Two masses (mA= 2 kg, mB= 6 kg) are attached to a (massless) meter stick, at the

Question

Two masses (mA= 2 kg, mB= 6 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively.

9. +14.25 points My Notes Ask Two masses (mA= 2 kg, mB= 6 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: ---Select-- 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:

2 * x = 6*(75 - x)

450 - 6x = 2x

x = 450/8

= 56.25 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:

2 * 0.5625 * 9.81 = F(1 - 0.5625)

F = 25.22 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 / 0.5625

= 17.44 rad / s

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