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Sholpan [36]
3 years ago
6

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

ω = 1.53 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 73 kg and velocity v = 4.2 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round.1)What is the magnitude of the initial angular momentum of the merry-go-round?kg-m2/sYour submissions:2)What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?kg-m2/sYour submissions:
Physics
1 answer:
Ulleksa [173]3 years ago
5 0

Answer:

299.88 kgm²/s

499.758 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.63 m

I = Moment of inertia = 196 kgm²

\omega_i = Initial angular velocity = 1.53 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.2 m/s

Initial angular momentum is given by

L=I\omega_i\\\Rightarrow L=196\times 1.53\\\Rightarrow L=299.88\ kgm^2/s

The initial angular momentum of the merry-go-round is 299.88 kgm²/s

Angular momentum is given by

L=mvR\\\Rightarrow L=73\times 4.2\times 1.63\\\Rightarrow L=499.758\ kgm^2/s

The angular momentum of the person 2 meters before she jumps on the merry-go-round is 499.758 kgm²/s

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In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
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Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

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Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

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