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Use conservation of energy or Work energy theorem.

M₁ g R q - M₂ g R sin q = (M₁ V² + M₂ V²) / 2 ------ (1)

This gives the value of V.

14/11/06

Two cases arise:

CASE 1: As seen by us V₀ = 2w₀R is to the right and the cylinder is rolling
clockwise. Then since V₀ > w₀R, we have the friction F in backward (left as seen by us) direction. Then the equation of motion of the cylinder is given by,

F = - m(V - V₀) / t = m R² ( W - w₀) / 2 R t --------- (1) For rolling with out sliding condition is WR = V ------(2)

Solving (1) and (2) we get the desired result.

CASE 2: As seen by us V₀ = 2w₀R is to the right and the cylinder is rolling anti-clockwise. Then we have the friction F in backward (left as seen by us) direction. Then the equation of motion of the cylinder is given by,

F = - m(V - V₀) / t = - m R² ( W - w₀) / 2 R t --------- (1)

For rolling with out sliding condition is WR = V ------(2)

Solving (1) and (2) we get the desired result.

Note that the problem can also be solved by using conservation of angular momentum about the bottom most point.

Using the relation L = I_cm w₀ + m ( R X V₀ ) = I_cm W + m ( R X V)
3mR² w₀/ 2 + m(2w₀ R) R = 3mR² V / 2R + mR V

When the smaller body is sliding on the larger block it has two velocities, one horizontal and the other vertical. Let them be w and v respectively. Since the smaller body is in contact with the bigger one both the bodies will have same horizontal velocity during the vertical climb.

Now conserve the linear momentum in the horizontal direction, because no external force is acting on them in the horizontal direction.

mu = (m + M) w ------- (1)

Now for the vertical velocity we have conservation of energy. As there is no loss of energy due to friction or other means ,

mu² /2 = (m + M) w² /2 + mv² /2 + mgh -------- (2)

Maximum height reached can be calculated from the following

mgH = mv² /2 ----------- (3)

After the smaller block leaves the bigger block it travels like a projectile with horizontal velocity ' w ' and vertical velocity ' v '. Since the horizontal components of both the masses are same the smaller block will land back on the bigger block again. The time of flight can be obtained from

T = 2v / g --------- (4)

During this time the larger block would have traveled a distance of ' S ' given by

S = w T --------- (5)

We can further enhance the questionnaire of this problem by the following two questions

1) What will be the velocities of both the bodies after smaller block reaches the horizontal portion of the bigger block?

2) What will be the center of mass velocity of the system during the entire motion in the horizontal direction?