Another Question----
A ball of clay of mass m travels with a velocity v in a path tangent to a disk of radius R and mass M. The clay collides with the disk tangentially to its outer rim (a totally inelastic collision) and the clay and disk begin to spin about the axis.
a.) What is the final angular speed of the clay and disk (DONT FORGET to include the mass m after the collision). (SOLVE IN TERMS OF VARIABLES)
b.)What if the clay had collided with the rim at an angle of 30 degrees above the horizontal? What would the angular speed be then???
Thanks for the help!!!!
The key to solving this problem is to recognize that the angular momentum measured about the axis of rotation remains constant. Initially the disk has no angular momentum. The angular momentum of the clay before it hits the deisc is m v r.
Thus
mvr = I w + m R^2 w
where I is the disc's moment of inertia and w is the final angular velocity.
If the disc is solid and uniform, I = (1/2)MR^2
Solve for w.
For (b), recognize that only the compnent of incident clay ball velocity that is tangent to the disc and perpendicular to the axis of rotation contributes to the angular momentum about the axis.
but you aren't given radius r of ball of clay
also on the right side of the equation what does Iw represent
I w is the product of the moment of inertia (which you need to calculate from M and R). w is the angular velocity (usually written as a Greek lower-case omega). The r in the equation should be R, the outer radius of the disc where the clay sticks.
ok ty
how did you get <mvr = I w + m R^2 w>
To solve part (a) of the problem, we need to calculate the final angular speed of the clay and disk after the collision.
We start by considering the conservation of angular momentum. The initial angular momentum of the system is zero since the disk is initially at rest. The final angular momentum of the system is the sum of the angular momentum of the disk and the angular momentum of the clay.
The angular momentum of the clay before the collision is given by mvr, where m is the mass of the clay, v is its velocity, and r is the radius of its path (which in this case is the same as the radius of the disk, denoted as R).
The angular momentum of the disk after the collision can be calculated using the moment of inertia (I) and the final angular velocity (w) of the system. Since the collision is totally inelastic, the clay sticks to the disk and they both rotate together.
The moment of inertia (I) of the disk can be calculated using the formula I = (1/2)MR^2, where M is the mass of the disk and R is its radius.
Therefore, the equation for the conservation of angular momentum becomes:
mvr = (1/2)MR^2w + mR^2w
We can now solve this equation for the final angular velocity (w). Rearranging the terms, we have:
w((1/2)MR^2 + mR^2) = mvr
w = mvr / ((1/2)MR^2 + mR^2)
This is the expression for the final angular velocity of the clay and disk after the collision, in terms of the given variables m, v, M, and R.
For part (b) of the problem, where the clay collides with the rim at an angle of 30 degrees above the horizontal, we need to consider the component of the clay's incident velocity that is tangent to the disk and perpendicular to the axis of rotation.
In this case, the initial angular momentum of the system is still given by mvr, where r is the radius of the disk. However, the final angular momentum of the disk will be different due to the change in the direction of the incident velocity.
To calculate the final angular velocity (w) in this case, we can use the same equation as before, but substitute v with the component of the velocity that is tangent to the disk and perpendicular to the axis of rotation. This can be calculated as v * sin(30 degrees).
Therefore, the expression for the final angular velocity (w) in this case becomes:
w = (m * v * r * sin(30 degrees)) / ((1/2) * M * R^2 + m * R^2)
This is the expression for the final angular velocity of the clay and disk after the collision, when the clay collides with the rim at an angle of 30 degrees above the horizontal.