F=ma 2025 Problem 07

Physics · mechanics

Problem

Question: A mechanism consists of three point masses, each of mass

m

,connected by two massless rods of length

l

and a torsion spring acting as a hinge. The potential energy of the torsion spring is given by

U_s

. This system is designed to "walk" down a set of stairs, as shown in the figure. The angles

\theta_1

and

\theta_2

(see figure) represent the orientation of the rods,and their rates of change,

\omega_1

and

\omega_2

, are the corresponding angular velocities. Assume that the mass on the surface is instantaneously at rest. Which equation correctly describes the total energy of the system? (A)

E = m{l}^2\omega_{1}^2 + \frac{1}{2}m{l}^2\omega_{2}^2 + m{l}^2\omega_{1}\omega_{2}\cos \left(\theta_{1} + \theta_{2}\right) + {2mgl}\sin \theta_{1} + {mgl}\sin \theta_{2} + {U}_{s}

. (B)

E = \frac{1}{2}m{l}^2\omega_{1}^2 + \frac{1}{2}m{l}^2\omega_{2}^2 + {2mgl}\sin \theta_{1} + {mgl}\sin \theta_{2} - {U}_{s}

. (C)

E = m{l}^2\omega_{1}^2 + \frac{1}{2}m{l}^2\omega_{2}^2 + m{l}^2\omega_{1}\omega_{2}\cos \left(\theta_{1} - \theta_{2}\right) + {2mgl}\sin \theta_{1} + {mgl}\sin \theta_{2} + {U}_{s}

. (D)

E = m{l}^2\omega_{1}^2 + \frac{1}{2}m{l}^2\omega_{2}^2 - m{l}^2\omega_{1}\omega_{2}\cos \left(\theta_{1} - \theta_{2}\right) + {2mgl}\sin \theta_{1} + {mgl}\sin \theta_{2} + {U}_{s}

. (E)

E = \frac{1}{2}m{l}^2\omega_{1}^2 + \frac{1}{2}m{l}^2\omega_{2}^2 + {2mgl}\sin \theta_{1} + {mgl}\sin \theta_{2} + {U}_{s}

. *Note: This problem references figure(s): `image_question/F=MA_2025_07_1.png`*

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