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 /...
I'm not sure about it, but as i read, n is number of beads.
I think because F = n.m.v
F has SI units of kg.m/s^2
on the right side, m.v is [kg.m/s] so i think n should be 1/s so that it has equal unit to force ?
or is it number of beads per second ?
F = Δp / Δt
As i said above, 100/s = 1/ Δt...
To find the mass in other pan, i need to find the force caused by beads on the pan
∴ KEinitial + PEinitial = KEfinal + PEfinal
0 + mgh = ½ mv^2
=> v = 3.13 m/s
∴ The change in momentum :
p2 - p1 = m ( v2-v1) = m( v - (-v)) = 2mv
∴ F = Δp / Δt = n. m. v
How can i apply the rate of 100...
This is the hint which i forgot to put in the question statement: It could be that you get more than one answer .
After considering your help and calculating again, i really got two possible maximum speed value like the hint stated
I want to ask how i can check if my results are correct in...
total time: t = 36 mins = 0.6h = t1 + t2
=> t2 = t - t1 = 0.6h - 0.1h = 0.5h
Vmax = a1 x t1
Vat C = Vmax + a2t2
substitute Vmax in Vat C we have : 0 = a1 x 0.1h + (-600 km/h²) x 0.5h => a1 = 3000km/h²
Vmax = a1 x t1 = 3000 x 0.1 = 300km/h
I check the result by:
x1 = ½ a1 t1² = ½ . (3000)...