Question

The system in the Figure has a linear frequency of $f = 6 \text{ Hz}$ for the following data: $m = 12 \text{ kg}$, $I_o = 6 \text{ kg.} m^2$, $r_1 = 15 \text{ cm}$, $r_2 = 30 \text{ cm}$. When the system is distributed by giving it an initial displacement, the amplitude of free vibration is reduced by 70 percent in 8 cycles. a) Find the equation of motion (EMO) for generic case by benefitting from the constraint $x = \theta r_2$. (Hint-1: Apply both linear Newton's 2nd law to the mass $m$ and the Moment equation to the pulley ($\sum M_o = I_o \alpha$) and then combine them to get the EMO; Hint-2: After the $j$ number of cycle, the decrease in the amplitude is given by $\frac{x_{j+1}}{x_j} = \exp(\frac{j 2\pi \zeta}{\sqrt{1-\zeta^2}}); Hint 3: \zeta = \frac{c}{2mw_n}$) 

          The system in the Figure has a linear frequency of $f = 6 \text{ Hz}$ for the following data:
$m = 12 \text{ kg}$, $I_o = 6 \text{ kg.} m^2$, $r_1 = 15 \text{ cm}$, $r_2 = 30 \text{ cm}$.
When the system is distributed by giving it an initial displacement, the amplitude of free vibration is reduced by 70
percent in 8 cycles.
a) Find the equation of motion (EMO) for generic case by benefitting from the constraint $x = \theta r_2$.
(Hint-1: Apply both linear Newton's 2nd law to the mass $m$ and the Moment equation to the pulley ($\sum M_o = I_o \alpha$)
and then combine them to get the EMO; Hint-2: After the $j$ number of cycle, the decrease in the amplitude is given
by $\frac{x_{j+1}}{x_j} = \exp(\frac{j 2\pi \zeta}{\sqrt{1-\zeta^2}}); Hint 3: \zeta = \frac{c}{2mw_n}$)
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The system in the Figure has a linear frequency of f = 6 Hz for the following data: m = 12 kg, Io = 6 kg. m^2, r1 = 15 cm, r2 = 30 cm. When the system is distributed by giving it an initial displacement, the amplitude of free vibration is reduced by 70 percent in 8 cycles. a) Find the equation of motion (EMO) for generic case by benefitting from the constraint x = θ r2. (Hint-1: Apply both linear Newton's 2nd law to the mass m and the Moment equation to the pulley (∑ Mo = Io α) and then combine them to get the EMO; Hint-2: After the j number of cycle, the decrease in the amplitude is given by (xj+1)/(xj) = ((j 2πζ)/(√(1-ζ^2))); Hint 3: ζ = (c)/(2mwn))

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University Physics with Modern Physics
Hugh D. Young 14th Edition
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The system in the Figure has a linear frequency of f = 6 Hz for the following data: m = 12 kg, Io = 6 kg.m^2, r = 15 cm, r = 30 cm. When the system is disturbed by giving it an initial displacement, the amplitude of free vibration is reduced by 70 percent in 8 cycles. a) Find the equation of motion (EMO) for the generic case by benefiting from the constraint x = Or^2. (Hint-1: Apply both linear Newton's 2nd law to the mass m and the Moment equation to the pulley (EMo = 1, a) and then combine them to get the EMO; Hint-2: After the j number of cycles, the decrease in the amplitude is given by x(t) = x(0) * (0.3)^(j/8)). Pulley, mass moment of inertia Jo
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Transcript

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00:01 So in this question, we have a system of a body and a polyasuring this diagram.
00:06 So the body has a mass m and the poly has a mass m capital m and has a radius r.
00:14 And the body is, you know, is connected to a spring with spring constant key.
00:23 And it's assumed that when the pulley rotates, there's no slipping way.
00:31 It does not slide, basically, this rate does not slide.
00:35 So you asked to find the frequency of small oscillations of the arrangement of the system, basically.
00:42 So first you asked what's the moment of inertia of eye of the podium? well, that's what's going by a half m r squared, right? and what's the equation motions? well, to find the equation motion, we need to analyze forces.
00:55 So on the body, you have two forces.
00:57 One is the force going down here, right? that is mg, g is that acceleration of graft in this and there's not the first which is probably going up i call it t1 okay and there's a force on this side and that is basically t2 but actually it's nothing but you know it's going to pull it down it's going to pull it down and that force i would call it minus k x so x obviously the stretch of the spring right albeit so now we can write down the equation.
01:30 First for the body, that's m .a.
01:35 So m .a must be equal to mg minus t1, right? and then for the poly we must have, the pulley is going to spin, right? so as going to stay up and it's going to spin, suppose it has acceleration, alpha.
01:49 And a is acceleration of the body, right? the acceleration of the body.
01:53 And alpha is the angle acceleration of this pulley, right? so i, the momentum, you know, at times the angle acceleration must be the torque.
02:02 But the torque is basically, you see the force t1, and they basically generate, you know, if you look at the force acting on the pole is going down, right? and it's going down.
02:16 So actually the torque generated by this to force are just opposite...
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