Question

Three small objects are arranged along a uniform rod of mass m and length L. one of mass m at the left end, one of mass m at the
center, and one of mass 2m at the right end. How far to the left or right of the rod's center should you place a support so that the rod
with the attached objects will balance there?

Answer 1

Answer:

See answer below

Explanation:

Hi there,

To get started, recall the Center of Mass formula for two masses:

$x_c_m = \frac{m_1x_1+m_2x_2}{m_1+m_2}$  where m is mass and x is displacement from the center of the shape.

Since masses at the center of a geometric shape have a displacement (x) value of 0, as the mass is already of the center, and does not affect Xcm. So, we can disregard the central mass, hence we use the above formula for two masses.

We can arbitrarily define left to be a negative (-) displacement, and vice versa for right direction. We proceed with the formula:$x_c_m=\frac{(-L/2)m+(L/2)2m}{m+2m} =\frac{(L/2)(-m+2m)}{3m} \\ x_c_m=\frac{L(m)}{6m} =\frac{L}{6}$

Since we defined left (-) and right (+), we notice the center of mass is (+) value. This makes sense, as there is slightly more mass on the right side. Hence, you should place a support 1/6 of the rod's length away from the rod's center.

Study well and persevere.

thanks,