susus
Dec 25, 2010, 01:26 AM
Spring Constant
A thin uniform rod has mass M = 0.5 kg and length L= 0.37 m. It has a pivot at one end and is at rest on a compressed spring as shown in (A). The rod is released from an angle θ1= 55.0o, and moves through its horizontal position at (B) and up to (C) where it stops with θ2 = 111.0o, and then falls back down. Friction at the pivot is negligible. Calculate the speed of the CM at (B).
(in m/s)
this is what I tried..
ranslational kinetic energy
TKE = 0.5mV^2
Angular kinetic energy
AKE = 0.5Iω^2
Where ω = Vsin(θ)/(L/2)
At (C) the rod has gravitational kinetic energy (GPE)
GPE = mgh
where h = L(cos(θ2) - cos(θ3))/2
ENERGY(B) = ENERGY(C)
0.5mV^2 + 0.5Iω^2 = mgh
0.5m(L(sin(θ2))^2/3 + 1)V^2 = 0.5mLg(cos(θ2) - cos(θ3))
V^2 = Lg(cos(θ2) - cos(θ3))/(L(sin(θ2))^2/3 + 1)
θ2 = 90
θ3 = 111
L = 0.37 m
g = 9.81 m/s^2
V^2 = 1.158
V = 1.076 m/s
but it's wrong :(:(:( I can not know why !
A thin uniform rod has mass M = 0.5 kg and length L= 0.37 m. It has a pivot at one end and is at rest on a compressed spring as shown in (A). The rod is released from an angle θ1= 55.0o, and moves through its horizontal position at (B) and up to (C) where it stops with θ2 = 111.0o, and then falls back down. Friction at the pivot is negligible. Calculate the speed of the CM at (B).
(in m/s)
this is what I tried..
ranslational kinetic energy
TKE = 0.5mV^2
Angular kinetic energy
AKE = 0.5Iω^2
Where ω = Vsin(θ)/(L/2)
At (C) the rod has gravitational kinetic energy (GPE)
GPE = mgh
where h = L(cos(θ2) - cos(θ3))/2
ENERGY(B) = ENERGY(C)
0.5mV^2 + 0.5Iω^2 = mgh
0.5m(L(sin(θ2))^2/3 + 1)V^2 = 0.5mLg(cos(θ2) - cos(θ3))
V^2 = Lg(cos(θ2) - cos(θ3))/(L(sin(θ2))^2/3 + 1)
θ2 = 90
θ3 = 111
L = 0.37 m
g = 9.81 m/s^2
V^2 = 1.158
V = 1.076 m/s
but it's wrong :(:(:( I can not know why !