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sesenic [268]
3 years ago
8

A 14 kilogram beam is hinged at one end. A 6.0-kilogram triangular object and a 7.5-kilogram capital I-shaped object are placed

as shown in the figure. The dots indicate the individual centers of gravity of the beam and objects. The center of mass of the entire system must be:

Physics
1 answer:
GuDViN [60]3 years ago
5 0

Answer: 1.3 m

Explanation:

From the figure attached, 3 materials are involved. A 14 kilogram beam hinged at one end, a 6.0-kilogram triangular object and a 7.5-kilogram capital I-shaped object.

Let take the moment at the hinge axis.

The moment for the beam will be

14 × ( 0.36 + 0.53 ) = 12.6

The moment of the triangle object will be 6 × 0.36 = 2.16

The moment of capital I shape will be

7.5 × ( 0.36 + 0.53 + 1.92 ) = 21.075

The total moment = 12.6 + 2.16 + 21.075

The total moment = 35.835

The total mass = 14 + 6 + 7.5 = 27.5

The center of mass of the entire system must be:

X = total moment ÷ total mass

Substitute both in the formula

X = 35.835/27.5

X = 1.3 m.

Therefore, The center of mass of the entire system must be 1.3m

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Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

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This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

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x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
2 years ago
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Choices 'C' and 'D' are both correct.

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Compare and Contrast Potential Energy with Thermal Energy
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ladessa [460]

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4 0
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Answer:

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4 0
3 years ago
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