* When a rigid body moves down an inclined plane of
height h, as per conservation of mechanical energy, mgh= mv^{2}/2 +Iω^{2}/2
= +( mk^{2})(v^{2}/r^{2})/2
where I is moment of inertia,ω is angular velocity, r is radius.
This leads to the equn, v^{2} = 2gh/
(1+ k^{2}/r^{2})=2g'h
where**g'=g/ (1+ k**^{2}/r^{2})
; For point mass, g' = g
**g''=g/ (1- k**^{2}/r^{2})
; For point mass, g' = g
v^{2} - u^{2}
=2as where u=initial velocity=0, s is inclined distance . Solving a
=g*cos**θ/**(1+ k^{2}/r^{2})=g'cos**θ**
where s =h/cos**θ where ****θ is the angle between g
vector and the inclined surface along which a vector acts.**
**Now s=ut+at**^{2}**/2 = at**^{2}**/2
or t =√(2s/a) =√(2h/g)*√**(1+ k^{2}/r^{2})*sec**θ ****
=√(2h/g')***sec**θ**
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=u^{2}
/ 2a and height reached h = s*cosθ
* Acceleration / deceleration along inclined plane is
independent of mass, dependent on g, angle of inclination with g, k^{2}/r^{2}
. 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, k^{2}/r^{2}.
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 , sec**θ
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-at^{2}/2 ....(1)
s2=at^{2}/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+ k^{2}/r^{2})*sec**θ ****
=√(8h/g')***sec**θ**
*******If a body is released from
rest from the top of an inclined plane, the vertical height covered
after time t is **
**cos****θ**** *at **
^{2}**/2 =t**^{2}***gcos**^{2}**θ/[2**(1+ k^{2}/r^{2})
] = **t**^{2}***g'cos**^{2}**θ/2**
So height left out is
**h0(t)=h -mt**^{2}
where
m=**gcos**^{2}**θ/[2**(1+ k^{2}/r^{2})
] = **g'cos**^{2}**θ/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+ k^{2}/r^{2})*sec**θ****]
=**
** [0,√(2h/g')***sec**θ****]**
***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+
k^{2}/r^{2}])***cosθ***v_{max}**
* t =√**(g'/(2h)***cosθ***v_{max}**
* t=vm***v_{max}** * t =**
**if vm < 1 , t > 1**
** vm =1, t =1**
** vm > 1, t < 1**
**to achieve **v_{max}**. **
For cases where body slides in stead of rolling, g' shall be
replaced by g. |