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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

  • Physics -

    how did you get <mvr = I w + m R^2 w>

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