 # Apparent Weight Riding in a elevator– why does our weight appear to change when we start up (increase) and slow down (decrease)? Our sensation of weight.

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Apparent Weight Riding in a elevator– why does our weight appear to change when we start up (increase) and slow down (decrease)? Our sensation of weight change is due to a force exerted on our feet by the elevator floor (normal force N). If force greater we feel heavier and vice versa.

Apparent Weight Eg. Upward accelerating elevator: As accelerating, there must be a net upward force. (2 nd law) F net = N – W = m a But our true weight: W = m g Apparent weight: N = W + ma N = m (g + a) (i.e. heavier) If lift accelerating downwards (or decreasing upwards): N = m (g – a) (ie. lighter) W N

Free-Falling When you jump off a wall, or throw a ball or drop a rock in a pool, the object is free-falling ie. falling under the influence of gravity. Question: What happens to our apparent weight in free-fall? Nasty Exp: Cut elevator wires so its downward acceleration a = g (i.e. free-fall)! Apparent weight N = m (g – a) But a = g, so N = 0 i.e. no normal force. “Weightless” is zero apparent weight. Everything is falling at same rate, so no normal force is needed to support your weight.

Free-Falling Ex: Aircraft flying in a parabolic path can create weightless conditions for up to 30 s! Spacecraft / astronauts in orbit are weightless as they (and the spacecraft) are continuously free- falling towards the Earth!!

Circular Motion (chapter 5) So far we have focused on linear motion or motion under gravity (free-fall). Question: What happens when a ball is twirled around on a string at constant speed? Ans: Its velocity continuously changes in direction. This implies: The velocity change is caused by an acceleration. By Newton’s 2 nd law an acceleration requires a force!

Circular Motion (chapter 5) By Newton’s 2 nd law an acceleration requires a force! Big questions: What is the nature of this force / acceleration? What is the relationship between the acceleration and the velocity of the ball and the radius of curvature?

In the absence of gravity, tension provides the only force action on the ball. This tension causes ball to change direction of velocity. Instantaneous velocity vector changing in direction but its magnitude stays constant. v1v1 v2v2 v3v3 v4v4 v5v5

Question: What happens if you let go of the string? Answer: Ball travels in direction of last instantaneous vector. (Newton’s 1 st law)

Let’s imagine you are on a kid’s “roundabout”… Question: Why do we feel an outward force if it’s not really there? You must pull inwards Your body naturally wants to move this way (Newton’s 1 st law) However, to keep you in circular motion you must apply a force inwards to change your direction.

Your pulling inwards creates the sensation that the roundabout is pushing you outwards! You must pull inwards Your body naturally wants to move this way (Newton’s 1 st law)

The force (tension) causes an acceleration that is directed inwards towards center of curvature. ie. The string is continuously pulling on the ball towards the center of curvature causing its velocity to constantly change. This is called centripetal acceleration (a c ). Centripetal Acceleration

 Centripetal acceleration is the rate of change in velocity of an object due to a change in its velocity direction only.  It is always perpendicular to the velocity vector and directed towards the center of curvature. There is NO such thing as centrifugal (ie outward) force. Centripetal Acceleration

Nature of ‘a c ’ |v 1 | = |v 2 | = same speed Accn. v2v2 v1v1  t vΔ a c  Acceleration is in direction of Δv. v1v1 v2v2 ΔvΔv v 1 + Δv = v 2

Dependencies of a c 1. As speed of object increases the magnitude of velocity vectors increases which makes Δ v larger. v1v1 v2v2 ΔvΔv v1v1 v2v2 ΔvΔv Therefore the acceleration increases.

Dependencies of a c 2. But the greater the speed the more rapidly the direction of velocity vector changes: Small angle change Slow Large angle change Fast increases

3. As radius decreases the rate of change of velocity increases – as vector direction changes more rapidly. d d Same distance (d) moved but larger angle change. Large radius Small radius Result: increases as radius decreases. t vΔ  Dependencies of a c

Points 1 and 2 indicate that the rate of change of velocity will increase with speed. Both points are independent of each other and hence a c will depend on (speed) 2. Summary

Point 3 shows that a c is inversely proportional to radius of curvature (i.e. ). Summary Thus: m/s 2 (towards center of curvature) i.e. Centripetal acceleration increases with square of the velocity and decreases with increasing radius. 2 c r v a  r 1 acac 

Example: Ball on a string rotating with a velocity of 2 m/s, mass 0.1 kg, radius=0.5 m. What forces can produce this acceleration? Tension Friction Gravitation attraction (planetary motion). Nuclear forces Electromagnetic forces ?

