1. Two 0.1 kg-masses are connected to a linear spring on a frictionless table. T
ID: 1557943 • Letter: 1
Question
1. Two 0.1 kg-masses are connected to a linear spring on a frictionless table. The center of mass of the 2-particle system is stationary. At the instant shown, the velocity is v¯1 = v¯2 = 0.1eˆr + 2.5eˆ m/s and r=0.5 m. The spring constant, K=10 N/m, and the spring applies no force when the masses are at the origin, so F=2Kr.
a. Determine the maximum and minimum distances to the origin (rmin and rmax)
b. Determine the speed at these points, |v|, when r = rmin and when r = rmax
mV yt mV Figure 1: Two masses connected by a springExplanation / Answer
a) When the distances are maximum and minimum the velocity in e^r direction will be zero.
Hence from the conservation of energy,
0.5*10*(1^2-4r^2) + 2*0.5*0.1*(0.1^2+2.5^2) = 2*0.5*0.1*v_theta^2 -------------(i)
the v_theta won't change because there is no impulse in e^theta direction hence this will remain same.
5*(1-4r^2) + 0.626 = 0.625
4r^2 = 1-0.001/5 = 0.9998
r = +- 0.499 m.
b) the velocity will be 2.5 m/s as the theta component will remain same and the r component will be zero.
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