Angular velocity of a Rod after a Ball hits it.

In summary, the problem involves a horizontal bar of mass 3.0 kg and length 1.0 m rotating freely about a vertical axis, colliding with a clay blob of mass 1.0 kg traveling at a velocity of 2.0 m/s. After the collision, the clay blob sticks to the end of the bar. To find the magnitude of the angular velocity of the rod after the collision, conservation of linear momentum and conservation of angular momentum can be used. The final equation is (\frac{m_hl^2}{12}+m_c\frac{l^2}{4})w_{h+c} = 2, with the final solution being w_{h+c} = 4 rad/s.
  • #1
cocoon
10
0

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?
 
Last edited:
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  • #2
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.
 
  • #3
cocoon said:
no external forces, so
[tex]K_i = K_f[/tex]
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.
 
  • #4
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, [tex]p_c + L_h = m_cv_c + I_hw_h = 1*2 + 0 = 2[/tex]
and after collision, [tex]L_{h+c} = I_{h+c}w_{h+c} = I_hw_{h+c} + m_c\frac{l^2}{4}w_{h+c} = 2[/tex]
so, [tex](\frac{m_hl^2}{12}+m_c\frac{l^2}{4})w_{h+c} = 2[/tex]
[tex](\frac{3*1^2}{12}+1*\frac{1}{4})w_{h+c} = 2[/tex]
[tex](\frac{1}{4} + \frac{1}{4})w_{h+c} = 2[/tex]
[tex]w_{h+c} = 4 rad/s[/tex]

yes?
 
  • #5
cocoon said:
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, [tex]p_c + L_h = m_cv_c + I_hw_h = 1*2 + 0 = 2[/tex]
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, [tex]L_{h+c} = I_{h+c}w_{h+c} = I_hw_{h+c} + m_c\frac{l^2}{4}w_{h+c} = 2[/tex]
so, [tex](\frac{m_hl^2}{12}+m_c\frac{l^2}{4})w_{h+c} = 2[/tex]
[tex](\frac{3*1^2}{12}+1*\frac{1}{4})w_{h+c} = 2[/tex]
[tex](\frac{1}{4} + \frac{1}{4})w_{h+c} = 2[/tex]
[tex]w_{h+c} = 4 rad/s[/tex]
Redo this.
 
  • #6
welp, guess i have to hit the books haha
 
  • #7
As a hint, you're trying to do too much at once. Maybe you could break the problem down into stages.
 

Related to Angular velocity of a Rod after a Ball hits it.

1. What is angular velocity?

Angular velocity is the measure of how fast an object is rotating around a fixed point. It is expressed in radians per second.

2. How is angular velocity calculated?

Angular velocity is calculated by dividing the change in angular displacement of an object by the time it took for that change to occur.

3. How does a ball hitting a rod affect its angular velocity?

When a ball hits a rod, the rod will experience a change in angular velocity. This change is determined by the mass and velocity of the ball, as well as the moment of inertia and angular velocity of the rod before the collision.

4. What factors can affect the angular velocity of a rod after a ball hits it?

The angular velocity of a rod after a ball hits it can be affected by the mass and velocity of the ball, the moment of inertia of the rod, the angle of impact, and the material properties of both the ball and the rod.

5. How does the angle of impact affect the angular velocity of a rod after a ball hits it?

The angle of impact can affect the angular velocity of a rod after a ball hits it by changing the direction and magnitude of the force exerted on the rod. A direct impact perpendicular to the rod's axis of rotation will result in a larger change in angular velocity compared to an oblique impact.

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