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earnstyle [38]
4 years ago
14

A thin, uniform rod is hinged at its midpoint. to begin with, one-half of the rod is bent upward and is perpendicular to the oth

er half. this bent object is rotating at an angular velocity of 9.3 rad/s about an axis that is perpendicular to the left end of the rod and parallel to the rod's upward half (see the drawing). without the aid of external torques, the rod suddenly assumes its straight shape. what is the angular velocity of the straight rod?

Physics
2 answers:
Verizon [17]4 years ago
6 0

The angular velocity of the straight rod is 4.65 rad/s

<h3>Explanation: </h3>

A thin, uniform rod is hinged at its midpoint. to begin with, one-half of the rod is bent upward and is perpendicular to the other half. this bent object is rotating at an angular velocity of 9.3 rad/s about an axis that is perpendicular to the left end of the rod and parallel to the rod's upward half (see the drawing). without the aid of external torques, the rod suddenly assumes its straight shape. what is the angular velocity of the straight rod?

The initial moment of inertia is

I = \frac{ML^2}{3}  + ML^2 = \frac{4ML^2}{3} because the vertical portion could be treated as a point mass when we rotate a parallel axis

where  is the length of each half of the rod.

When it straightened,

I' = \frac{ML^2}{3} + \frac{ML^2}{12} + M \frac{3L}{2}^2 =  \frac{ML^2}{12} (4 + 1 + 27) = \frac{8ML^2}{3}

by the parallel axis theorem on the distant segment.

Then after the conservation of angular momentum:

\frac{4ML^2}{3} * 9.3 rad/s = \frac{8ML^2}{3} * \omega

\omega = 4.65 rad/s

Learn more about the angular velocity brainly.com/question/12618480

#LearnWithBrainly

elena-14-01-66 [18.8K]4 years ago
5 0
<span>Answer: initial I = (m/2)L²/3 + (m/2)L² where L = ½ the length of the rod, and the vertical half can be treated as a point mass. initial I = mL²(1/6 + 1/2) = 2mL²/3 final I = m(2L)²/3 = 4mL²/3 Since I has doubled and momentum is conserved, ω has halved. ω = 3.9 rad/s. By formulae: 2mL²/3 * 7.8rad/s = 4mL²/3 * ω</span>
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The addition of vectors allows finding that the arrival airport is 1506.02 km and 15.02º North of the East, from Dalas airport.  

Displacement is a vector quantity that must be calculated with vector algebra. Its magnitude can be calculated with the Pythagorean Theorem and using trigonometry its direction can be found

Let's calculate the magnitude of the vector from the departure airport (Dalas) to the arrival airport, in the attached we can see a diagram of this displacement  

 

             d² = x² + y²

             d = \sqrt{1454.55^2 + 390.27^2}

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The direction of this airport can be calculated using trigonometry

            tan θ = y / x

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angle can be written 15.02º North of East

In conclusion with the addition of vectors we can find that the arrival airport is 1506.02 km and 15.02º North of the East of Dalas airport.

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2 years ago
A car travels 35.0 mph north for 1.00 hour, then 40.0 mph east for 0.0500 hour, and finally 50.0 mph northeast for 2.00 hours. C
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Hello!

A car travels 35.0 mph north for 1.00 hour, then 40.0 mph east for 0.0500 hour, and finally 50.0 mph northeast for 2.00 hours. Calculate the average speed.

We have the following data:

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We apply the data to the weighted average formula to find the average speed, let us see:

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Weight\:Average = \dfrac{35.0*1.00+40.0*0.0500+50.0*2.00}{1.00+0.0500+2.00}

Weight\:Average = \dfrac{35+2+100}{3.05}

Weight\:Average = \dfrac{137}{3.05}

Weight\:Average = 44.91803279...

\boxed{\boxed{Weight\:Average = 44.92\:mph}}\Longleftarrow(Average\:Speed)\:\:\:\:\:\:\bf\purple{\checkmark}

Answer:

The Average Speed is approximately 44.92 mph

_______________________

\bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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