Suppose you’re riding a Ferris wheel that’s slowing down at a steady rate. As the wheel slows down, does your acceleration point more and more towards the center of the wheel, or more and more away from it? Explain.
you do not have to down shift, use your brakes to stop you mostly the front when you stop ,gear down for take off,no it will not hurt the bike,,,,,,,,,,,,you need to have a lot more training if you are asking qestions like this
Pull in the clutch and brake. then shift down. You might have to let the clutch out just a little other wise the gears wont shift. So it would be like this Clutch - brake down gear, down gear, clutch (half way) down gear, down gear, clutch (half way) neutral Motorcycle gears are different from car gears. They WON'T shift unless the internals are spinning. That is why you let the clutch out a little and then press it back to shift more gears. Good Luck.
Lights do not change “all of a sudden”, they are entirely predictable, but yes you can brake without downshifting but you will need to apply the clutch. I see little benefit in being able to pull away if you think you are going to be hit from behind, unless you can be certain that you won't get t-boned by a driver legally obeying their own green light.
Your acceleration point will be towards the center of the wheel. Here's why, an object in motion will keep in motion until an outer force exerts on it. So, when the ferris wheel turns slow, the gravity force will pull the ferris wheel down. Since you are being pulled down, usually you will fall straight to the ground, however; since the ferris wheel was in motion, the momentum of the ferris wheel to keep turning makes the wheel to go away from the center but since the wheel is fixed at the center then it is pulling the momentum to the center of the ferris wheel.
Consider you are a point on a vertically upright wheel that is spinning. If the spin is at a constant w constant angular velocity, there are but two forces acting on you. One is your weight due to gravity; W mg. The other is centripetal force, which is counteracted by centrifugal force, F m(wr)^2/r mw^2r ma(r); where r is the radius of the wheel and a(r) is the acceleration along the radius (the spokes) of the wheel. This a(r) is the acceleration that points toward the center of the wheel. Now, add the slowing down of the wheel; so there is now one more force; that's f ma(T); where a(T) is the tangential deceleration.your slowing down. But note that this deceleration is always tangent to the wheel's rim. Meanwhile as the wheel slows, we find that F mw^2r, the force pointing towards the center of the wheel, is becoming smaller because w, the angular velocity, is diminishing. But a(r) w^2r is still pointing towards the center along the spokes. It has changed magnitude, gotten smaller, but it still points the same direction as before. So, as you see, the magnitude of the radial acceleration lessens, but its direction does not change. In fact, if the wheel were totally stopped and caused to rotate in the opposite direction, a(r) would still point towards the center of the wheel. But a(T) would become an acceleration rather than a deceleration as the wheel started in the opposite rotation.
