## Apparent weight

### Fictitious Forces

Newton's first law, also called the law of inertia, defines a special class of reference frames, called inertial frames.  It states that, when viewed in an inertial reference frame, an object at rest remains at rest and an object in motion continues in motion with constant velocity unless it is acted on by an external net force.  Newton's second law and Newton's third law correctly describe the motion of objects as viewed by observers in inertial reference frames.  In an inertial frame F = ma, where the net force F is the vector sum of all the real known forces acting on an object of mass m.  If observers at rest in an accelerating frame want to use F = ma to predict the motion of an objet in their reference frame, then F has to include not only the vector sum of all the real known forces acting on the object but also a fictitious force.  Fictitious forces appear in accelerating frames.

You just got your new car.  You want to experience it.  You want to "feel" its power.  You floor the gas pedal, and you experience a force pressing you back into your seat.  Where does this force come from?

This force is a fictitious force.  Fictitious forces appear in accelerating reference frames.  Such frames are NOT inertial reference frames.  The accelerating reference frame in the above example is accelerating with your car.  In this frame your car and you are at rest.  But you are feeling a force pushing you against the back of your seat.  To your friend observing you from the sidewalk things look different.  The motor is responsible for the forward acceleration of the car.  Because of your mass, you have inertia.  Without a force acting on you, you would remain at rest with respect to the sidewalk.   To keep you accelerating forward, the back of the seat has to push on you.  (You will be pushing on the seat with a force equal in magnitude, but opposite in direction.)  The fictitious force appearing in the accelerating frame is the negative of the real force responsible for maintaining your acceleration and keeping you at rest in the accelerating frame.  The real force acting on you is in the forward direction, while the fictitious force experienced in the accelerating frame is in the backward direction.

In the accelerating frame of the car, you experience the fictitious force in the backward direction and your weight, pointing down.  The net force experienced is the vector sum of these two forces.  This force becomes your apparent weight, which points in a direction backward and down.

The apparent weight of an accelerating object is the vector sum of its real weight and the negative of all the forces that produce the object's acceleration a = dv/dt.

wapparent = wreal - ma.

Assume you are riding on a merry-go-round.  A reference frame in which you are stationary, i.e. a frame that is moving with you as you are moving along a circular path, is also an accelerating frame.  This frame is moving with constant speed, but the direction of its velocity is constantly changing.  You are sitting still on your seat while the merry-go-round is turning.  But something seems to be pulling you towards the outside, away from the center.  You experience a fictitious force.  To your friend on the ground things again look different.  You are moving in a circle.  The direction of your velocity is constantly changing.  You are accelerating.  The direction of your acceleration is towards the center of the circle, so there must be a force pushing or pulling you toward the center.  If you are sitting in a seat, the wall of the seat will be pushing against you, pushing you towards the center.

As an object moves along a circular path, the direction of its velocity is constantly changing.  The direction of the acceleration is towards the center of the circle.  The magnitude of the acceleration, v2/r, depends on the speed and the radius of the circle along which the object traveling.   In an accelerating frame, which is moving with the object, the object is at rest.  But it experiences the fictitious centrifugal force, constantly trying to throw it towards the outside of the circle.

In many Science Fiction books, humans live in space in a space station that is rotating about a central axis.  Their real weight is close to zero.  The acceleration of a person with mass M at rest with respect to the space station near the rim is a = v2/r directed towards the axis.  The apparent weight of the person is wapparent = wreal - Ma = Mv2/r directed towards the rim.

If the rim is the floor of some room without windows, and the rotation speed and the radius of the station are adjusted so that a = v2/r = g, then there is no way a human or a scientific instrument in the room can distinguish between the apparent weight and the force of gravity.  If the human steps on a scale, the scale will read the same "weight" as it does on the surface on earth.  If the human throws a ball near the "surface", the ball will follow the same trajectory it would on earth.

#### Problem:

Engineers are trying to create artificial gravity in a ring-shaped space station by spinning it like a centrifuge.  The ring is 100 m in radius.  How quickly must the space station turn in order to give the astronauts inside it apparent weights equal to their real weights at the earth's surface

Solution
We want a = v2/r = g , or v2 = gr = (9.8 m/s2)(100 m) = 980 (m/s)2.  Therefore v = 31.3 m/s.
The circumference of the space station is 2πr = 628 m.
The space station therefore must completes a turn in (628 m)/(31.3 m/s) = 20 s.

### Roller Coasters

In an amusement park the most thrilling rides let the rider experience the largest fictitious forces.  Fictitious forces are experienced in accelerating reference frames.  The seat or cart in which the rider is sitting is accelerating, because it is speeding up, slowing down, or turning rapidly.  A ride that is just moving you along with constant velocity will probably not excite you.  You might as well be in your car on the freeway with the cruise control engaged.

