→ rθ(t) ω 2m, r θ(t) k → 2rθ(t) g → x(t) F(t) ϕ(t) y(t) m L L

Name: → rθ(t) ω 2m, r θ(t) k → 2rθ(t) g → x(t) F(t) ϕ(t) y(t) m L L
Uploaded: 2023-07-27T08:31:48-08:00
Duration: 2 min 9 s
Channel: Numerade
Description: → rθ(t) ω 2m, r θ(t) k → 2rθ(t) g → x(t) F(t) ϕ(t) y(t) m L L

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→ rθ(t) ω 2m, r θ(t) k → 2rθ(t) g → x(t) F(t) ϕ(t) y(t) m L...

$\rightarrow r\theta(t)$ \newline $\omega$ \newline $2m, r$ \newline $\theta(t)$ \newline $k$ \newline $\rightarrow 2r\theta(t)$ \newline $g$ \newline $\rightarrow x(t)$ \newline $F(t)$ \newline $\phi(t)$ \newline $y(t)$ \newline $m$ \newline $L$ \newline $L$

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Consider the system shown below. The disk rolls without slip and has a radius of ð‘Ÿ and a mass of 2ð‘š. The pulley is massless and frictionless. The rod is assumed to be rigid, homogeneous, and uniform such that it has a mass of ð‘š, a length of 2ð¿, and its center of mass is located directly at its center (i.e., a distance ð¿ from either end). The left end of the rod is constrained to move in the vertical direction only whereas the right end is constrained to move in the horizontal direction only. As such, there is a kinematic constraint between ð‘¥ and ðœ™ as well as ð‘¦ and ðœ™ (i.e., you can express ð‘¥ and ð‘¦ in terms of ðœ™). An external, vertical force, ð¹(ð‘¡), is applied to the left end of the rod. Gravity is included. Determine the equations of motion in terms of ðœƒ(ð‘¡) and ðœ™(ð‘¡). You will need to express ð‘¥(ð‘¡) and ð‘¦(ð‘¡) in terms of ð¿ and ðœ™(ð‘¡). Determine the total energy at an arbitrary time assuming that the system begins with zero initial energy in terms of ðœƒ(ð‘¡) and ðœ™(ð‘¡).