 shape id radius ...(radius)2 k2 1 +  k2/r2 1 -  k2/r2 m axis ofrotation I N=          h /2πr secθ for N(min) θ in °  for N(min) Point mass 0 ----- Sphere (solid) 1 ----- through cm Sphere(hollow) 2 ----- through cm cylinder(solid) 3 ----- through center and along the length cylinder(hollow) 4 ----- ----do----- circular disc 5 ----- perpendicular to discand through cm circular ring 6 ----- ----do----- acceleration due to gravity    g: m/s2 g' g'' Height from which mass is rolled  h: m angle of rolling surface with g ,θ : (in deg) id of object The Object is released from an inclined surface at a height h and having angle θ with g vector. velocity near ground, v acceleration a: time period T: Time to reach the bottom (sec) inclined distance S velocity as % of free fall acceleration as% of g : After time sec velocity : velocity:*vmax*t=vmax inclined distance s1=(1/2)at2 height(s-s1)cosθ height(t)h-{gcos2θ/[2(1+ k2/r2)]}t2 The Object is pushed from the bottom of inclined plane upward with velocity: acceleration: -a time taken to come to rest (sec) inclined distance height The Object is pushed from the bottom of inclined plane upward with velocity as above, After time sec, velocity inclined distance height The Object is pushed from the bottom of inclined plane upward with velocity as above and at the same time, another body with same spatial configuration is released from rest from the top, After how much time, they shall hit each other ? t :sec * When a rigid body moves down an inclined plane of height h, as per conservation of mechanical energy, mgh= mv2/2 +Iω2/2 = +( mk2)(v2/r2)/2 where I is moment of inertia,ω is angular velocity, r is radius. This leads to the equn, v2 = 2gh/ (1+ k2/r2)=2g'h whereg'=g/ (1+ k2/r2)  ; For point mass, g' = g g''=g/ (1- k2/r2) ; For point mass, g' = g v2 - u2 =2as where u=initial velocity=0, s is inclined distance . Solving a =g*cosθ/(1+ k2/r2)=g'cosθ where s =h/cosθ where θ is the angle between g vector and the inclined surface along which a vector acts. Now s=ut+at2/2 = at2/2 or t =√(2s/a) =√(2h/g)*√(1+ k2/r2)*secθ =√(2h/g')*secθ Similarly, when a body is pushed up an inclined plane with velocity u, time taken to reach zero speed is t= u/a, inclined distance travelled is s=u2 / 2a and height reached h = s*cosθ * Acceleration / deceleration  along inclined plane is independent of mass, dependent on g, angle of inclination with g, k2/r2 . a=gcosθ for dropped bodies and a=a=g'cosθ for bodies which have rolling motion along with translational motion. * Velocity is also independent of mass, independent of angle of inclination but dependent on g, height h, k2/r2. Hence When a body is released from a height, its sliding  velocity at any instant is equal to its dropping velocity and the translational velocity due to rolling is less than sliding/dropping velocity since some amount of energy is consumed in rolling. Velocity =√(2g'h) for rolling bodies and Velocity =√(2gh) for dropped bodies. Translational velocity of rolling body is less than that of sliding/dropped body. * time of travel is dependent on the (inclined path length) / (height along with angle of inclination). with same height , secθ is deciding factor.   t=√(2h/g) * secθ for bodies without rolling and t=√(2h/g') * secθ for bodies with rolling . Other things remaining same, t for rolling body is more than a non rolling body. * Suppose a body of some configuration is pushed up an inclined plane with initial velocity u and at the same time another body of same configuration (not necessarily same material ) is released from rest from the top. after what time, they shall meet each other ? s1= ut-at2/2 ....(1) s2=at2/2 s=s1+s2=ut or t=s/u =h/(ucosθ) * If a body is released from rest from the top of an inclined plane and at the bottom, there is an perfectly elastic wall, the body bounces to the top.If T is the time period, T=2*t =√(8h/g)*√(1+ k2/r2)*secθ =√(8h/g')*secθ *If a body is released from rest from the top of an inclined plane, the vertical height covered after time t is cosθ *at 2/2 =t2*gcos2θ/[2(1+ k2/r2) ] = t2*g'cos2θ/2 So height left out is h0(t)=h -mt2 where m=gcos2θ/[2(1+ k2/r2) ] = g'cos2θ/2 (t√m )2+(√h0)2 =(√h)2 or This is the equation of a circle with origin (0,0) and radius √h and t = [0,√(2h/g)*√(1+ k2/r2)*secθ] =  [0,√(2h/g')*secθ] *If a body is released from rest from the top of an inclined plane, the velocity after time t is   v(t) =√(g/(2*h*[1+ k2/r2])*cosθ*vmax * t =√(g'/(2h)*cosθ*vmax * t=vm*vmax * t = if  vm < 1 , t > 1     vm =1,  t =1    vm > 1, t < 1 to achieve  vmax. For cases where body slides in stead of rolling, g' shall be replaced by g. Integer Number of Rollings down an inclined plane: no. of rolling N = s / 2πr = h/(2πrcosθ) =(h/2πr) * secθ Minimum value of secθ is 1 and maximum value infinity. (a) If (h/2πr) is fixed , N can be made an integer say n by adjusting secθ. n (Nmin ) =Math.ceil N       secθ = Math.ceil N /N  ..... for  Nmin  .       secθ = (Math.ceil N)+1 /N  ..... for  1+Nmin  . And so on. (b) If angle (θ) of inclination to g is fixed,For rolling to become integer, (Nmin ) =Math.ceil N =N*secφ ;      and (Nmin ) =(h/2πcosθ) / (r*secφ) =(h/2πcosθ) / (r1)      or   (Nmin ) =(1/2πr*cosθ) / (h/secφ) =(1/2πr*cosθ) / (h1)      For altering the number of rotations , change r1 or h1 accordingly. For rolling to be at least 1      N h/2π ⋜ nr where n is an integer.