Angular Momentum - Find angular velocity of a hanging bar after a collision

In summary, a 1.0kg object moving horizontally at 10m/s makes a collision with the lower end of a 1.2m, 3kg bar that is suspended about its upper end. The bar then swings freely up to 90 degrees with respect to the vertical, while the object deflects to an angle below the horizontal at a velocity v. Using the equations for conservation of kinetic energy and angular momentum, the angular velocity of the bar, speed of the object, magnitude of the angular momentum of the object, and the angle \theta can be determined. The results are: angular velocity of the bar = 5rad/s, speed of the object = 8m/s, magnitude of angular momentum of the
  • #1
I think I've gotten it but figured it would be best to get a second opinion because I feel like I made a few leaps of faith. I sincerely appreciate your time. <3

Homework Statement


A 1.0 kilogram object is moving horizontally with a velocity of 10m/s, when it makes a collision with the lower end of a bar that is hanging vertically at rest. For the system of the bar and object, linear momentum is not conserved but kinetic energy is. The bar, of length l = 1.2m and mass m = 3kg, is pivoted about its upper end. Immediately after the collision, the object moves with speed v at an angle (theta) relative to its original direction. The bar swings freely, reaching a maximum angle of 90 degrees with respect to the vertical. The moment of inertia for the bar about the pivot is [itex] I = (ml^2) / 3 [/itex]. Ignore all friction.

Synopsis:
1.0kg object moving horizontally at 10m/s.
1.2m, 3kg bar [itex] I = (ml^2) / 3 [/itex] suspended about its upper end.
Object hits the bottom of the bar in a glancing collision.
Bar then pivots up to 90 degrees with respect to the vertical.
Object then deflects to an angle [itex]\theta[/itex] below the horizontal at a velocity v.

Diagram:
29QQA.jpg


Questions:
1. Determine the angular velocity of the bar immediately after the collision

2. Determine the speed v of the 1-kilogram object immediately after the collision

3. Determine the magnitude of the angular momentum of the object about the pivot just before the collision

4. Determine the angle [itex]\theta[/itex].

Homework Equations


The all-important: [itex]F = ma [/itex]
Conservation of angular momentum: [itex]L_i = L_f[/itex]
Conservation of kinetic energy: [itex]K_o = K_ol + K_b[/itex]
Angular momentum around a point: [itex]L = mvr sin(\theta)[/itex]
[itex]\tau = F(lever arm) = I\alpha [/itex]
[itex]L = I\omega[/itex]

The Attempt at a Solution


Enclosed in quotes to make it easier to see.
1. Determine the angular velocity of the bar immediately after the collision
[tex]K_t + K_r = K_f + U_f [/tex] -- [itex]K_t[/itex] and [itex]K_f[/itex] are both zero
[tex]K_r = U_f[/tex]
[tex](1/2)I\omega^2 = mgh[/tex]
[tex](1/2)[(ml^2)/3]\omega^2 = 3kg * 10m/s * 1.2m/2[/tex] -- On the right is a leap of faith. I'm pretty sure the center of mass is what I need to account for here, and that's only moving 0.6m, so I used that.

[tex][3kg * (1.2m)^2 / 6]\omega^2 = 18J[/tex]
[tex]0.72\omega^2 = 18J[/tex]

[tex]\omega = 5rad/s[/tex]

2. Determine the speed v of the 1-kilogram object immediately after the collision
Total kinetic energy is conserved.
Subscript o represents object, b represents bar.

[tex]K_i = K_f[/tex] -- Initial kinetic energy equals final kinetic energy.
[tex](1/2)mv^2_o = (1/2)mv^2_o + (1/2)I\omega^2[/tex]
[tex](1/2) * 1kg * (10m/s)^2 = (1/2) * 1kg * V^2_f + (1/2)[(1/3)ml^2]\omega^2[/tex]
[tex]50J = (1/2)V^2_f + (1/6) ml^2\omega^2[/tex]
[tex]50J - [(1/6)3kg * (1.2m)^2] * [5rad/s]^2 = (1/2)V^2_f[/tex]
[tex]32m^2/s^2 = (1/2)v^2_f[/tex]

[tex]v_f = 8m/s[/tex]

3. Determine the magnitude of the angular momentum of the object about the pivot just before the collision
[tex]L = mvr sin(\theta)[/tex]
[tex]L = 1kg * 10m/s * 1.2m * 1[/tex]

[tex]L_i = 12kg m^2/s[/tex]

4. Determine the angle [itex]\theta[/itex].
[tex]L = mvr sin(\theta)[/tex]
[tex]L = 1kg * 8m/s * 1.2m * sin(\theta)[/tex]
[tex]L = 9.6sin(\theta)[/tex]

[tex]L_b = I\omega[/tex]
[tex]L_b = [(ml^2)/3] * 5rad/s[/tex]
[tex]L_b = 7.2kg m^2/s[/tex]

[tex]12kg m^2/s - 7.2kg m^2/s = 4.8kg m^2/s[/tex] -- Plug this into the equation above...

[tex]4.8kg m^2/s = 9.6kg m^2/s sin(\theta)[/tex]
[tex]sin(\theta) = 0.5[/tex]

[tex]\theta = 30 degrees[/tex]

Thanks so much!
 
Last edited:
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  • #2
I haven't checked all your numbers, but your method seems good to me.
 

1. What is angular momentum?

Angular momentum is a property of a rotating object that is determined by its mass, velocity, and distance from the axis of rotation.

2. How is angular velocity related to angular momentum?

Angular velocity is the rate of change of angular displacement, and it is directly proportional to angular momentum. This means that as angular velocity increases, so does angular momentum.

3. What is the formula for calculating angular momentum?

The formula for calculating angular momentum is L = Iω, where L is angular momentum, I is the moment of inertia, and ω is angular velocity.

4. How is angular momentum conserved in a collision?

In a collision, angular momentum is conserved, meaning that the total angular momentum before the collision is equal to the total angular momentum after the collision. This is because there are no external torques acting on the system.

5. How can I find the angular velocity of a hanging bar after a collision?

To find the angular velocity of a hanging bar after a collision, you can use the formula L = Iω. Calculate the moment of inertia of the bar and the initial angular momentum before the collision, then solve for ω to find the final angular velocity.

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