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Prove that result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v^(2)=(2gh)/((1+(k^(2))/(R^(2)))) using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane. |
Answer» Solution :Suppose a BODY of mass M and radius R is rolling on inclined plane of height h, length l and angle of inclination `theta` is show as in figure. The forces acting on body are shown as in figure.  Suppose a is acceleration in downward direction. Equation of motion this body `N-Mgcostheta=0` and `F=Ma-f` `therefore F=Mgsintheta-f....(1)` Torque NECESSARY against friction force f `tau=Rxxf` `Ialpha=Mk^(2)((a)/(R))` where `alpha=(a)/(R )` `therefore RF=(Mk^(2))/(R)a` `therefore f=(Mk^(2))/(R^(2))a` `therefore` From eqn. (1), `Ma=Mgsintheta-(Mk^(2))/(R^(2)).a [F=Ma]` `(M+(Mk^(2))/(R^(2)))a=Mgsintheta` `therefore a=(Mgsintheta)/(M((R^(2)+k^(2))/(R^(2))))` `therefore a=(gsintheta)/((R^(2)+k^(2))/(R^(2)))=(gsintheta)/(1+(k^(2))/(R^(2)))` Let the height of SLOPE is h and its length is l. `therefore` From figure `(h)/(l)=sinthetaimpliesl=(h)/(sintheta)` The velocity at the bottom of slope is v and at maximum height is `u=0` `therefore` In eqn. `v^(2)-u^(2)=2al,u=0 and a=(gsintheta)/(1+(k^(2))/(R^(2)))` `v^(2)-0=2xx(gsintheta)/(1+(k^(2))/(R^(2)))XX(h)/(sintheta)` `therefore v^(2)=(2gh)/(1+(k^(2))/(R^(2)))` is proved. |
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