# Q. Portion AB of the wedge shown in figure is rough and BC is smooth.A solid cylinder rolls without slipping from A to B.The ratio of translational kinetic energy to rotational kinetic energy,when...

Q. Portion AB of the wedge shown in figure is rough and BC is smooth.A solid cylinder rolls without slipping from A to B.The ratio of translational kinetic energy to rotational kinetic energy,when the cylinder reaches point C is:-

A) 3/4

B) 5

C) 7/5

D) 8/3

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The initial vertical height of point A is `H` . Point B lies at the height `H/2` .

Between A and B the variation in potential energy transforms into rotational energy + kinetic energy because there are friction forces that place in rotation the cylinder. The speed of the cylinder at point B is `V_1` and its angular speed is `omega_1` .

`(I*omega_1^2)/2 +(m*V_1^2)/2 =m*g*H/2`

with the condition `V_1 =omega_1*R`

For a solid cylinder the momentum of inertia is

`I =(m*R^2)/2`

Thus the first relation becomes

`(m*R^2)/2 *(V_1/R)^2 +m*V_1^2 =m*g*H`

`(3/2)*V_1^2 =g*h` or equivalent `V_1^2 =2/3*g*H`

Between points B and C there is no friction to further increase the rotational speed of the cylinder. The variation of potential energy transforms only into kinetic energy. The speed of cylinder at point C is `V_2` and its angular speed stay unchanged `omega_1` .

`(m*V_2^2)/2 -(m*V_1^2)/2 = m*g*H/2`

`V_2^2 -V_1^2 =g*H`

`V_2^2 -2/3*g*H =g*H`

`V_2^2 =5/3*g*H = 5/2*V1^2`

Now we can compute the ratio of translational kinetic energy to rotational energy at point C:

`(m*V_2^2)/(I*omega_1^2) = (m*(5/2)*V_1^2)/((mR^2)/2*(V_1^2/R^2)) =5`

**The correct answer is B) 5.**