Question

A slender rod is 80.0 cm long and has mass 0.290 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.0500 kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest. What is the linear speed of the 0.0500 kg sphere as it passes through its lowest point?

Answer 1

Answer:

V= 1.2m/s

Explanation:

Since both spheres fall to an equidistant center, both spheres travel a distance of 0.4m

This is:

The 0.05kg sphere falls 0.4m while the 0.02kg sphere goes up 0.4m

The decrease in Potential Energy is given by:

$\Delta PE = (m2-m1) * g * h = (0.05-0.02) * 9.8 * 0.4 = 0.1176 J$

The moment of inertia for the entire system is given by:

$I = \frac{mL ^ 2}{12}+(m1 + m2) (\frac{L}{2} )^2$

$I = 0.290kg * (0.8m) ^ 2/12 + 0.07kg * (0.4m)^2 = 0.026Kg * m^2$

In this way the kinetic energy is given by

$KE = \frac{1}{2} Iw^2$

$0.1176J = \frac{1}{2}* 0.026Kg *m^2 * w^2$

$w = 3 rad / s$

$v = wr = 3 * 0.4 = 1.2m / s$