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UNO [17]
1 year ago
10

Please help me with this rotation and linear motion problem

Physics
1 answer:
kirill115 [55]1 year ago
5 0

The initial angular velocity of the wheel would be 15.5 rad/s

<h3>What is Velocity?</h3>

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.

The relation between the linear velocity and the angular velocity

V= ωr

as given the initial  linear velocity is 8.22 m/s

the radius of the wheel is 0.530 m

By substituting the value of the linear velocity and the radius

8.22 = 0.530ω

ω = 8.22/0.530

ω = 15.5 rad/s

Thus, the initial angular velocity of the wheel comes out to be  15.5 rad/s

Learn more about Velocity from here

brainly.com/question/18084516

#SPJ1

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A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass
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Answer:

The shortest distance is  S = 24.86 ft

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is v_b = 22\ mph = 22 * \frac{5280}{3600}   =  32.26 ft/s

     The distance between the axial is  d = 42 \ in

The mass center of the cyclist and the bicycle is m_c = 26 \ in  behind the front axle

The mass center of the cyclist and the bicycle is m_h = 40 \ in above the ground

   For the bicycle not to be thrown over the

     Momentum about the back wheel must be zero so

                \sum _B = 0

=>             mg (26) = ma(40)

=>             a = \frac{26}{40} g

Here  g = 32.2 ft/s^2

     So     a =  \frac{26}{40} (32.2)

             a =  20.93 ft/s^2

Apply the equation of motion to this motion we have

       v^2 = u^2 + 2as

 Where  u = 32.26 ft /s

             and v = 0 since the bicycle is coming to a stop

        v^2 = (32.26)^2 - 2(20.93) S

=>      S = 24.86 ft        

                 

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Answer:

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Explanation:

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