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Lorico [155]
3 years ago
6

Block A is released from rest and slides down the frictionless ramp to the loop. The maximum height h of the loop is the same as

the initial height of the block. Will A make it completely around the loop without losing contact with the track?

Engineering
1 answer:
Zolol [24]3 years ago
4 0

Answer:No

Explanation:

Given

Block A is at height h and released from rest

Initial Energy possessed by block A is equal to Potential energy of Block which is given by

E_i=mgh

where m=mass of block

After releasing the block, block first reaches to the bottom of the circle and then uses this energy to reach a top point of the loop.

But as soon as block reaches the top point of the loop it acquires the energy which is equal to Initial energy i.e. all the energy is stored in the form of potential energy and there is no kinetic energy so the block will not able to move further and fall from the top point.

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An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit
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Answer:

equivalent stiffness is 136906.78 N/m

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Explanation:

given data

mass = 120 kg

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frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

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here  ω(n) is natural frequency i.e = √(k/m)

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and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

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0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

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so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

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∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

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damping = 2∈ √(mk)

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damping = 2(0.08869) √(120×136906.78)

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The lead time of the actual batch will be in

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