A system consisting of two elastically constrained masses m1 and m2 is free to s

Question

A system consisting of two elastically constrained masses m1 and m2 is free to slide on a frictionless rod that rotates about the pivot point O. Mass m1 is attached to O along the rod by spring of stiffness k1 with a free length of l1, and m1 and m2 are connected by a spring of stiffness is length k2 and l2, respectively. The angle ? describes the rod's orientation relative to the horizontal, and the position of m1 and m2 along the rod at given by L1 and L2, respectively. Find the expressions for the system's total kinetic and potential energies.

Correct answer:

KE=.5m1[(L1)^2+(L1(thetadot))^2]+.5m2[(L2)^2+(L2(thetadot))^2]

PE=m1gL1sin(theta)+m2gL2sin(theta)+.5k1(L1-l1)^2+.5k2(L2-L1-l2)^2

Explanation / Answer

Vertical height of m1 is h1 = L1*Sin theta

Hence, PE due to m1 is PE1 = m1*g*h1 = m1*g*L1*Sin theta

Vertical height of m2 is h2 = L2*Sin theta

Hence, PE due to m2 is PE2 = m2*g*h2 = m2*g*L2*Sin theta

Extension in Spring1 is x1 = L1 - l1

PE due to spring k1 is PE3 = 0.5*k1*x1^2 = 0.5*k1*(L1 - l1)^2

Extension in Sprin21 is x2 = L2 - L1 - l2

PE due to spring k2 is PE4 = 0.5*k2*x2^2 = 0.5*k2*(L2 -L1 - l1)^2

Hence Total PE = PE1 + PE2 + PE3 + PE4

TotalPE=m1gL1sin(theta)+m2gL2sin(theta)+.5k1(L1-l1)^2+.5k2(L2-L1-l2)^2

Linear velocity of m1 is L1dot

Angular velocity of m1 is L1*(thetadot)

Hence net velocity of m1 is v1 = [(L1dot)^2 + (L1*thetadot)^2]^0.5

Hence KE due to m1 is KE1 = 1/2*m1*v1^2

KE1 =.5m1[(L1)^2+(L1(thetadot))^2]

Linear velocity of m2 is L2dot

Angular velocity of m2 is L2*(thetadot)

Hence net velocity of m2 is v2 = [(L2dot)^2 + (L2*thetadot)^2]^0.5

Hence KE due to m2 is KE2 = 1/2*m2*v2^2

KE2 =.5m2[(L2)^2+(L2(thetadot))^2]

Total KE = KE1 + KE2

KE=.5m1[(L1)^2+(L1(thetadot))^2]+.5m2[(L2)^2+(L2(thetadot))^2]

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