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Mashutka [201]
2 years ago
10

four particles connected by rigid rods of negligible mass where y1 = 5.70 m. the origin is at the center of the rectangle. The s

ystem rotates in the xy plane about the z axis with and angular speed of 5.60 rad/s. a) calculate the moment of inertia of the system about the z axis. b) Calculate the rotational kinetic energy
Physics
1 answer:
Sophie [7]2 years ago
8 0

The definition of moment of inertia and kinetic energy we can find the results for the questions are:  

          a) The moment of inertia is: I = 32.49 kg m²

          b) The kinetic energy is: K = 509.4 J

a) The moment of inertia is a scalar quantity that has the same function as mass, it is given by

             I = m r²

       

They indicate that the length of the bar is y₁ = 5.70 m, if the two rigid bars form the square that supports the masses, the distance from the center to the end where each mass is:  

            r = y₁ / 2

             r = 5.70 / 2 = 2.85 m

The moment of inertia of the system is.

         I = 4 m r²

Let's calculate.

        I =4 m 2.85²

        I = 32.49m kg m²

If we assume that the mass of each point particle is:

        m = 1 kg

   

         I = 32.49 kg m²

b) They ask for the rotational kinetic energy, if the angular velocity is.

           w = 5.60 rad / s

The kinetic energy for rotational motion is.

         K = ½ I w²

Let's calculate.

          K = ½ 32.49 5.60²

          K = 509.4 J

In conclusion with the definition of moment of inertia and kinetic energy we can find the results for the questions are:

          a) The moment of inertia is: I = 32.49 kg m²

          b) The kinetic energy is: K = 509.4 J

Learn more here:  brainly.com/question/14838079

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Answer:

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Explanation:

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3 years ago
Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c
Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is  v_A= 4m/s

Explanation:

From the question we are told that

    The length of the string is  L = 1m

     The initial speed of block A is u_A

     The final speed of block A is  v_A = \frac{1}{2}u_A

      The initial speed of block B is u_B = 0

      The mass of block A  is  m_A = 7kg  gh

      The mass of block B is  m_B  = 2 kg

According to the principle of conservation of momentum

       m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}

Since block B at initial is at rest

       m_A u_A  = m_Bv_B + m_A \frac{u_A}{2}

      m_A u_A  - m_A \frac{u_A}{2} = m_Bv_B

          m_A \frac{u_A}{2} = m_Bv_B

  making v_B the subject of the formula

             v_B =m_A \frac{u_A}{2 m_B}

Substituting values

               v_B =\frac{7 u_A}{4}  

This v__B is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity  of v__B' at  the top of the vertical circle  

 The angular centripetal acceleration  would be mathematically represented

                   a= \frac{v^2_{B}'}{L}

Note that  this acceleration would be toward the center of the circle

      Now the forces acting at the top of the circle can be represented mathematically as

         T + mg = m \frac{v^2_{B}'}{L}

    Where T is the tension on the string

  According to the law of energy conservation

The energy at  bottom of the vertical circle   =  The energy at the top of

                                                                                the vertical circle

   This can be mathematically represented as

                 \frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L

From above  

                (T + mg) L = m v^2_{B}'

Substitute this into above equation

             \frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L  + mg 2L  

             \frac{49 mv_A^2}{16}  = \frac{1}{2} (T + mg) L + mg 2L

          \frac{49 mv_A^2}{16}  = T + 5mgL

The  value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is  zero

        This is mathematically represented as

                      \frac{49 mv_A^2}{16}  = 5mgL

making  v_A the subject

            v_A = \sqrt{\frac{80mgL}{49m} }

substituting values

          v_A = \sqrt{\frac{80* 9.8 *1}{49} }

              v_A= 4m/s

     

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Explanation:

I hope this helps! :D

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Answer:

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And we know that at t = 5s, the car was brought to stop, so the velocity must be zero.

v(5s) = 0 = a*5s + 110km/h.

a = (110km/h)*(1/5s)

now we have that:

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then:

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Then we have:

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Now the velocity equation is:

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To write the positon equation we must integrate over time again, we can get:

p(t) = (1/2)*(-6.11m/s^2)*t^2 + (30.56m/s)*t + p0

Where p0 is the initial position, here i will assume that is zero, because it does no really mater.

The total displacement of the car will be equal to p(5s)

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