with the frame being inertial.

However, one free-particle trajectory does not by itself prove that the frame is inertial. An inertial frame is identified by the behavior of free particles generally, together with the absence of frame acceleration or rotation relative to another established inertial frame.

Solution 2: Galilean position and velocity transformation

For coincident origins at t = 0,

′ x = x − V t.

(30)

Here V = 5 m∕s, so

x′ = 20 + 8t − 5t.

(31)

Therefore

x′ = 20 + 3t m.

(32)

Differentiating,

v′ = 3 m ∕s.

(33)

At t = 4 s,

x = 20 + 8(4) = 52 m,

(34)

and

′ x = 20 + 3(4) = 32 m.

(35)

The position coordinates are different because the two frame origins have separated by V t = 20 m.

Solution 3: Acceleration invariance

The Galilean position transformation is

′ x = x − 7t.

(36)

Thus

′ 2 x = 4 − 5t + t m.

(37)

In S,

v = 2 + 2t m ∕s,

(38)

while in S′,

v′ = − 5 + 2t m ∕s.

(39)

Differentiating again gives

a = a′ = 2 m ∕s2.

(40)

For m = 3 kg,

F = ma = 6 N.

(41)

The velocities differ by the constant relative frame velocity, but the acceleration is identical.

Solution 4: Correct sign for an accelerating observer

The observer velocity is

u(t) = 2t.

(42)

The relative velocity is

v′ = v − u (t) = 12 − 2t m ∕s.

(43)

Therefore

a′ = − 2 m ∕s2.

(44)

This minus sign is essential: an observer accelerating in +x sees a free particle accelerate in −x relative to the observer.

The real net force on the particle is zero. In the accelerated frame, to write

ma ′ = Freal + Finertial,

(45)

we need

Finertial = − mA.

(46)

With m = 4 kg and A = 2 m∕s²,

Finertial = − 8 N.

(47)

The apparent force points opposite the frame acceleration.

Solution 5: Apparent force in an accelerating elevator

In the inertial ground frame, upward is positive. Newton’s second law gives

N − mg = mA.

(48)

Hence

N = m (g + A ).

(49)

Substitution gives

N = 5(9.81 + 3.00) = 64.05 N.

(50)

In the elevator frame the mass is at rest, so a′ = 0. The elevator frame accelerates upward, so the translational inertial force is downward:

Finertial = − mA = − 15.0 N.

(51)

The elevator-frame force balance is

N − mg − mA = 0,

(52)

which again gives

N = 64.05 N.

(53)

Both descriptions agree when the inertial force is handled consistently.

Solution 6: Projectile inside a uniformly moving train

Take the launch point as the origin at t = 0. In the train frame,

x ′ = 0,

(54)

1 y′ = 10t − -gt2. 2

(55)

The ball returns to the launch height when

1- 2 10t − 2 gt = 0.

(56)

For the nonzero root,

20 T = ---= 2.039 s. g

(57)

In the ground frame, the horizontal train speed is added:

x = 20t,

(58)

y = 10t − 1-gt2. 2

(59)

Thus the ground-frame horizontal distance is

Δx = 20T = 40.77 m.

(60)

The train frame sees a vertical path; the ground frame sees a parabola. Both are inertial because their relative velocity is constant.

Solution 7: Vector velocity addition

The Galilean velocity relation is

′ v = V + v .

(61)

Therefore

v = 12e + 5e m ∕s. x y

(62)

The speed is

-------- |v | = √ 122 + 52 = 13 m ∕s.

(63)

Solution 8: Newton’s second law in two inertial frames

Newton’s second law in S gives

F a = -- . m

(64)

Thus

2 a = 3ex − 2ey m ∕s .

(65)

Because S′ moves at constant velocity relative to S,

′ a = a.

(66)

Hence

a ′ = 3e − 2e m ∕s2. x y

(67)

No inertial force is needed because S′ is also inertial.

Solution 9: Prove that uniformly translating frames form a family of inertial frames

Start with

r = R + r′.

(68)

Since

R = R0 + Vt,

(69)

we have

dR V = ---- dt

(70)

and

dV--= 0. dt

(71)

Differentiating the position relation once,

v = V + v′.

(72)

Differentiating again,

a = a′.

(73)

If a free particle has a = 0 in S, then it also has

′ a = 0

(74)

in S′. Therefore a nonrotating frame translating uniformly relative to an inertial frame is also inertial in Newtonian mechanics.

Solution 10: A rotating frame is not inertial

A point fixed at radius r on a disk rotating at constant angular speed has centripetal acceleration

a = ω2r.

(75)

Thus

a = (2.0)2(1.0) = 4.0 m ∕s2

(76)

directed inward.

