__Linear acceleration__

At t = 0 a 100 g ball is thrown upward with initial speed v
= 2 m/s in an open-platform elevator which at that time is moving downward with
v = 3 m/s and accelerating downward with an acceleration of magnitude a = 3 m/s^{2}.
A drag force with magnitude F_{drag} = 0.8 N acts on the ball.

(a) What is the net force **F**_{net} acting on the ball in the
frame of the elevator just after it has been thrown?

(b) What is the net force **F**_{net} acting on the ball in the
frame of a person standing on the ground just after it has been thrown?

Solution:

- Concepts:

Motion in an accelerating frame - Reasoning:

Fictitious forces appear in an accelerating frame.

In the inertial frame the ball moves downward just after it is thrown, the drag force which opposes the relative motion of the ball and the air therefore points upward. - Details of the calculation:

Let the upward direction be the positive y –direction.

(a)**F**_{net}= m**g**- m**a**+**F**_{drag }= (-mg + ma + F_{drag})**j**= (-0.1*6.8 N + 0.8 N)**j**

= 0.12 N**j**. The ball accelerates upward in the frame of the elevator.

(b)**F**_{net}= m**g**+**F**_{drag }= (-mg + Fd_{rag})**j**= (-0.1*9.8 N + 0.8 N)**j**= 0.18 N**j**.

The ball accelerates downward in the frame of the person on the ground.

A hauling truck is traveling on a level road. The driver suddenly applies the
brakes, causing the truck to decelerate by an amount g/2. This causes a box in
the rear of the truck to slide forward. If the coefficient of sliding friction
between the box and the truck bed is ⅓, find the acceleration of the box
relative to

(a) the truck and

(b) the road.

Solution:

- Concepts:

Motion in an accelerating frame. - Reasoning:

In an accelerating frame fictitious forces appear. The net force in such a frame is**F**=**F**_{inertial}– m**a**, where**a**is the acceleration of the frame. - Details of the calculation:

This is a one-dimensional problem. Let the positive direction be the direction of the trucks initial velocity.

(a) F = -mg/3 + mg/2 = +mg/6. The acceleration of the box relative to the truck is g/6.

(b) F = -mg/3. The acceleration of the box relative to the road is -g/3.

(a) An elevator in which a woman is standing moves upward at 4 m/s. If the
woman drops a coin from a height 1.4 m above the elevator floor, how long does
it take the coin to strike the floor? What is the speed of the coin relative to
the floor just before impact?

(b) Now assume that the elevator is moving downward with zero initial velocity
and acceleration of 1 m/s^{2} at t = 0, the women releases the coin at t
= 1 s. How long does it take the coin to strike the floor? What is the speed
of the coin relative to the floor just before impact?

Solution:

- Concepts:

The equations of motion in an inertial and in an accelerating frame. - Reasoning:

For part (a) the elevator is an inertial frame and for part (b) it is an accelerating frame. - Details of the calculations:

(a) The elevator is an inertial frame. The time it takes the coin to reach the floor is

t = (2h/g)^{½}= 0.53 s.

The speed just before impact is v = gt = 5.24 m/s.

(b) Now the elevator is an accelerating frame.

In non-inertial frames fictitious forces appear.

Consider a particle moving with velocity**v**in a reference frame K which moves with velocity**V**(t) relative to the inertial frame K_{0}.

The equations of motion are

md**v**/dt = m**g**- md**V**/dt .

Here dv/dt = g - 1 m/s^{2}= 8.8 m/s^{2}= g', downward.

t = (2h/g')^{½}= 0.56 s.

The speed just before impact is v = g't = 4.96 m/s.

__Rotating frames__

Write **F **= m**a** in a rotating coordinate system and identify
Coriolis and centrifugal terms.

Solution:

- Concepts:

Motion in a non-inertial frame - Reasoning:

A rotating coordinate system is a non-inertial frame. Fictitious forces appear. - Details of the calculation:

The relationship between the time derivatives of any vector**A**in a fixed, inertial frame and in a frame rotating with constant angular velocity**Ω**is

(d**A**/dt)_{fixed}= (d**A**/dt)_{rotating}+**Ω**× A.

Therefore:

(d**r**/dt)_{fixed}= (d**r**/dt)_{rotating}+**Ω**×**r**.The relationship between the velocity

**v**_{i}in the inertial frame and the velocity**v**in a frame rotating with constant angular velocity**Ω**is

**v**_{i}=**v**+**Ω**×**r**.

Differentiating again we have

(d**v**_{i}/dt)_{fixed}= (d**v**/dt)_{fixed}+**Ω**× (d**r**/dt)_{fixed}.

