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प्रश्न
Answer in Brief:
A rigid object is rolling down an inclined plane derive the expression for the acceleration along the track and the speed after falling through a certain vertical distance.
उत्तर
The figure shows a rigid object of mass M and radius R, rolling down an inclined plane, without slipping. Inclination of the plane with the horizontal is θ. As the objects starts rolling down, its gravitational P.E. is converted into K.E. of rolling. Starting from rest, let v be the speed of the centre of mass as the object comes down through a vertical distance h. If the frictional force on the body is large enough, the body rolls without slipping.
v = Linear speed of the centre of mass
R = Radius of the body
ω = Angular speed of rotation of the body, `therefore omega="v"/R` for any particle
M = Mass of the body
K = Radius of gyration of the body `therefore I=MK^2`
Total kinetic energy of rolling = Translational K.E. + Rotational K.E.
`E = 1/2M"v"^2+1/2Iomega^2 = 1/2 M"v"^2 [1 + K^2/R^2]`
`therefore E=Mgh = 1/2 M"v"^2 [1 + K^2/R^2]`
`therefore "v" = sqrt((2gh)/[1 + K^2/R^2])`
Linear distance travelled along the plane is `s = h/sintheta`
During this distance, the linear velocity has increased from zero to v. If `a` is the linear acceleration along the plane,
`2as = "v"^2-u^2 therefore 2a(h/sintheta) = (2gh)/[1 + K^2/R^2]-0`
`therefore a=(gsintheta)/[1 + K^2/R^2]`
For pure sliding, without friction, the acceleration is `gsintheta` and final velocity is `sqrt(2gh)`. Thus, during pure rolling, the factor `[1 + K^2/R^2]` is effective for both the expressions.
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