The real inward force is

F = ma = (0.50)(4.0) = 2.0 N.

(77)

In the rotating frame the mass is stationary, yet the real inward force remains. To write an equilibrium equation in the rotating frame, a centrifugal apparent force of

Fcf = 2.0 N

(78)

must be introduced outward. The need for this inertial force shows directly that the rotating frame is non-inertial.

Solution 11: How non-inertial is the Earth-fixed frame?

The equatorial rotational acceleration is

arot = Ω2R.

(79)

Substitution gives

arot = 0.03388 m ∕s2.

(80)

Relative to standard gravity,

arot = 0.00345. g

(81)

Thus the rotational acceleration is about 0.345 percent of g at the equator.

The Earth-fixed frame rotates, so it is not exactly inertial. For many short-duration introductory problems the resulting Coriolis and centrifugal corrections are small enough to neglect, so the ground is treated as approximately inertial.

Solution 12: Center-of-mass frame of an isolated system

The total mass is

M = 2 + 3 = 5 kg.

(82)

The center-of-mass velocity is

m1v1-+-m2v2-- VCM = M .

(83)

Thus

2(4) + 3(− 1) VCM = ------------- = 1.0 m ∕s. 5

(84)

The velocities in the center-of-mass frame are

′ v1 = 4 − 1 = 3 m ∕s,

(85)

v′2 = − 1 − 1 = − 2 m∕s.

(86)

The total momentum there is

p′ = 2(3) + 3(− 2) = 0.

(87)

For an isolated system, total momentum is constant, so V _CM is constant. Therefore the center-of-mass frame translates at constant velocity relative to the inertial laboratory frame and is itself inertial, provided its axes do not rotate.

Solution 13: Momentum conservation under a Galilean transformation

For a system of total mass M, the total momentum transforms as

p′ = p − M V.

(88)

Numerically,

p′ = 10 − 5(3) = − 5 kg m ∕s.

(89)

The two frames assign different momentum values to the same system.

Differentiate the transformation:

′ dp- dp- dV- dt = dt − M dt .

(90)

For a Galilean inertial transformation, V is constant, so

dV- dt = 0.

(91)

If p is conserved, dp∕dt = 0, and therefore

′ dp- = 0. dt

(92)

Momentum conservation holds in both inertial frames even though the numerical momenta differ.

Solution 14: An elastic collision in the center-of-mass frame

For equal masses,

6-+-0 VCM = 2 = 3 m ∕s.

(93)

The initial center-of-mass-frame velocities are

′ u1 = 6 − 3 = 3 m ∕s,

(94)

′ u2 = 0 − 3 = − 3 m ∕s.

(95)

For a one-dimensional elastic collision of equal masses in the center-of-mass frame, the velocities reverse:

′ v1 = − 3 m ∕s,

(96)

v′ = 3 m ∕s. 2

(97)

Transform back by adding V _CM = 3 m∕s:

v = 0, 1

(98)

v2 = 6 m ∕s.

(99)

The familiar laboratory result is recovered: the moving equal mass stops and the initially stationary mass leaves with the original speed.

Solution 15: A freely falling frame in a uniform gravitational field

The acceleration transformation for a translating frame is

a′ = a − A.

(100)

Here

a = − g

(101)

and

A = − g.

(102)

Therefore

a ′ = 0.

(103)

The free particle is at rest or moves uniformly relative to the freely falling frame.

The real gravitational force is

Fg = − mg.

(104)

The translational inertial force is

Finertial = − mA = +mg.

(105)

Hence

Fg + Finertial = 0.

(106)

The falling frame is still non-inertial in Newtonian mechanics because its origin accelerates relative to the ground-frame approximation.

Solution 16: What Galilean invariance does not mean

In frame S, the displacement over Δt = 3 s is

Δx = v Δt = 10 (3 ) = 30 m.

(107)

In S′, the particle speed is

v′ = v − V = 10 − 4 = 6 m∕s.

(108)

Thus

Δx ′ = v′Δt = 6 (3 ) = 18 m.

(109)

The displacements are not equal.

What is invariant under an ordinary Galilean transformation is the time interval,

Δt ′ = Δt,

(110)

and, because V is constant, the acceleration:

a ′ = a.

(111)

Consequently Newton’s second law retains the same form in both inertial frames for Newtonian force laws. Galilean invariance does not mean that position, displacement, velocity, momentum, or kinetic energy have the same numerical value in every inertial frame.

References

[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[2] S. T. Thornton and J. B. Marion, Classical Dynamics of Particles and Systems, 5th ed., Brooks/Cole, 2004.

[3] D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.

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