A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00-meter-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 1.30-kilogram rod has a light basket at the other end to catch the 0.163-kilogram puck. The puck slides across the ice with velocity →v (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates.If the rod makes one revolution every 0.736s after the puck is caught, what was the puck's speed just before it hit the rod?

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DevonFen DevonFen · Answer 1 · 2019-11-28T06:51:45+01:00

Explanation:

Let the mass of the rod be [tex]m_{r}[/tex]

Let the length of the rod be [tex]l[/tex]

Let the mass of the puck be [tex]m_{p}[/tex]

Let moment of inertia of the rod about pivot be [tex]I_{r}[/tex]

Let moment of inertia of the puck about one end be [tex]I_{p}[/tex]

Let [tex]v[/tex] be the velocity of the puck before impact.

Let ω be the angular velocity of the rod+puck system about pivot after collision.

Let [tex]T[/tex] be the time period of the rod+puck system about pivot after collision.

Given,

[tex]l_{r}=2m\\m_{r}=1.3Kg\\m_{p}=0.163Kg\\T=0.736s[/tex]

ω=[tex]\frac{2\pi}{T}[/tex]=[tex]\frac{2\pi}{0.736}[/tex]=[tex]8.53s^{-1}[/tex]

[tex]I_{p}=ml_{r}^{2}=0.163\times 4=0.652[/tex]

[tex]I_{r}=\frac{1}{3}m_{r}l_{r}^{2}=\frac{1}{3}\times 1.3\times 4=1.73[/tex]

Let us conserve the angular momentum of the rod and puck system before and after the impact.

The puck sticks to the rod end.

So,

[tex]m_{p}vl_{r}=(I_{r}+I_{p})w[/tex]

[tex]0.163\times 2\times v=(0.652+1.73)\times 8.53^{2}\\v=531.64ms^{-1}[/tex]