# 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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## Expert Answers

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...

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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.

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