three particles of masses 1 kg 2 kg and 3 kg kept along a straight line in the same order with equal separation of 1 M
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Fix A as your reference point. So your 1 kg mass is at A (0, 0), 2 kg mass is at B (0.5, 0.866) and the 3 kg mass is at C (1, 0)
By definition Centre of Mass is the point where the whole mass of the system appears to be concentrated and it is the weighted mean of coordinates along with their masses
ie X (CM) = (m1x1+m2x2+m3x3...+mnxn)/(m1+m2+m3...+mn)
Likewise, Y (CM) is obtained by using the Y coordinates.
(X (CM), Y (CM)) gives you the coordinates of the centre of mass.
In this case
m1 = 1kg (x1, y1)=(0, 0)
m2 = 2kg (x2, y2)=(0.5, 0.866)
m3 = 3kg (x3, y3)=(1, 0)
Now apply the formula
X (CM) =( (1*0)+(2*0.5)+(3*1))/(1+2+3) = 0.667.
Y (CM)= ((1*0)+(2*0.866)+(3*0))/(1+2+3) = 0.286
Now, A (0, 0) CM (0.667, 0.286)
Use the distance between two points formula.. The answer will be 0.7267 metres.
By definition Centre of Mass is the point where the whole mass of the system appears to be concentrated and it is the weighted mean of coordinates along with their masses
ie X (CM) = (m1x1+m2x2+m3x3...+mnxn)/(m1+m2+m3...+mn)
Likewise, Y (CM) is obtained by using the Y coordinates.
(X (CM), Y (CM)) gives you the coordinates of the centre of mass.
In this case
m1 = 1kg (x1, y1)=(0, 0)
m2 = 2kg (x2, y2)=(0.5, 0.866)
m3 = 3kg (x3, y3)=(1, 0)
Now apply the formula
X (CM) =( (1*0)+(2*0.5)+(3*1))/(1+2+3) = 0.667.
Y (CM)= ((1*0)+(2*0.866)+(3*0))/(1+2+3) = 0.286
Now, A (0, 0) CM (0.667, 0.286)
Use the distance between two points formula.. The answer will be 0.7267 metres.
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