Question

If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod? (See details).

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-m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 1.90-kg rod has a light basket at the other end to catch the 0.163-kg 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.

Answer 1

We know that the linear velocity `v` of the puck is related to its angular velocity `omega` through the expression

`v = rω`
`=> ω = v/r`
where `r` is radius of rotation.

Also angular momentum

`L=I*omega`
where `I` is moment of inertia

From the Law of Conservation of angular momentum we know that

`"Initial Angular Momentum"\ =\ "Final Angular Momentum"`
`(I*ω)_i = (I*ω)_f`

LHS of above expression can be rewritten as

`(I*v/r)_i = (I*ω)_f`

Considering angular momentum around the pivot.
At the time of impact, only the puck is moving. As the rod is stationary it does not contribute to initial angular momentum.

Moment of inertia of puck about the pivot `I_p=m_pr^2`

`LHS_"puck only"=m_pr^2*v/r=m_pvr`

After the impact puck-rod system rotates. Moment of inertia of rod around its end is `I_r=1/3M_rr^2`

`:.RHS_"puck rod"=(m_pr^2+1/3M_rr^2)omega`

Equating both sides and solving for `v` we get

`v = ([m_pr^2+(1/3M_rr^2)] ω )/(m_pr)`

Noting that `omega=(2pi)/T=(2pi)/0.736.`
Inserting given values we get

`v = ([0.163xx(2.00)^2+(1/3xx1.90xx(2.00)^2)] (2pi)/0.736 )/(0.163xx2.00)`
`=>v = 83.41\ ms^-1`