Let’s consider the ball on a string again… If no gravity: T T Center of motion m acac Ball rotates in a horizontal plane. T = m a c = m v 2 r

Let’s consider the ball on a string again… With gravity: T T W = mg ThTh TvTv String and ball no longer in the same horizontal plane. The horizontal component of tension (T h ) provides the necessary centripetal force. (T h = ma c ) The vertical component (T v ) balances the downward weight force (T v = mg).

Stable Rotating Condition T h = T cos θ T v = T sin θ = mg T h = = T cos θ m v 2 r As ball speeds up the horizontal, tension will increase (as v 2 ) and the angle θ will reduce. T W=mg ThTh TvTv θ

Stable Rotating Condition Thus, as speed changes T v remains unaltered (balances weight) but T h increases rapidly. T W=mg ThTh TvTv θ High speed T W=mg ThTh TvTv θ Low speed

Unstable Condition T v no longer balances weight. Centripetal force F c = ma c = 0.1 x 8 = 0.8 N T ThTh TvTv Ex. again: Ball velocity 2 m/s, mass 0.1 kg, radius=0.5 m. Thus, horizontal tension (T h ) = 0.8 N. Now double the velocity… Centripetal F c = ma c = 0.1 x 32 = 3.2 N Thus, the horizontal tension increased 4 times! The ball can’t stay in this condition.

A centripetal force F c is required to keep a body in circular motion: This force produces centripetal acceleration that continuously changes the body’s velocity vector. Thus for a given mass the needed force: increases with velocity 2 increases as radius reduces. r vm a m F 2 c c   Summary

Example: The centripetal force needed for a car to round a bend is provided by friction. If total (static) frictional force is greater than required centripetal force, car will successfully round the bend. The higher the velocity and the sharper the bend, the more friction needed! FfFf FfFf As F s = μ s N - the friction depends on surface type (μ s ). Eg. If you hit ice, μ becomes small and you fail to go around the bend. Note: If you start to skid (locked brakes) μ s changes to its kinetic value (which is lower) and the skid gets worse! Moral: Don’t speed around tight bends! (especially in winter) F s > mv 2 r

The normal force N depends on weight of the car W and angle of the bank θ. There is a horizontal component (N h ) acting towards center of curvature. This extra centripetal force can significantly reduce amount of friction needed… Motion on a Banked Curve N v = mg NhNh W=mg θ N If tan θ = then the horizontal N h provides all the centripetal force needed! v 2 rg FcFc FnFn Ice skaters can’t tilt ice so they lean over to get a helping component of reaction force to round sharp bends. In this case no friction is necessary and you can safely round even an icy bend at speed…

Vertical Circular Motion Total (net) force is thus directed upwards: N > W Feel pulled in and upward W=mg N W T T > W Ferris Wheel Ball on String Bottom of circle: Thus: N = W + ma c i.e. heavier/larger tension F net = N - W = ma c N=apparent weight (like in elevator) Centripetal acceleration is directed upwards.

T > W Feel thrown out and down N < W N > W W=mg N N W W T T Component of W provides tension Top of circle: N = W – m a c i.e. lighter / less tension If W = m a c → feel weightless (tension T=0) Weight only force for centripetal acceleration down. (larger r, higher v) gror v r v ga i.e. 2 c  W

Example: The centripetal force needed for a car to round a bend is provided by friction. If total (static) frictional force is greater than required centripetal force, car will successfully round the bend. The higher the velocity and the sharper the bend, the more friction needed! FfFf FfFf As F s = μ s N - the friction depends on surface type (μ s ). Eg. If you hit ice, μ becomes small and you fail to go around the bend. Note: If you start to skid (locked brakes) μ s changes to its kinetic value (which is lower) and the skid gets worse! Moral: Don’t speed around tight bends! (especially in winter) F s > mv 2 r

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