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denis23 [38]
3 years ago
11

Amazon rectangular bar of low carbon steel A-36 is exposed to an axial strees of 150 MPa. What is the original length of the bar

in m, if it elongates 1.35 mm under these conditions?
Physics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

1.8m

Explanation:

Let the Elastics of the steel ASTM-36 E = 200000 MPa

The strain of the bar when subjected to 150 MPa is

\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

\epsilon = \frac{\Delta L}{L}

L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm or 1.8m

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A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro
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Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

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Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

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