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3 years ago

7

lock of mass m2 is attached to a spring of force constant k and m1 . m2. If the system is released from rest, and the spring is

initially not stretched or com- pressed, find an expres- sion for the maximum displacement d of m2

1 answer:

patriot [66]3 years ago

7 0

Answer:

The maximum displacement of the mass m₂ $= \frac{2(m_1-m_2)g}{k}$

Explanation:

Kinetic Energy (K) = 1/2mv²

Potential Energy (P) = mgh

Law of Conservation of energy states that total energy of the system remains constant.

i.e; Total energy before collision = Total energy after collision

This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.

$m_1gd = m_2gd+\frac{1}{2}kd^2\\\\m_1g = m_2g+\frac{1}{2}kd\\\\d = \frac{2(m_1-m_2)g}{k}$

d = maximum displacement of the mass m₂

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A cast-iron flywheel has a rim whose OD is 1 m and whose ID is 0.8 m. The flywheel weight is to be such that an energy fluctuati

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A.Coefficient of speed fluctuation of the flywheel = 0.222

B. The width <em>(thickness)</em> of the rim should be 0.131 m (131 mm)

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A.

Coefficient of speed fluctuation $(C_{s})$ $= \frac{N_{2}-N_{1}}{N}$

$N_{1}$ = minimum speed = 200 rpm

$N_{2}$ = maximum speed = 250 rpm

$N$ = average speed $= \frac{N_{2}+N_{1}}{2} = \frac{250+200}{2} = 225 rpm$

∴ $Cs = \frac{250-200}{225}=0.222$

Hence the coefficient of speed fluctuation of the flywheel = 0.222

B.

The moment of Inertia , $I=\frac{E_{2}-E_{1}}{C_{s}\times\omega^{2}}$

Where

$E_{2}-E_{1}=$ energy fluctuation of flywheel = 6.75 J

$\omega^{2}=$ angular velocity of flywheel = $\frac{2\pi N}{60} = \frac{2\pi \times 225}{60}= 23.56 rad/sec$

$C_{s}=$ coefficient of speed fluctuation of the flywheel = 0.222

Hence,

$I=\frac{6.75\times10^{3}}{0.222\times(23.56)^{2}}=54.78 Nms^{2}$

Similarly,

I = $\frac{m}{8}\times(d_{o}^{2}- d_{i}^{2})$

From the moment of Inertia, we can get the weight of the flywheel as

$m=\frac{8I}{(d_{o}^{2}+ d_{i}^{2})}= \frac{8\times 54.78}{(1^{2}+0.8^{2})}=267.22kg$

From this weight, we will be able to calculate the volume of the flywheel and hence, estimate the thickness. But to do this, we need to know its density first. this can be got from standard tables.

Specific weight of cast iron = $70.6KN/m^{3}$ ( from standard material property table)

density of cast iron , $\rho =\frac{70.6\times 10^{3}}{9.81}= 7,197kg/m^{3}$

Volume of cast iron flywheel = $\frac{m}{\rho}= \frac{267.22}{7197}= 0.03713 m^{3}$

similarly, the volume of the flywheel can also be obtained through the formula :

$V= \frac{\pi t(d_{o}^{2}-d_{i}^{2})}{4}$

we can easily estimate the thickness of the flywheel from here by solving for t as shown below

$0.03713=\frac{\pi t(1^{2}-0.8^{2})}{4}$

$0.03713=0.2827t$

$t=\frac{0.03713}{0.2827}$

$\therefore t= 0.131m \approx 131mm$

The width of the rim = 131 mm

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3 years ago