Two masses (mA= 2 kg, mB= 4 kg) are attached to a (massless) meter stick, at the

Question

Two masses (mA= 2 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*?
N, (DOWN OR UP)

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

Explanation / Answer

Solution: a) Center of mass, Xcm = ( mA x1 + mB x2 )  / ( mA + mB )                                  = ( 2 * 0 + 4 * 75 ) / ( 2 + 4 )                                  = 50 cm Ans: Distance of center of mass, d = 50 cm (from A) b) Let Distance of point of suspension from A = x Distance of point of suspension from B = 75 - x About the point of suspension, Clockwise moments = Anticlockwise moments mB * g * (75 - x) = mA * g * x 4 ( 75 - x ) = 2 x 300 - 4 x = 2x 6x = 300 x = 50 cm Ans: Distance of point of suspension, x = 50 cm c) Let Force at 100 cm mark = F Distance of point of suspension from A = 50 cm Distance of point of suspension from 100 cm mark = 100 - 50                                                                              = 50 cm About the point of suspension, Clockwise moments = Anticlockwise moments F * 50 = mA * g * 50 F * 50 = 2 * 9.8 * 50 F = 19.6 N Ans: Force, F = 19.6N ( down )    d) Angular acceleration,  =  / I                                       = (mA g x) / (mA x^2)                                       = g / x                                        = 980 / 50                                        =19.6 rad/s^2 Ans: Angular acceleration,  = 19.6 rad/s^2       
Let Force at 100 cm mark = F Distance of point of suspension from A = 50 cm Distance of point of suspension from 100 cm mark = 100 - 50                                                                              = 50 cm About the point of suspension, Clockwise moments = Anticlockwise moments F * 50 = mA * g * 50 F * 50 = 2 * 9.8 * 50 F = 19.6 N Ans: Force, F = 19.6N ( down )    d) Angular acceleration,  =  / I                                       = (mA g x) / (mA x^2)                                       = g / x                                        = 980 / 50                                        =19.6 rad/s^2 Ans: Angular acceleration,  = 19.6 rad/s^2       

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