Inserting (d**r**/dt)_{fixed}from above we have

m(d**v**_{i}/dt)_{fixed}=**F**_{i}= m(d**v**/dt)_{fixed}+ m**Ω**×**v**+ m**Ω**×**Ω**×**r**).

Now we use

(d**v**/dt)_{fixed}= (d**v**/dt)_{rotating}+**Ω**× v

to obtain the equations of motion in the rotating frame.

The equations of motion in the rotating frame therefore are

md**v**/dt =**F**_{i}- 2m**Ω**×**v**- m**Ω**× (**Ω**×**r**).

If**F**_{i}is derivable from a potential then**F**_{i}= -∂U/∂**r**.

2m(**v**×**Ω**) = -2m(**Ω**×**v**) = Coriolis force

m(**Ω**×**r**) ×**Ω**= -m**Ω**× (**Ω**×**r**) = centrifugal force

A disc of radius R is spinning in the horizontal plane with a constant
angular speed **Ω**. A ladybug walks along the radius of the spinning disc,
traveling from the center of the disc toward the edge. The ladybug maintains a
constant speed v relative to the disc.

What is the acceleration of the ladybug at the instant it reaches the edge of
the disc?

Solution:

- Concepts:

Motion viewed from an inertial and a non-inertial frame - Reasoning:

In non-inertial frames fictitious forces appear. - Details of the calculation:

Consider a particle moving with velocity**v**in a reference frame K which moves with velocity**V**(t) relative to the inertial frame K_{0}and rotates with angular velocity**Ω**(t). The equations of motion are in the non-inertial frame are

md**v**/dt =**F**_{inertial}- md**V**/dt + m**r**× d**Ω**/dt - 2m**Ω**×**v**- m**Ω**× (**Ω**×**r**).

Here

-md**V**/dt = fictitious force due to acceleration of frame,

m**r**× d**Ω**/dt = fictitious force due to non-uniform rotation of frame,

-2m**Ω**×**v**= Coriolis force,

-m**Ω**× (**Ω**×**r**) = Centifugal force.

In the inertial frame we have**F**_{inertial}= md**v**/dt + md**V**/dt - m**r**× d**Ω**/dt + 2m**Ω**×**v**+ m**Ω**× (**Ω**×**r**).

Details of the calculation:

For the bug d**v**/dt = d**V**/dt = d**Ω**/dt = 0.

**F**_{inertial }= 2m**Ω**×**v**+ m**Ω**× (**Ω**×**r**).

In the inertial frame the magnitudes of the tangential and radial force are

F_{tangential}= 2mvΩ, F_{rad}= mΩ^{2}r.

In the inertial frame the magnitudes of the tangential and radial acceleration are therefore given by

a_{tangential}= 2vΩ, a_{rad}= Ω^{2}r

At the instant the bug reaches the edge of the disk

a = (a_{tangential}^{2}+ a_{rad}^{2})^{½}= Ω(4v^{2}+ Ω^{2}R^{ 2})^{½}.

An object on a planar platform moves in an elliptical trajectory described by

x(t) = x_{1} + x_{2} cos(αt), y(t)
= y_{1} + y_{2} sin(αt),

where x and y are measured with respect to a coordinate system fixed on the
platform. The platform is rotating with respect to an inertial coordinate
system XYZ. The two coordinate systems XYZ and xyz are the same at time t = 0.
The axis of rotation is the Z-axis, but the angular speed of rotation is
fluctuating with time and is given by

ω(t) = ω_{1} + ω_{2} sin(βt).

Find an expression for the velocity components V_{x} and V_{y}
of the object in the inertial frame at time.

Solution:

- Concepts:

Rotating coordinate systems,**v**_{fixed}=**v**_{rot}+**ω**×**r** - Reasoning:

The position of an object as a function of time is given in a non-inertial (rotating) frame. We are asked to find the velocity of the object in an inertial frame. - Details of the calculation:

Assume the platform is rotating counterclockwise. Then, .

Given : x(t) = x

_{1}+ x_{2}cos(αt), y(t) = y_{1}+ y_{2}sin(αt)

Therefore: dx/dt = -α x_{2 }sin(αt), dy/dt = α y_{2 }cos(αt)

Given : ω(t) = ω_{1}+ ω_{2}sin(βt), θ(t= 0) = 0

Therefore: θ(t) = ω_{1}t + (ω_{2}/β)(1 – cos(βt))

V_{x}= dX/dt = d(cosθ x - sinθ y)/dt

= cosθ dx/dt - x sinθ dθ/dt - sinθ dy/dt - y cosθ dθ/dt

= cosθ dx/dt - sinθ dy/dt – (x sinθ + y cosθ)dθ/dt

V_{y}= dY/dt = d(cosθ y + sinθ x)/dt

= cosθ dy/dt - y sinθ dθ/dt + sinθ dx/dt + x cosθ dθ/dt

= sinθ dx/dt + cosθ dy/dt + (x cosθ - y sinθ)dθ/dtInserting:

