Question

In the figure below, two identical containers of sugar are connected by a cord that passes over a frictionless pulley. The cord and pulley have negligible mass, each container and its sugar together have a mass of 970 g, the centers of the containers are separated by 50 mm, and the containers are held fixed at the same height. (a) What is the horizontal distance between the center of container 1 and the center of mass of the two-container system initially? mm (b) Now 35 g of sugar is transferred from container 1 to the container 2. What is the horizontal distance between the center of container 1 and the center of mass of the system? mm (c) The two containers are now released. In what direction does the center of mass move? (d) At what acceleration does the center of mass move? Magnitude m/s2 Direction

Answer 1

(b.) Each container is originally 25 mm from the center

Let x = distance of the larger mass from the new COM
50 - x = distance of the lesser mass from the COM

Larger mass = M = 970 + 35 = 1005
Lesser mass = m = 970 - 35 = 935

M*x = m*(50 - x)
x*(M + m) = 50*m
x = 50 *m / (M + m)
x = 50*935/1940) = 24.09 mm

The distance from the COM for the lesser mass is 50 - 24.09 = 25.902 mm. Ans

​(​​​​​d.)  Find the acceleration of the masses.

T = tension in the string.
a = downward acceleration so it is positive.
Force on M: F_{​​​​​​M}= M*g - T = M*a

Force on m: F_{​​​​​​m} = m*g - T = -m * a
Subtract the second equation from the first:

M*g - m*g = M * a + m *a

a = g * [(M - m)/(M + m)]

a = 9.8* [ (1005 - 935)/ (1005+935)]

a = 9.8*[ 70/1940]

a = 0.3536 m/s^{​​​​​​2}​

So the heavier mass is accelerating downwards at 0.3536 m/s²and the lighter is accelerating upward at -0.3536 m/s^{​​​​​​2}

For the COM we can now write :
A_{​​​​​​Com} = (M*a - m*a)/(M + m) = a*[(M - m)/(M + m)]
A_{​​​​​​Com} = 0.3536* [ 70/1940]

A_{​​​​​​Com} = 0.0127 m/s^{​​​​2}Ans