A uniform rod of length b stands vertically upright on a rough floor and then tips over. What is the rod's angular velocity when it hits the floor?
To find the rod's angular velocity when it hits the floor, we can use the principle of conservation of energy.
Let's assume that the rod tips over without any external forces acting on it. In that case, the only form of energy that changes during the motion is the potential energy.
Initially, when the rod is standing upright, its center of mass is at a height of b/2 above the floor. The potential energy at this point is given by mgh, where m is the mass of the rod, g is the acceleration due to gravity, and h is the height.
When the rod hits the floor after tipping over, its center of mass is at the same level as the floor, so its potential energy is zero.
According to the conservation of energy, the initial potential energy should be equal to the final potential energy:
mgh = 0
We can rearrange this equation to solve for the height h:
h = 0/(mg)
Since h = b/2, we can substitute this into the equation:
b/2 = 0/(mg)
Simplifying further, we get:
b = 0
This means that the height is zero, which implies that the rod is already at the floor when it tips over. Therefore, the angular velocity when it hits the floor is zero.