Given that
intial angular acceleration ω0
final angular accleration ω = which is constant
α = angular accleration
by using kinematic eq = 0 = ω0 +αt
= α = -ω0/t
= - 39/32 = - 1.21 rev/s
α = -1.21 *2π = - 7.66 rad/s2
(B) the moment of inertia of rod = I = Ml2/12
the contribution of each ball = M(l/2)2
total moment of inertia = I = Ml2/12 + 2 ml2/4
plug values and do caliculations for I
(c) since the system comes to rest the mechnical energy is converted to thermal energy which is simply intial kinetic energy
Ki = 1/2 Iω02
plug values docaliculations for Ki----J
(d) applying kinematic eq =
θ = ω0t + 1/2 αt2
2π ( 39) (32) +1/2 (-7.66) (32)
2
docaliculations and that value gives angular displacement
(e) only mechnial energy is converted into thermal energy can still be computed without any additional information