Angular velocity of a Rod after a Ball hits it.

[cocoon]

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 $$\frac{1}{12}*m*l^2$$.

Homework Equations

Let w be the angular velocity, $$I$$ be the moment of intertia, $$r$$ be the radius of the bar, and $$K_{rot}$$ 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:
$$K_i = .5 * m_c * v_c^2 = .5 * 1 * 2^2 = 2$$

no external forces, so
$$K_i = K_f$$

after the collision the system's Kinetic energy is
$$K_f = K_{rot} = .5 * I_h * w_h^2 + .5 * m_c * r^2 * w_c^2$$
let $$w = w_h = w_c$$
$$K_f = .5 * (I_h + m_c) * w^2 = 2$$

$$(m_h * l^2 / 12 + m_c) * w^2 = 4$$
$$(3 * 1^2 / 12 + 1) * w^2 = 4$$
$$\frac{5}{4} * w^2 = 4$$
$$w= \sqrt{\frac{16}{5}} rad/s$$
is the angular velocity of the rod after the collision.

Is this correct? Is there an easier way?

 

[Mindscrape]

I'm sorry, it's not correct way to solve the problem.

What conservation law is always true? What conservation law is always true, but has lots of different forms that we can't always account for, particularly in the case of inelastic collisions? Additionally, the way the problem is described, it seems like you should count the ball as being a part of the cylinder once it hits.

 

[Doc Al]

no external forces, so
$$K_i = K_f$$

The collision is perfectly inelastic (they stick together) so KE is not conserved. (But what is?)

The lack of external forces is not a criterion for KE to be conserved. (But it is for other conserved quantities.)

Also: There are external forces acting--the axis of the rod is fixed.

 

[cocoon]

I remember conservation of linear momentum now from earlier this semester. I think I got it now. If not, then please just tell me what to do haha...

let p be linear and L be angular momentum.
before collision, $$p_c + L_h = m_cv_c + I_hw_h = 1*2 + 0 = 2$$
and after collision, $$L_{h+c} = I_{h+c}w_{h+c} = I_hw_{h+c} + m_c\frac{l^2}{4}w_{h+c} = 2$$
so, $$(\frac{m_hl^2}{12}+m_c\frac{l^2}{4})w_{h+c} = 2$$
$$(\frac{3*1^2}{12}+1*\frac{1}{4})w_{h+c} = 2$$
$$(\frac{1}{4} + \frac{1}{4})w_{h+c} = 2$$
$$w_{h+c} = 4 rad/s$$

yes?

 

[Doc Al]

I remember conservation of linear momentum now from earlier this semester. I think I got it now. If not, then please just tell me what to do haha...

Is linear momentum conserved in this collision? Consider the last line of my previous post.

let p be linear and L be angular momentum.
before collision, $$p_c + L_h = m_cv_c + I_hw_h = 1*2 + 0 = 2$$

Linear and angular momentum are two different things (with different units!)--they cannot be added together in a physically meaningful way. Only one of them is conserved in this case.

and after collision, $$L_{h+c} = I_{h+c}w_{h+c} = I_hw_{h+c} + m_c\frac{l^2}{4}w_{h+c} = 2$$
so, $$(\frac{m_hl^2}{12}+m_c\frac{l^2}{4})w_{h+c} = 2$$
$$(\frac{3*1^2}{12}+1*\frac{1}{4})w_{h+c} = 2$$
$$(\frac{1}{4} + \frac{1}{4})w_{h+c} = 2$$
$$w_{h+c} = 4 rad/s$$

Redo this.

 

[cocoon]

welp, guess i have to hit the books haha

 

[Mindscrape]

As a hint, you're trying to do too much at once. Maybe you could break the problem down into stages.