The law of conservation of angular momentum
I₁•ω₁ = I₂•ω₂
I₁ =mR²/2, ω₁ = 5.1 rad/s
I₂ = mR²/2 +m₀•R². ω₂ = ?
(mR²/2) • ω₁=(mR²/2 +m₀•R²)•ω₂
Solve for “ ω₂ “
A solid, horizontal cylinder of mass 11.0 kg and radius 1.70 m rotates with an angular speed of 5.10 rad/s about a fixed vertical axis through its center. A 1.21 kg piece of putty is dropped vertically onto the cylinder at a point 0.600 m from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.
3 answers
The value I am getting is saying its too large, how large is your value?
The law of conservation of angular momentum
I₁•ω₁ = I₂•ω₂
I₁ =mR²/2, ω₁ = 5.1 rad/s
I₂ = mR²/2 +m₀•r². ω₂ = ?
(mR²/2) • ω₁=(mR²/2 +m₀•r²)•ω₂
ω₂ =(mR²/2) • ω₁/(mR²/2 +m₀•r²)=
=m•R²•ω₁/(m•R²+2 •m₀•r²)=
=11•1.7²•5.1/(11•1.7² +2•1.21•0.6²) =
=4.96 rad/s
I₁•ω₁ = I₂•ω₂
I₁ =mR²/2, ω₁ = 5.1 rad/s
I₂ = mR²/2 +m₀•r². ω₂ = ?
(mR²/2) • ω₁=(mR²/2 +m₀•r²)•ω₂
ω₂ =(mR²/2) • ω₁/(mR²/2 +m₀•r²)=
=m•R²•ω₁/(m•R²+2 •m₀•r²)=
=11•1.7²•5.1/(11•1.7² +2•1.21•0.6²) =
=4.96 rad/s