Group # 14MembersRabia Aftab Firdous BatoolZubaria Aamir

Submitted to : Mam

Topics The RotorThe Banked Curve Equations Of Motion

The RotorThe rotor is a cylinderical shell capable of rotation about its vertical cylindrical axis. A man enter the rotor ,closes the door and stand up against the wall.The rotor is then set into rotation.

The forces on the man standing on the rotor are.Its weight mg acting vertically downward2) Force of static friction between the man and the rotor wall acting upward 3) The normal force N of the wall.

F = f mg = ma = 0 f = mg there is no acceleration along z-axis The normal force N provide the centripital force F = -N = ma = - mv/R where r is the radius

f = mg = u N = u mv/R

g = uv/R or v = (gR/u)

The Banked CurveDefination : A curve in which the outside edge is higher than the inside edge

Overview : Consider a car moving at constant speed V on a level road around a curved path of radius R. The forces acting on the car are :

Its weight mg acting vertically downward on the roadNormal force N acting vertically upward on the car by the road bed.A horizontal force F on the car which provides centripetal force necessary to keep the car along the circular path.

However ,these forces are not large enough at all times and also cause unnecessary wear. For safe turn around the curved path , the road bed or the railway track is banked .Raising outer edge of thr track with respect to the inner edge is called Banking.

When the road bed or the railway track is banked at angle , the normal force has vertical and horizontal component N cos and N sin respectively . The vertical component N cos balances the weight of the car . That is : Fz = N cos mg = maz = 0Ncos = mgThere is no acceleration along the vertical .The horizontal or radical component of normal force N sin supplies the centripetal acceleration and hence centripetal force again , the second law of motion gives :

F = -N sin = mar = -mvROr N sin = mvRN sin N cos =mvmgROrTan = vgR

Equation Of MotionFrom Newton Laws of Motion we define acceleration time rate of change of velocity.That is. a = lim v/t = dv/dt Rewriting it in the form dv = adt and integerate we get

dv = adt Where v = velocity at t= 0 when we start counting time v = a dt = a t v-v = at v(t) = v + at which give the velocity at time t Now

V = lim x/t = dx/dt dx = vdt And integrating we get dx = vdt = (v + at)dt = vdt + at dt x-x = vt + at If a is a constant x-x = x + vt + at which gives the postion of object at any time..