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## Homework Statement

A horizontal bar of mass 3.0 kg and length 1.0 m is free to rotate about a vertical axis through its center. A clay blob of mass 1.0 kg traveling at a velocity of 2.0 m/s in the direction shown collides with the end of the bar and sticks to it. Find the magnitude of the angular velocity of the rod after the collision. The moment of inertia of a bar about a perpendicular axis through the center is [tex]\frac{1}{12}*m*l^2[/tex].

## Homework Equations

Let w be the angular velocity, [tex]I[/tex] be the moment of intertia, [tex]r[/tex] be the radius of the bar, and [tex]K_{rot}[/tex] be the rotational kinetic energy (K is linear kinetic energy). Subscripts c stand for the clay blob and h stand for the horizontal bar.

## The Attempt at a Solution

before collision, the system's Kinetic energy is:

[tex]K_i = .5 * m_c * v_c^2 = .5 * 1 * 2^2 = 2[/tex]

no external forces, so

[tex]K_i = K_f[/tex]

after the collision the system's Kinetic energy is

[tex]K_f = K_{rot} = .5 * I_h * w_h^2 + .5 * m_c * r^2 * w_c^2[/tex]

let [tex] w = w_h = w_c[/tex]

[tex]K_f = .5 * (I_h + m_c) * w^2 = 2[/tex]

[tex](m_h * l^2 / 12 + m_c) * w^2 = 4[/tex]

[tex](3 * 1^2 / 12 + 1) * w^2 = 4[/tex]

[tex]\frac{5}{4} * w^2 = 4[/tex]

[tex]w= \sqrt{\frac{16}{5}} rad/s[/tex]

is the angular velocity of the rod after the collision.

Is this correct? Is there an easier way?

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