Two masses (m_a= 3 kg, m_b= 5 kg) are attached to a (massless) meter stick, at t

Question

Two masses (m_a= 3 kg, m_b= 5 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively. Where is the center of mass of this system? The system is then hung from a string, so that it stays horizontal. Where should the string be placed? 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*? 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*. *For parts c-d, assume the string remains attached at the same location you found in part b.

Explanation / Answer

a) location of center of mass is Xcm = (m1*x1)+(m2*x2)/(m1+m2)

Xcm = ((3*0)+(5*0.75))/(3+5) = 0.46875 m = 46.875 cm

Yxm = 0 m

b) it should hunh at center of mass

xcm = 0.46875 m = 46.875 cm

c) net torque acting is zero

m1*g*46.875 = F*(100-46.875)

3*9.8*46.875 = F*(100-46.875)

Force = F = 25.94 N

D) Torque T = I*alpha = r*F = r*m1*g = 0.46875*3*9.8 = 13.78

I = m*r^2 = 3*0.46875^2 = 0.659 kg m^2

then alpha = 13.78/0.659 = 20.9 rad/s^2

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