| * 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
where g'=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. |