Problem involving angular momentum

In summary, the problem posted involves finding the angular speed immediately after impact in a perfectly inelastic collision between a snowball and a sign. The initial angular speed before impact is calculated using the conservation of linear momentum, while the final angular speed after impact is calculated using the conservation of angular momentum. However, it is important to note that the center of mass is not free to translate due to external forces at the pivot, which must be considered when conserving angular momentum.
  • #1
issacnewton
998
29
Hi

I have posted the problem in files 1 and 2. Its the same problem split in two files. The problem asks to find the angular speed immediately after impact. The problem is from the chapter
of angular momentum but I am trying to solve it using conservation of linear momentum.
In part a, I calculated the angular speed immediately before the impact. Its
[itex]\omega_i=2.35 \,\mbox{rad/s} [/itex]. Now when the snowball hits the sign, we have a perfectly inelastic collision. So for calculating the speed immediately after the impact, we need to calculate the center of mass of the sign+snowball system.

[tex]r_{cm}=\frac{(0.4)(0.5)+(2.4)(0.25)}{0.4+2.4}=0.2857\, \mbox{m}[/tex]

So we set up the following equation for the conservation of momentum.

[tex]-m_{snow}v_{snow}+m_{sign}v_i \omega_i=(m_{snow}+m_{sign})r_{cm}\omega_f[/tex]

where [itex]v_i[/itex] is the velocity of the center of mass of the sign just before the impact. Since vi is the product of the distance of the center of mass from the axis of rotation and the initial angular velocity, we have

[tex]v_i=(0.25)(2.35)[/tex]

so we get, after plugging in the numbers,

[tex]-(0.4)(1.6)+(2.4)(0.25)(2.35)=(0.4+2.4)(0.2857)\omega_f[/tex]

[tex]\omega_f=0.9625\, \mbox{rad/s}[/tex]

But using the methods of angular momentum conservation, I get

[tex]\omega_f=0.498\, \mbox{rad/s}[/tex]So where am I going wrong ?

Thanks
 

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  • #2
IssacNewton said:
So where am I going wrong ?
Linear momentum is not conserved. The center of mass is not free to translate because of the external forces at the pivot. You need to conserve angular momentum about the pivot.
 
  • #3
Kuruman, thanks. I just overlook that fact.
 

What is angular momentum?

Angular momentum is a measure of an object's rotational motion. It is defined as the product of an object's moment of inertia and its angular velocity.

What is the conservation of angular momentum?

The conservation of angular momentum states that the total angular momentum of a system remains constant unless an external torque is applied. This means that if the initial angular momentum of a system is zero, it will remain zero unless acted upon by an external force.

How is angular momentum related to torque?

Angular momentum and torque are directly proportional to each other. This means that as torque is applied to an object, its angular momentum will change. The direction of the change in angular momentum is determined by the direction of the torque.

How is angular momentum calculated?

Angular momentum is calculated by multiplying an object's moment of inertia by its angular velocity. The moment of inertia depends on the mass and distribution of the object, while angular velocity is the rate at which the object is rotating.

What are some real-life applications of angular momentum?

Angular momentum is used in many fields, such as engineering, physics, and astronomy. It is important in designing machines that involve rotational motion, such as turbines and propellers. Angular momentum is also used to explain the motion of planets and other celestial bodies in space.

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