Question

A barbell is 1.5 m long. Three weights, each of mass 20 kg, are hung on the left and two weights of the same mass, on the right. The width of each weight is 4 cm and each group of weights is placed 4 cm from the ends. Where is the center of mass of the barbell as measured from the mid-point, M, of the bar? The bar is of uniform mass and has mass 5 kg, and the retaining collars are of negligible mass. Take to the right as positivea. -5.90cmb. -11.6cmc. +13.7cmd. +5.90cme. none of the above

Answer 1

Answer:

b. -11.6 cm

Explanation:

We have given parameters:

Length, l = 1.5 m = 150 cm

Mass of weight, $m_1$ = 20 kg

Width, x = 4 cm

Distance d = 4 cm

Mass of bar, $m_{bar}$ = 5 kg

We are asked to find the center of mass from the mid-point, $X_{CM} = ?$

Since 3 weights are on the left and 2 weights are on the right, we know:

$m_{left}$ = 3 * 20 = 60 kg

$m_{right}$ = 2 * 20 = 40 kg

And also we know that, $M = \frac{l}{2}$ = 150/2 = 75 cm

For the left side, center of mass is:

$x_{left} = \frac{3 * 4}{2} = 6$ cm

From the midpoint, the distance to the left is:

$X_{left} = -(M - 4 - x_{left}) = -(75 - 4 -6) = -65$ cm

For the right side, center of mass is:

$x_{right} = \frac{2 * 4}{2} = 4$ cm

From the midpoint, the distance to the right will be:

$X_{right} = (M - 4 - x_{right}) = (75 - 4 - 4) = 67$ cm

Hence,

$X_{CM} = \frac{m_{right}*x_{right} + m_{left}*x_{left} }{m_{right} + m_{left} + m_{bar}} = \frac{40 * 67 - 60 * 65}{40 + 60 + 5} = -11.62$ cm