Rod rotating and hitting a mass

In summary, the rod falls, rotates around A, and hits another mass which gains velocity. The rod's angular velocity is ω1 immediately after the hit. The energy lost is πMgL.
  • #1
Karol
1,380
22

Homework Statement


Snap1.jpg
A is the pivot at the upper edge of the rod.
The rod of length L and mass M stays horizontally. it can pivot round it's edge. suddenly it falls, rotates round A and hits a mass m which gains velocity v.
At which angular velocity the rod reaches the mass
What is the rod's angular velocity immediately after it hits?
How much energy is lost during the hit

Homework Equations


moment of inertia of a rod round it's edge: ##I_A=\frac{1}{3}ML^2##
Conservation of momentum: ##m_1v_1+m_2v_2=m_1v_1'+m_2v_2'##
Kinetic energy of a solid body: ##E_k=\frac{1}{2}I\omega^2##

The Attempt at a Solution


The energy of height of the COM in the horizontal position transforms to kinetic:
$$\frac{1}{2}MgL=\frac{1}{2}MV^2~~\rightarrow~~V^2=gL$$
Conservation of momentum, V1 is the rod's COM velocity after the hit: ##MV=MV_1+mv##
Angular velocity, ω1 is the rod's angular velocity after: ##\frac{1}{2}L\omega_1=V_1##
$$\rightarrow~~\omega_1=\frac{2}{ML}(M\sqrt{gL}-mv)$$
The energy loss is:
$$\frac{1}{2}MgL-\left( \frac{1}{2}I_A\omega_1+mv^2 \right)$$
 
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  • #2
Well now, interesting. Is there a question ? Or do you just want a stamp of approval (which really isn't PF business). You wouldn't get it from me anyway, because I miss things like conservation of angular momentum. I also have difficulty with your ##V^2 = gL##: it's not as if the COM is dropping over ##L/2## only: there is also rotational kinetic energy to consider !
 
  • #3
As BvU observes, you have ignored the rotational energy acquired by the rod.
Also, linear momentum will not be conserved. Can you see why?

The problem as stated does not provide enough information. You need to know the coefficient of restitution in the impact. Probably should assume it is completely inelastic.
 
  • #4
Conservation of energy between the initial condition and the vertical one:
$$\frac{1}{2}MgL=\frac{1}{2}I_A\omega_0^2~~\rightarrow~~\omega_0^2=\frac{3g}{L}$$
Conservation of angular momentum: ##\vec L=I\times \vec{\omega}##
$$I_A\omega_0=I_A\omega+mLv~~\rightarrow~~\omega=\omega_0-\frac{mLv}{I_A}$$
 
  • #5
Karol said:
Conservation of energy between the initial condition and the vertical one:
$$\frac{1}{2}MgL=\frac{1}{2}I_A\omega_0^2~~\rightarrow~~\omega_0^2=\frac{3g}{L}$$
Conservation of angular momentum: ##\vec L=I\times \vec{\omega}##
$$I_A\omega_0=I_A\omega+mLv~~\rightarrow~~\omega=\omega_0-\frac{mLv}{I_A}$$
That all looks right, except for a possible flaw in the question. The block mass m is shown as having significant height, so its angular momentum about A will be less than mLv. Leaving that aside...

Next, you need to represent the fact that the rod does not penetrate the block. That puts a constraint on the relationship between ω and v. If we assume a fully inelastic collision then you can take that relationship as equality.
 
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  • #6
we don't assume assume inelastic collision since v, the block's velocity, is given.
 
  • #7
Karol said:
we don't assume assume inelastic collision since v, the block's velocity, is given.
Good point!
 
  • #8
The energy that was lost:
Initial energy is ##E_i=\frac{1}{2}MgL##
$$\Delta E=\frac{1}{2}MgL-\frac{1}{2}I_A\omega_0^2-\frac{1}{2}mv^2$$
 
  • #9
Karol said:
The energy that was lost:
Initial energy is ##E_i=\frac{1}{2}MgL##
$$\Delta E=\frac{1}{2}MgL-\frac{1}{2}I_A\omega_0^2-\frac{1}{2}mv^2$$
ω, not ω0, right?
 
  • #10
right, Haruspex, thank you very much...
 

FAQ: Rod rotating and hitting a mass

1. How does the rotation of a rod affect its impact on a mass?

The rotation of a rod can greatly affect its impact on a mass. The speed, angle, and direction of rotation can all impact the force and direction of the impact. Additionally, the length and weight of the rod can also play a role in the impact.

2. What factors determine the force of impact when a rod hits a mass?

There are several factors that can determine the force of impact when a rod hits a mass. These include the speed and direction of the rod's rotation, the weight and length of the rod, and the density and composition of the mass being hit. Other factors such as air resistance and friction may also play a role.

3. Can the angle of rotation affect the outcome of the impact?

Yes, the angle of rotation can have a significant impact on the outcome of the impact. For example, a rod rotating perpendicular to the mass will have a different impact than a rod rotating parallel to the mass. The angle can also affect the distribution of force and the direction of the impact.

4. What happens to the momentum of the rod and mass during the impact?

During the impact, the momentum of both the rod and the mass will change. The magnitude and direction of this change will depend on the factors mentioned before, such as the speed and angle of rotation. The total momentum of the system will remain constant, following the law of conservation of momentum.

5. How can the impact of a rotating rod on a mass be calculated?

The impact of a rotating rod on a mass can be calculated using principles of physics, such as Newton's laws of motion and the conservation of momentum. Factors such as the speed, angle, and weight of the rod, as well as the density and composition of the mass, can be used to determine the force and direction of the impact. Computer simulations and experiments can also be used to calculate and analyze the impact in more complex scenarios.

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