Solve Roatational Motion of Cylinder on Block Edge

[prabhat rao]

a rectangular rigid fixed block has a long horizontal edge. a solid homogenous cylinder of radius r is placed horizontally at rest wid its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane. there is sufficient friction present at the edge so that a very small displacement causes the cylinder to roll off the edge widout slipping. determine:
(a) the angle theta through which de cylinder rotates b4 it leaves contact wid de edge
(b) de speed of de center of mass of de cylinder b4 leaving contact wid de edge
(c) de ratio of de transitional nd rotational kinetic energies of de cylinder when its center of mass is in horizontal line wid de edge

MY SHOT AT THE PROBLEM:

It would have better had u told the source of the problem or showed your work regarding the question. Well this is very simple; I can do it involving higher degree of mathematics. But don’t I won’t do that. Now when a body is pure roll the body would describe a cycloid function and now take a look at your notes for verification. Let's us assume right now that the center of mass is initially parallel to the x-axis. As we know that the center of mass moves along this the same direction. Now consider that there is a point at the top of the cylinder of mass dm. As I told u earlier in my explanation that the body describes a cycloid curve, the reference point in our problem on its downward movement seems to be describing a circle of radius r (though not true). In this problem till now we have assumed that the center of mass has rotated through an angle theta. Now there is centripetal force acting towards the center of mass. The cylinder leaves the edge only when the normal force is zero. Our reference point has also no normal force exerted on it. Writing the force equation we get:

dm*v^2/r = dmg cos theta
Now consider the kinetic energy of the particle
The particle has both rotation and transition
K_e due to rotation = 1/2I*omega^2
K_e due to transition = 1/2dm*v^2
Remember that the above equation can be used, but one has to remember the fact that we are not considering the center of mass and thereby we cannot use the equation v_cm = r*omega
. But we can write omega in the above equation as omega = r/v_particle. But there is a jinx over here: we cannot consider the kinetic energy due to rotation why??. Because the axis of rotation in the body doesn’t pass through the point we are considering in the body.

But after this I am not able to get to the answer and and i think my method is wrong

Somebody please solve it

 

[member 553083]

. ANSWER:(a) The angle theta through which the cylinder rotates before it leaves contact with the edge is equal to the angle at which the center of mass moved relative to the edge. This angle can be calculated using the equation: theta = arctan(r/v_cm), where v_cm is the velocity of the center of mass.(b) The speed of the center of mass of the cylinder before leaving contact with the edge can be calculated using the equation: v_cm = sqrt(2*g*r). (c) The ratio of the transitional and rotational kinetic energies of the cylinder when its center of mass is in horizontal line with the edge can be calculated using the equation: K_e (transitional)/K_e (rotational) = m*v_cm^2 / (I*omega^2), where m is the mass of the cylinder, v_cm is the velocity of the center of mass, I is the moment of inertia of the cylinder, and omega is the angular velocity of the cylinder.