QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
# MECE 3338: Funwork 10 – Due Date Monday 11 / 13 / 2023 11:59 PM
## Problem 1:
Consider the following system that consists of an inverted pendulum OB with mass **m = 6 kg** and length ** 2α** where **α = 0.5 m**. The pendulum rotates about its bottom end point O with rotational viscous friction coefficient **b_r = 8 N m s/rad** and is supported by two horizontal springs of equal stiffness **k = 100 N/m** attached at its middle point C, as shown. A horizontal spring of same stiffness **k = 100 N/m** is attached at the pendulum's top point **D**. A **displacement input z(2$)** at the free end point A of the spring sets the system to motion. Small angular displacements from the vertical equilibrium position of the pendulum are considered.
a) Derive the transfer function **G(s)** of the system from the **displacement input z(2$)** to the **angular displacement output θ(t)** of the pendulum. Use the values of the system parameters **α, m, k** and **b_r** as stated above in the calculation of **G(s)**.
b) Considering a **sinusoid displacement input z(2$) = A sin(2$)** find the **worst-case excitation frequency α'** that results in the **maximum steady-state angular displacement amplitude |θ|_max** of the pendulum from its vertical equilibrium position.
c) Find this **maximum angular displacement amplitude |θ|_max** if the amplitude of the sinusoid displacement input excitation is **A = 0.1 m**.
d) Sketch the **magnitude Body diagram** of the transfer function **G(s)** from the displacement input **z(2$)** to the angular displacement output **θ(t)** of the pendulum.
e) We want to reduce the steady-state angular displacement amplitude **|θ|_max** to half of the value calculated in c), that is, **|θ'| = |θ|_max / 2 **, by changing the rotational viscous friction coefficient to a new value **b_r**. What should be this **new value of the rotational viscous friction coefficient b_r'** such that at the excitation frequency **α'** calculated in b) **the steady-state angular displacement amplitude is reduced by half?**
Fig (1): Vertical pendulum with pulling spring on top
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Answer
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Step 1: Determine the equations of motion for the system
\tau\_s = k \cdot (z(t) - \alpha \cdot \theta(t)) \cdot \alpha
The pendulum rotates about its bottom end point O, and there are two torques acting on it: the torque due to the rotational viscous friction, and the torque due to the displacement input z(2$). The equation of motion for the angular displacement θ(t) is given by: where I is the moment of inertia of the pendulum about point O, and τ\_net is the net torque acting on the pendulum. The net torque can be expressed as: where b\_r is the rotational viscous friction coefficient, and τ\_s is the torque due to the displacement input z(2$). The torque τ\_s can be expressed as: where k is the stiffness of the springs, α is the length of the pendulum, and z(2$) is the displacement input.
Step 2: Find the transfer function G(s)
G(s) = \frac{\theta(s)}{z(s)} = \frac{k \cdot \alpha}{I \cdot s^2 + b\_r \cdot s + k \cdot \alpha^2}
To find the transfer function G(s), we need to express the angular displacement θ(s) in terms of the displacement input z(s). First, we find the equation for θ(s) by taking the Laplace transform of the equation of motion: Next, we solve for θ(s): Now, we can express the transfer function G(s) as:
Final Answer
a) The transfer function G(s) is given by: G(s) = \frac{k \cdot \alpha}{I \cdot s^2 + b\_r \cdot s + k \cdot \alpha^2} where k = 100 N/m, α = 0.5 m, I = m \* α^2 = 0.75 kg \* m^2, and b\_r = 8 N m s/rad. b) The worst-case excitation frequency α' is the frequency that maximizes the magnitude of the transfer function G(iω). c) The maximum angular displacement amplitude |θ|\_max is the angular displacement amplitude at the worst-case excitation frequency α'. d) The magnitude Body diagram of the transfer function G(s) is a plot of the magnitude of the transfer function |G(iω)| as a function of the frequency ω. e) The new value of the rotational viscous friction coefficient b\_r' such that at the excitation frequency α' the steady-state angular displacement amplitude is reduced by half is given by: b\_r' = \frac{1}{2} \cdot \sqrt{\frac{(k \cdot \alpha^2 - I \cdot \omega'^2)^2 + (b\_r \cdot \omega')^2}{k \cdot \alpha^2 - I \cdot \omega'^2}} \cdot b\_r where ω' is the worst-case excitation frequency.
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