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jek_recluse [69]
2 years ago
12

An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ. Find an expression for

the angular velocity ω in terms of g, L and angle θ
Physics
2 answers:
VikaD [51]2 years ago
8 0
We know that
g = LcosΘ 
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>
Semenov [28]2 years ago
3 0

The angular velocity of the object moving in the horizontal circle is given as \boxed{\omega = \sqrt {\frac{g}{{L\sin \theta }}}} .

Further Explanation:

When the object moves in the horizontal circle about a fixed axis, then the vertical component of the tension in the wire is balanced by the weight of the object and the horizontal component of the tension in the wire is balanced by the centripetal force acting in the outward direction as shown in the figure below.

Let there be tension T developed in the string due to the motion of the block in the horizontal circle and the mass of the block is m.

The centripetal force acting on the block due to its motion in the circular path of radius r is:

{F_c} = m{\omega ^2}r

The force balancing equation on the box in the vertical direction is:

T\sin \theta  = mg             .......(1)

The force balancing equation on the box in horizontal direction is:

T\cos \theta  = m{\omega ^2}r            ......(2)

Divide equation (1) by (2).

\dfrac{{T\sin \theta }}{{T\cos \theta }} = \dfrac{{mg}}{{m{\omega ^2}r}}

Substitute L\cos \theta for r in above expression.

\begin{aligned}\tan \theta &= \frac{g}{{{\omega ^2}L\cos \theta }}\\\omega &= \sqrt {\frac{g}{{L\cos \theta \tan \theta }}} \\&=\sqrt {\frac{g}{{L\sin \theta }}}\\\end{aligned}

Thus, the angular velocity of the object moving in the horizontal circle is given as \boxed{\omega = \sqrt {\frac{g}{{L\sin \theta }}}}.

Learn More:

1. If forces acting on an object are unbalanced, the object could experience a change in , direction, or both brainly.com/question/2720955

2. A toy train rolls around a horizontal 1.0-m-diameter track. The coefficient of rolling friction is 0.10 brainly.com/question/9575487

3. A 30.0-kg box is being pulled across a carpeted floor by a horizontal force brainly.com/question/7031524

Answer Details:

Grade: College

Subject: Physics

Chapter: Uniform Circular Motion

Keywords:  Angular velocity, horizontal circle, tilts downward, angle theta, acceleration due to gravity, weight, centripetal force, outward, and force balancing.

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Ans: 
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Explanation:

The figure is attached down below.

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We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

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8 0
2 years ago
Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w
ioda

Answer:

Q_{week} = 458884.6\, BTU

Explanation:

The weekly water consumption of Kimonoski is:

m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

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m_{others, week} = 1346.218\,lbm

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3 0
1 year ago
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vampirchik [111]

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How high above the earth's surface is g reduced to 8.80m/^2?
Sladkaya [172]
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</span>

<span>Hope my answer would be a great help for you.    
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6 0
2 years ago
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