m = 60kg, ω0 = 2.094 rad/s, I of disk = 130 kgm^2 , outer position ro = 1.5m, inner position ri = 0.3m
∴Fifth object :
Ffriction = m.ac
μ.m.g = m. v^2 / R
=> vmax = √ 3. (1.5m) . (9.81 m/s^2 ) = 6.64 m/s => ωmax = 4.43 rad/s
so when the fifth object move with greater speed than vmax...
I think it should reach the same height because there is no outside force. we can proof it by using conservation of energy
PE initial = PE final => mgh1 = mgh2 => h1 = h2
Does it reach greater height than original height ? I'm not sure about this
- for part b). yes you're right. As i rounded...
mball = 2 kg, mputty = 0.05 kg, L = 0.5 m, v = 3m/s
a) Moment of inertia : I = (2mball + mputty ). ¼ L^2 = 0.253125 kg.m^2
Linitial = Lfinal => mputty. v. r = I.ω => ω = (4.mputty.v.r) / I = 0.148 rad/s
b) K initial = 1/2 m v^2 = 0.225 J
K final = 1/2 Iω^2 = 2.85.10^(-3) J => Kfinal /...
Oh i missed it, the height should be 0.5 m.
and yes the speed does not change that's why the Δp = m( v -(-v) ) but it's not really clear written as you said