Question

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.35 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed v0, so that the rod begins to rotate upward about the pivot. What must be the value of v0 such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to it initial orientation

Answer 1

Answer:

5.24 m/s

Explanation:

So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.

Since the rod is uniform in mass, we can treat the body as 1 point, at its center of mass, or geometric center, aka 0.35 / 2 = 0.175 m from the pivot.

For the rod to swing from bottom to top, the center must have moved a distance of h = 0.175 * 2 = 0.35 m, vertically speaking.

Since we neglect friction and air resistance, according to the law of energy conservation then:

$E_k = E_p$

$mv^2/2 = mgh$

Where v is the speed at the center of mass, g = 9.81 m/s2 is the gravitational acceleration, and m is the mass. We can divide both sides by m

$v^2 = 2gh = 2*9.81*0.35 = 6.867$

$v = \sqrt{6.867} = 2.62 m/s$

As this is only the speed at the center of mass, the speed at the bottom end would be different, to calculate this, we need to find the common angular speed:

$\omega = v / r = 2.62 / 0.175 = 14.97 rad/s$

Where r is the rotation radius, or the distance from pivot point to the center of mass

$v_0 = \omega R = 14.97*0.35 = 5.24 m/s$

Where R is the distance from the pivot to the bottom end of the rod