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
Oh if so "v" in n.m.v is the change in velocity
i just showed the full formula with the part n.m.v but didn't know how to deal with this part so i used instead Δp / Δt
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...
n is the number of particles or in this case glass beads and its unit is 1/s i think ?
and so i think 100/s = 1/ Δt => Δt = 1s/100
=> F = 0.313 N
=> M = 0.313/ 9.81 = 31.9 g
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...
So work done equal negative change of potential energy
and in case of kinetic energy, work done is equal to the change of kinetic energy
am i correct ?
Yes i have calculated all the quantities you mentioned, and find each distance x1,x2,x3 which add up 60km. So i think this should be correct. Thank you very much for your help
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)...