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
What do you mean? I'm confused... You need to put the rest of the question
Answer:
(A) It will take 22 sec to come in rest
(b) Work done for coming in rest will be 0.2131 J
Explanation:
We have given the player turntable initially rotating at speed of 
Now speed is reduced by 75 %
So final speed 
Time t = 5.5 sec
From first equation of motion we know that '

(a) Now final velocity 
So time t to come in rest 
(b) The work done in coming rest is given by
