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.
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
To find the angular velocity of the rod as it passes through the vertical orientation, we can make use of the principle of conservation of energy.
The initial energy of the system consists of two parts: potential energy and rotational kinetic energy.
The potential energy at the horizontal position is zero since there is no vertical displacement. Therefore, the total initial energy is given by the rotational kinetic energy of the rod.
The rotational kinetic energy of the rod is given by the expression:
KE = (1/2) * I * ω^2
Where:
KE is the rotational kinetic energy,
I is the moment of inertia of the rod about its pivot point, and
ω is the angular velocity of the rod.
Given:
L = 1.2 m (length of the rod),
I = (1/3) * m * L^2 (rotational inertia of the rod, where m is the mass of the rod)
Since the rod is initially held horizontally and released with no initial velocity, the potential energy at the vertical position is also zero. Therefore, at the vertical position, the total energy is entirely converted into rotational kinetic energy.
Setting the initial energy equal to the energy at the vertical position, we have:
(1/2) * I * ω^2 = 0
Simplifying the equation, we get:
(1/2) * (1/3) * m * L^2 * ω^2 = 0
Solving for ω, we find:
ω = 0 rad/s
Hence, the angular velocity of the rod as it passes through the vertical orientation is 0 rad/s.