A solid cylinder is released from rest from top of an inclined plane time taken

Text Solution

Solution : Let the mass of the cylinder be m and its radilus r. Suippose the linear spee of the cylinder when it reaches the bottom is v. As the cylinder rolls without slilpping, its angular speed about its axis is `omega=v/r`. The kinetic energy at the bottom will be <br> `K=1/2Iomega^2+1/2mv^2` <br> `=1/2(1/2mr^2)omega^2+1/2mv^2=1/4+1/2mv^2=3/4mv^2` <br> This should be equal to the loss of potential energy mgl sintheta`. Thus, `3/4mv^2=mglsintheta` <br> `or v=sqrt(4/3gl sin theta)`

A solid cylinder is released from rest from the top of an inclined plane of inclination θ and length ' 1'. If the cylinder rolls without slipping, then find it's speed when it reaches the bottom of inclined plane.

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