Two pieces of lead are placed on a half-meter stick. Where: X1 = 2.00 cm, X2 = 3

Question

Two pieces of lead are placed on a half-meter stick. Where: X1 = 2.00 cm, X2 = 37.00cm, XR = 25.00 cm / M1 = 5.0 kg, M2 = 15.0 kg, MR = 0.50 kg.

(1) Where is the center of mass of this configuration Xcofg located along the ruler?
(2) Which mass must move to get Xcofg to the center of the ruler? What is the new position of this mass?

Two pieces of lead are placed on a half-meter stick as shown on the diagranm below: 2. m, 0 50 XR X2 X1 ol Cm Where: x 2.00cm X2 37.00cm XR = 25.00cm 37cm mi = 5.0 kg m2 15.0 kg = 0.50 kg MR

Explanation / Answer

Although unclear but if mass of the ruler is given as mr,

Considering xr to be reference point, and left side of it as negative and right side of it as positive,

1) x= m1r1+ m2r2+0/m1+m2+mr= 3.17cm from xr.

2)0= m1r1+ m2r2

Keeping r2 same, r1= -36cm which is not possible as its max value is -25 cm.

So, keeping r1 same, r2= -7.66 cm which is possible.

So, the new position of this mass m2 is 17.34 cm from the left end of the stick.

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