Two masses (m A = 3 kg, m B = 4 kg) are attached to a (massless) meter stick, at

Question

Two masses (mA= 3 kg, mB= 4 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively.

a.) Where is the center of mass of this system?
____________________cm, from mass A.

b.) The system is then hung from a string, so that it stays horizontal. Where should the string be placed?
____________________cm, from mass A.

c.) Now, if mass B was removed, how much force would need to be exerted at the 100 cm mark in order to keep the meter stick level*?
size: ___________ N, dir: upwards or downwards.

d.) Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant*.
__________________________ rad/s2

*For parts c-d, assume the string remains attached at the same location you found in part b.

Explanation / Answer

Let:
T be the tension in the string,
x cm. be the distance of the centre of mass from the 0cm. mark.

(a)
Moments about the centre of mass:
3 * x = 4(75 - x)
300 - 4x = 3x
x = 300 / 7
= 42.85 cm.

(b)
As the masses have no moment about the C of M, that is where the string should be attached.

(c)
Let:
F be the downward force needed at 100cm,
x cm. be the new string position,
9.81 m/s^2 be the acceleration due to gravity.

Moments about the end of the string:
1 * 0.4285 * 9.81 = F(1 - 0.4285)
F = 7.355 N.

(d)
Let:
a be the acceleration (initially downwards) of the mass.
g be the acceleration due to gravity.

mg = ma
a = g.

The angular acceleration is:
9.81 / 0.4285
= 22.89 rad / s.

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