A thin uniform rod of length l and mass
m is swinging freely about a horizontal
axis passing through its end. Its
maximum angular speed is . Its
centre of mass rises to a maximum
height of :
Answers
Expalaination
The moment of inertia of the rod about O is 1 /3 ml² .
The maximum angular speed of the rod is when the rod is instantaneously vertical.
The energy of the rod in this condition is 1/2 Iw² where I is the moment of inertia of the rod about O.
When the rod is in its extreme portion, its angular velocity is zero momentarily.
In this case, the energy of the rod is mgh where h is the maximum height to which the centre of mass (C.M) rises
mgh= 1/2 Iw² =1/2 (1/3m l²) w²
Therefore h = l²w² /6g
Answer:
1/6 l^2w^2/g
Explanation:
) : The uniform rod of length l and mass m is swinging about an axis passing through the end. When the centre of mass is raised through h, the increase in potential energy is mgh. This is equal to the kinetic energy = (1/2)Iω2. ⇒ mgh = 1/2(m(l2/3)).ω2 Therefore, h = (l2 . ω2)/6g.Read more on Sarthaks.com - https://www.sarthaks.com/40325/thin-uniform-rod-length-mass-swinging-freely-about-horizontal-axis-passing-through-its-end