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A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sphere of mass 0.25 kg is thrown horizontally with a speed v1 = 12.5 m/s to hit the rod at the point a one-fifth of the way up from the bottom of the rod. the sphere bounces back horizontally with a speed v2 = 9.50 m/s, while the rod swings to the right through an angle θ before swinging back toward its original position. what is the angular speed of the rod immediately after the collision?

Respuesta :

shinmin

Solution:


initial sphere mvr = final sphere mvr + Iω 
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m² 
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω 
where: ω = 2.87 rad/s 

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)² 
E = 12.64 J becomes PE = mgh, so 
12.64 J = 2.3 kg * 9.8m/s² * h 
h = 0.29 m 

h = L(1 - cosΘ) → where here L is the distance to the CM 
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ 
Θ = arccos((1-0.29)/1) = 44.77 º 

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