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 1.70-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.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?
Right now I am trying to relate angular momentum to linear momentum because I think it is conserved and I keep getting 60 m/s but that isn't right.
L
To solve this problem, we can apply the principle of conservation of angular momentum. When the puck collides with the basket, angular momentum is conserved before and after the collision since there is no external torque acting on the system.
The formula for angular momentum is given by:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Initially, before the collision, the rod is stationary, so its initial angular momentum is zero. However, the puck has linear momentum, which can be related to angular momentum as:
L_initial = m_puck * v_puck * r
Where m_puck is the mass of the puck, v_puck is its velocity, and r is the distance from the pivot point to the puck's trajectory.
After the collision, the puck is caught by the basket, and the system starts rotating. The final angular momentum can be calculated as:
L_final = I_rod * ω
According to the problem, the rod makes one revolution every 0.736 s. One revolution corresponds to 2π radians, so the angular velocity can be calculated as:
ω = (2π radians) / (0.736 s)
Now, we need to determine the moment of inertia of the rod, which is given by:
I_rod = (1/3) * m_rod * L^2
Where m_rod is the mass of the rod, and L is the length of the rod.
Plugging in the given values:
m_rod = 1.70 kg
L = 2.00 m
We can calculate I_rod.
With the final angular momentum determined, we can equate the initial and final angular momenta:
L_initial = L_final
m_puck * v_puck * r = I_rod * ω
Substituting the known values and solving for v_puck:
0.163 kg * v_puck * r = (1/3) * 1.70 kg * 2.00 m^2 * (2π radians) / (0.736 s)
Now, we can solve this equation to find the puck's speed just before it hit the rod. Remember to multiply the final answer by the radius "r" to get the linear speed:
v_puck = (1/3) * 1.70 kg * 2.00 m^2 * (2π radians) / (0.736 s * 0.163 kg * r)
Simplifying this equation will give you the correct answer.