Hollow cylinder B will reach the bottom of the slope first if?
- No friction on slope
- They roll without slipping
Both the cylinders have the same mass and radius. A is a solid cylinder.
In this question we have 2 cylinders A is solid and B is hollow.
Let the momentf of interia of A and B be !!I_A!! and !!I_B!! respectively. Then !!I_A=I_B/2!!.
When it is slipping down if there is no friction both A and B behave identically, they slip down the plane without rolling and will reach bottom of the plane at nearly the same.
If there is friction and it executing pure rolling without slipping. Then when it reach the bottom of the plane with say velocity !!v!!.
We know energy is conserver so !!mgh = 1/2mv^2+1/2Iomega^2!!
!!gh = 1/2v^2+1/2 I/m v^2/r^2!!
!!v^2 = 2gh/(1+I/(mr^2))!!
Since I is higher for hollow cylinder its velocity will lower as compared to the velocity of the solid cylinder when it reaches the bottom of the plane.
Now consider the case where there is rolling and slipping. In that case we will have.
!!v^2 = (2g(h-mul))/(1+I/(mr^2))!!
So even now the velocity of the hollow cylinder will be lower as compared to velocity of solid cylinder.
And since they are travelling the same amount we get the !!B!! will always reach after !!A!!