Two moving blocks, find small blocks final velocity

[toastie]

Homework Statement

A small block (mass m) slides down a circular path (radius R) which is cut into a large block (mass M), as shown. M rests on a table and both blocks slide without friction. The blocks are initially at rest, and m starts at the top of the path. Determine the velocity of m when it loses contact with the large block.
Data: M = 3.4 kg; m = 1.2 kg; R = 0.7 m.

Homework Equations

Using basic geometry I have solved a = (angular velocity^2)*r*cos(phi)cos(theta)

The Attempt at a Solution

Using the above equation though, I believe to be in the wrong refrence frame since both blocks are moving. Thus, the 90 degree refrence angles will not be 90 when the small block is no longer touching the large block.

Thank you in advance!

 

[LowlyPion]

Maybe consider the energy that's in the system?

Don't you have a total of m*g*R available?

And won't the Center of Mass in the absence of external forces need to be in the same position when the small block exits?

 

[nlightner]

I would approach this problem by first analyzing the forces acting on the small block as it moves down the circular path. Since there is no friction, the only forces acting on the block are its weight (mg) and the normal force from the large block (N). Since the blocks are initially at rest, we can assume that the normal force is equal to the weight of the small block (N=mg).

Next, we can use Newton's Second Law (F=ma) to analyze the motion of the small block. We know that the acceleration of the block is directed towards the center of the circular path, and we can use the equation a=(v^2)/r to relate the acceleration to the velocity of the block.

Plugging in the known values for the masses and radius, we can solve for the velocity of the small block at any point along the circular path. However, we are interested in the velocity at the point where the small block loses contact with the large block. This occurs when the normal force becomes zero, and the small block is no longer experiencing a centripetal force.

Therefore, we can set N=0 in our equation and solve for the velocity at this point. This will give us the final velocity of the small block as it leaves the circular path.

Finally, we can use the conservation of energy to check our answer. Since there is no friction, the total mechanical energy (kinetic + potential) of the system should remain constant. We can calculate the initial and final energies and make sure they are equal.

In summary, the velocity of the small block when it loses contact with the large block can be found by setting the normal force to zero and solving for the velocity using Newton's Second Law. This approach takes into account the motion of both blocks and is not limited by a specific reference frame.