V_{x}= -cos(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))α x_{2 }sin(αt) – sin(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))α y_{2 }cos(αt)

- [(x_{1}+x_{2}cos(αt)) sin(ω_{1}t + (ω_{2}/β)(1 – cos(βt))) + (y_{1}+y_{2}sin(αt))(cos(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))]

*( ω_{1}+ ω_{2}sin(βt))V

_{y}= -sin(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))α x_{2 }sin(αt) + cos(ω_{1}t + (ω_{2}/β)(1 – cos(βt))) α y_{2 }cos(αt)

+ [(x_{1}+x_{2}cos(αt)) cos(ω_{1}t + (ω_{2}/β)(1 – cos(βt))) - (y_{1}+y_{2}sin(αt))(sin(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))]

*( ω_{1}+ ω_{2}sin(βt))

Verify that**v**_{fixed}=**v**_{rot}+**ω**×**r**.**ω**= ωk.**ω**×**r**= -ω(t)y(t)**i**+ ω(t)x(t)**j**

v_{fixed}= (-α x_{2 }sin(αt) -ω(t)y(t))**i**+ (α y_{2 }cos(αt) + ω(t)x(t))**j**Here

**v**_{fixed}is expressed in terms of the coordinates in the rotating system.

Change to the coordinates of the inertial system:

V_{x}= cosθ(-α x_{2 }sin(αt) -ω(t)y(t)) - sinθ(α y_{2 }cos(αt) + ω(t)x(t))

= -cosθ(α x_{2 }sin(αt)) - sinθ(α y_{2 }cos(αt)) - [cosθ y(t)) + sinθ x(t)]ω(t)

= -cos(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))α x_{2 }sin(αt) – sin(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))α y_{2 }cos(αt)

- [(x_{1}+x_{2}cos(αt)) sin(ω_{1}t + (ω_{2}/β)(1 – cos(βt))) + (y_{1}+y_{2}sin(αt))(cos(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))]

*(ω_{1}+ ω_{2}sin(βt))

Similarly

V_{y}= sinθ(-α x_{2 }sin(αt) -ω(t)y(t)) + cosθ(α y_{2 }cos(αt) + ω(t)x(t))

= -sinθ(α x_{2 }sin(αt)) + cosθ(α y_{2 }cos(αt)) - [sinθ y(t)) - cosθ x(t)]ω(t)

= -sin(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))α x_{2 }sin(αt) + cos(ω_{1}t + (ω_{2}/β)(1 – cos(βt))) α y_{2 }cos(αt)

+ [(x_{1}+x_{2}cos(αt)) cos(ω_{1}t + (ω_{2}/β)(1 – cos(βt))) - (y_{1}+y_{2}sin(αt))(sin(ω_{1}t + (ω_{2}/β)(1 – cos(βt)))]

*(ω_{1}+ ω_{2}sin(βt))

A brave physics student (an undergraduate, of course) climbs aboard a high
powered merry-go-round and goes to the center, at r = 0. At time t = 0,
the platform starts from rest (Ω = 0) and begins to spin about its vertical axis
with constant angular acceleration α. Also at time t = 0, the student
begins to crawl radially outward at constant speed v, relative to the platform.

Assuming the student does not slip, find the student's acceleration in the
inertial frame of an outside observer.

Solution:

- Concepts:

Motion viewed from an inertial and a non-inertial frame - Reasoning:

In non-inertial frames fictitious forces appear.

Consider a particle moving with velocity**v**in a reference frame K which moves with velocity**V**(t) relative to the inertial frame K_{0}and rotates with angular velocity**Ω**(t).

The equations of motion are in the non-inertial frame are

md**v**/dt =**F**_{inertial}- md**V**/dt + m**r**× d**Ω**/dt - 2m**Ω**×**v**- m**Ω**× (**Ω**×**r**).

Here -md**V**/dt = fictitious force due to acceleration of frame.

m**r**× d**Ω**/dt = fictitious force due to non-uniform rotation of frame.

-2m**Ω**×**v**= Coriolis force.

-m**Ω**× (**Ω**×**r**) = Centifugal force.

In the inertial frame we have**F**_{inertial}= md**v**/dt + md**V**/dt - m**r**× d**Ω**/dt + 2m**Ω**×**v**+ m**Ω**× (**Ω**×**r**). - Details of the calculation:

For our student d**v**/dt = d**V**/dt = 0.**F**_{inertial }= -m**r**× d**Ω**/dt + 2m**Ω**×**v**+ m**Ω**× (**Ω**×**r**).

