center of mass of meter stick?

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, ---Select--- upwards. 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.

answer my other questions posted.

Answer 1

(a)
Sum of moments =0
r1 x 2 = r2 x 4 so r1=2 r2
also

r1+r2=75
2r2+r2=75 then
r2=25cm
and r1 =2r2=50cm

b) on the 50cm mark ( or .5m )

c) Again sum moments =0
r1 x 2= r2 x m
r1=50 cm ; r2=100-50=50cm
m=r1 x 2/r2= 50 2/50= 2kg
F=mg= 2 x 9.81=19.62 N downward

d) alpha = T/I
for a point mass I=mr^2
T=r x F= r mg
alpha = r mg/ m r^2=9.81/r=9.81/.5=19.62 rad/sec^2

Answer 2

a) x_cm=(mA*xA+mB*xB)/(mB+mB) --it really is, the gap of the middle of mass out of your reference factor (for this reason x=0) is mA circumstances its displacement from the reference factor (0cm) plus mB circumstances its displacement (75cm) everywhere in the completed mass of the device. b) The string should be positioned the position the torques exerted by technique of each mass are equivalent. t=mgL --torque equals mass circumstances the acceleration of gravity circumstances the gap between the stress (the a lot weighing down on the ruler) and the point of rotation (lever arm), that's the position your string will be positioned. the following, mA*g*L1=mB*g*L2 the position L1+L2=75cm and g cancels out. L1 is what you're trying to resolve for, so set L2=seventy 5-L1 and change: mA*L1=mB*(seventy 5-L1). because that mA and mB are given, you are able to extremely opt for for L1. c) the mandatory stress equals the torque from mA, mA*g*L1, divided by technique of the gap between the string and the 100cm mark, 100-L1. d) Angular acceleration equals tangential acceleration divided by technique of radius, for this reason g/L1 on the instantaneous of launch.

Answer 3

a♠ for center of mass mA*xA=mB*xB and xA+xB=L=75cm, where xA is arm from center of mass to mA=2kg, and xB is arm from center of mass to mB=4kg, thence mA*xA=mB*(L-xA), hence xA=L*mB/(mA+mB) = 75*4/(2+4) = 50 cm;
b♠ on mark 50 cm!! since it is center of mass, i.e. torques are equal, so
TA=TB; or; FA*xA=FB*xB; or; (mA*g)*xA = (mB*g)*xB,
hence deleting g we get the equation above;
c♣ removing B we must save center of mass by compensating with mass mC determined from equation mA*50=mC*(100-50), hence mC=mA=2kg; and force =mC*g = 19.6N;
d♦ Newton says: T=mA*g*xA = I*w’, where I=mA*xA^2,
hence w’= T/I = mA*g*xA / (mA*xA^2) =g/xA = 980/50 = 19.6 rad/s^2;