Problem involving angular momentum

[issacnewton]

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
$\omega_i=2.35 \,\mbox{rad/s}$. 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.

$$r_{cm}=\frac{(0.4)(0.5)+(2.4)(0.25)}{0.4+2.4}=0.2857\, \mbox{m}$$

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

$$-m_{snow}v_{snow}+m_{sign}v_i \omega_i=(m_{snow}+m_{sign})r_{cm}\omega_f$$

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

$$v_i=(0.25)(2.35)$$

so we get, after plugging in the numbers,

$$-(0.4)(1.6)+(2.4)(0.25)(2.35)=(0.4+2.4)(0.2857)\omega_f$$

$$\omega_f=0.9625\, \mbox{rad/s}$$

But using the methods of angular momentum conservation, I get

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

Thanks

 

[kuruman]

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.

 

[issacnewton]

Kuruman, thanks. I just overlook that fact.