Roller coasters offer thrilling rides.  Most roller coaster rides start with a steep "vertical" drop.  The cart descends on a track down a very steep slope.
Assume the top of the hill is 50 m above the ground.  The gravitational potential energy of a object on top of the hill is Mgh.  Neglecting friction, this potential energy has been converted into kinetic energy when the object reaches the bottom of the hill.
K = ½Mv2=  Mgh.
v2 = 2gh = 2*(9.8 m/s2)*(50 m) = 980 (m/s)2.
v = 31.1 m/s = 70 mph.
At the bottom of the hill the object (the cart, you, …) moves with a speed of 70 mph.

The forces on the object as it descends are the force of gravity and the support from the track.  The component of the gravitational force perpendicular to the track is canceled by the support from the track, and the component of the gravitational force tangential to the track accelerate the object.  However if you are sitting in the accelerating cart, then you are experiencing a fictitious force in the direction opposite to the direction of the velocity in addition to the other forces.  This force cancels the tangential component of gravity.  You are left with an apparent weight only due to the perpendicular component of the force of gravity, much less than your normal weight.

if the track pointed straight down, the perpendicular component of the gravitational force would be zero, and your apparent weight would be zero.  You would feel weightless.  Feeling weightless means that your apparent weight (not necessarily your real weight) is zero.  A friend on the ground observes you accelerating with acceleration g because of your real weight.  The weight of the cart also accelerates the cart at 9.8 m/s2.  Your velocity and the velocity of the cart change at the same rate, your relative velocity stays zero.  You are both in free fall, not subject to any forces other than gravity, just as a skydiver, who stepped out of an airplane.

Since the cart of a roller coaster is attached to a track, it is possible, using motors, to produce accelerations in the direction of the velocity greater than g.

Many roller coasters have a vertical loop, 20 m high.  Assume a cart enters the loop at a speed of 25 m/s or 56 mph. As the cart moves through the loop, it is moving in a circle and therefore it is accelerating.  An observer on the ground concludes that the forces on the cart are the force of gravity and the support from the track.  The support from the track has to cancel the component of the weight perpendicular to the track.  But if it just did that, the cart would keep on moving tangential to the track, since there would be no force pointing towards the center of the circle, providing the centripetal acceleration.  So the force that the track exerts on the cart must be the sum of the perpendicular component of mg and the centripetal force mv2/r.

A rider in the accelerating cart again experience a fictitious force in addition to the real forces.  At the bottom of the loop the fictitious centrifugal force has magnitude mass*(25 m/s)2/(10 m) = mass*62.5 m/s2 = mass*6.4 g in the downward direction.  This force is added to the rider's weight to yield an apparent weight of mass*7.4 g.

On the top of the loop the sum of the cart's kinetic energy has been converted into potential energy.
½mvtop2 = ½mvbottom2 - mgh.
vtop2 = vbottom2 - 2gh = (25 m/s)2 - 2*(9.8 m/s2)*(20 m) = 233 (m/s)2.
v = 15.3 m/s.
The fictitious centripetal force the rider experience is now directed upward and has magnitude
mv2/r = mass*(15.3  m/s)2/(10 m) = mass*(23.4 m/s2) = mass*2.4 g.
Subtracting the real weight of the rider, which points in the downward direction, we obtain an apparent weight = mass*1.4 g in the upward direction.
Things seem to "fall" upward, towards the bottom of the inverted car.

Halfway up the loop the weight points straight down, but the fictitious centrifugal force points horizontally toward the outside of the loop.  The apparent weight is the vector sum of these forces.

### Gravity

Locally, near the surface of the earth, the gravitational force is approximately constant.  In general, the gravitational force exerted by point object 1 on point object 2 is given by

F12 = (-G m1m2/r122) (r12/r12).

Here r12 is the distance between particles 1 and 2, and (r12/r12) is the unit vector pointing from particle 1 to particle 2.
G is the gravitational constant, G = 6.67*10-11 Nm2/kg2.
The above formula is also valid for spherical objects with r12 denoting the distance between their centers.  The formula is always approximately valid as long as the distance between the objects is much larger than their respective sizes.  When the distance between the objects changes, the gravitational force does work.  The change in the gravitational potential energy is

Uf - Ui = -∫r12ir12f F12·dr12 = -G m1m2r12ir12f (1/r122) dr12 = -G m1m2(1/r12f - 1/r12i).

The difference in the gravitational potential energy is uniquely defined.  To agree on the value of the gravitational potential energy we need to agree on a reference point r12i.  Often this reference point is chosen to be at infinity.  Then

Ug(r12) = -Gm1m2/r12.

The gravitational potential energy of two objects (stars, planets, satellites) separated by a distance d then is negative, Ug = -Gm1m2/d.  Its absolute value is equal to the amount of work that needs to be done to completely separate the objects, i.e. to let d go to infinity.

#### Problem:

How much energy is required to move a 1000 kg mass from the Earth's surface to an altitude twice the Earth's radius?

Solution:
W = Uf - Ui = -G m1m2(1/r12f - 1/r12i).
G = 6.67*10-11 Nm2/kg2.
m1 = mearth = 5.98*1024 kg.
m2 = 1000 kg.
r12i = 6.37*106m.
r12f = 2r12i.
W = 3.13*1010 J.