In the inertial frame

F_{tangential}= (mrα + 2mvαt), F_{rad}= -m(αt)^{2}r,

F_{tangential}= 3mrα, F_{rad}= -m(αt)^{2}r.

In the inertial frame the acceleration is therefore given by

a_{tangential}= 3rα, a_{rad}= -(αt)^{2}r.

Assume that the direction of**Ω**is the z-direction.

Then**a**_{tangential}= 3rα(**ф**/ф) and**a**_{rad}= -(αt)^{2}**r**.

Transforming from polar to Cartesian coordinates we have

(**r**/r) = cosф**i**+ sinф**j**, (**ф**/ф) = -sinф**i**+ cosф**j**.

Therefore a_{x}= -(αt)^{2}cosф - 3rα sinф, a_{y}= -(αt)^{2}sinф + 3rα cosф.

r = vt, ф = ½αt^{2}.

a_{x}= -(αt)^{2}cos(½αt^{2}) - 3vtα sin(½αt^{2}), a_{y}= -(αt)^{2}sin(½αt^{2}) + 3vtα cosф(½αt^{2}).

On the surface of the earth an object is given an initial speed v on a friction less surface at latitude λ. Show that the object will move in a circle and find the radius of the circle for velocities small enough that the radius is much smaller than the earth radius.

Solution:

- Concepts:

Motion in an accelerating frame - Reasoning:

Assume an observer at mid latitudes λ = 90^{o}- θ.

The object is observed by an observer in a uniformly rotating coordinate system.

The equations of motion in a rotating coordinate system contain fictitious forces.

md**v**/dt =**F**_{inertial}- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v**= -m**g**+**N**- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v**

-m**Ω**× (**Ω**×**r**) = centrifugal force

-2m**Ω**×**v**= Coriolis force

The force of gravity and the centrifugal force, giving m**g**_{eff}≈ m**g**, are balanced by the force of constraint from the ground.

Let the z-axis of the coordinate system point into the -**g**_{eff}direction. Then the first two terms have no component ⊥to the z-axis.

The Coriolis force, however, has a component ⊥to the z-axis. - Details of the calculation:
**F**_{coriolis⊥}= (2m Ω_{z}dy/dt)**i**- (2m Ω_{z}dx/dt)**j**.

We therefore have m d^{2}x/dt^{2}= 2m Ω_{z}dy/dt, m d^{2}y/dt^{2}= -2m Ω_{z}dx/dt.

dv_{x}/dt = Ω_{z}v_{y}, dv_{y}/dt = Ω_{z}v_{x}.

Solutions: v_{x}= A cos(ωt + φ), v_{y}= -A sin(ωt + φ), ω = 2Ω_{z}.

x = (A/(2Ω_{z})) sin(2Ω_{z}t + φ), y = (A/(2Ω_{z})) cos(2Ω_{z}t + φ).

This is motion in a circle, clockwise when looking down in the northern hemisphere.

Ω_{z}= (2π/day) cos(θ), v^{2}= v_{x}^{2}+ v_{y}^{2}= A^{2}, radius r = v/(2Ω_{z}).

A particle moves in a horizontal plane on the surface of the Earth. Show that the magnitude of the horizontal component of the Coriolis force is independent of the direction of the motion of the particle.

Solution:

- Concepts:

Motion in an accelerating frame - Reasoning:

Assume an observer at mid latitudes.

The object is observed by an observer in a uniformly rotating coordinate system.

The equations of motion in a rotating coordinate system contain fictitious forces.

md**v**/dt =**F**_{inertial}- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v**= -m**g**+**N**- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v**

-m**Ω**× (**Ω**×**r**) = centrifugal force

-2m**Ω**×**v**= Coriolis force

The first two terms sum to zero. The force of gravity and the centrifugal force, giving m**g**_{eff}≈ m**g**, are balanced by the force of constraint from the ground.

Let the z-axis of the coordinate system point into the -**g**_{eff}direction. Then the first two terms have no component ⊥to the z-axis.

The Coriolis force, however, has a component ⊥to the z-axis. - Details of the calculation:

**F**_{coriolis⊥}= (2m Ω_{z}dy/dt)**i**- (2m Ω_{z}dx/dt)**j**.

|**F**_{coriolis⊥}| = [(2m Ω_{z}v_{y})^{2}+ (2m Ω_{z}v_{x})^{2}]^{½}= 2m Ω_{z}v

The magnitude of the horizontal component of the Coriolis force, |**F**_{coriolis⊥}| = 2m Ω_{z}v, is independent of the direction of**v**.