A rod of length L = 1.2 m pivots freely about one end. The rod is held horizontally and released, with no intial velocity. Find the angular velocity of the rod as it passes through the vertical orientation. (Use conservation of energy. The rotational inertia of a rod like this is I = (1/3)mL2.) The units are rad/s.
2 answers
The answer I get is 12.24 rad/s. Can anyone confirm this?
PE =KE =KE(trans)+ KE(rot) =
mgL=mv²/2 +Iω²/2=
=m ω²L²/2 + m ω²L²/3•2=
=2 m ω²L²/3,
ω=sqrt(3g/2L)=sqrt(3•9.8/2•1.2) =
=sqrt(12.24)=3.5 rad/s
mgL=mv²/2 +Iω²/2=
=m ω²L²/2 + m ω²L²/3•2=
=2 m ω²L²/3,
ω=sqrt(3g/2L)=sqrt(3•9.8/2•1.2) =
=sqrt(12.24)=3.5 rad/s
1 answer