## Apparent weight

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 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.

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.