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Consider a spinning plate is dropped onto a stationary plate (which is itself at rest on a frictionless surface). Both plates have a radius of 30cm and a mass of 1kg. The spinning plate is initially spinning at a rate of 0.7 revolutions per second. Hint: This is like a totally-inelastic collision.

Required:
a. After a sufficiently long time, what is the angular velocity of the initially-spinning plate? What about the initially-stationary plate?
b. Assume that the period of velocity matching happens over a course of 2 seconds. Further, assume that the torque exerted by each plate on the other is constant over time. In that case, what is the magnitude of the acceleration that each plate feels during those two seconds? Hint: Use the rotational impulse-momentum theorem.

Respuesta :

okpalawalter8

Answer:

The final angular velocity is  [tex]w_f = 2.1994 rad/sec[/tex]

The angular acceleration is  [tex]\alpha = 1.099 \ rad/sec^2[/tex]

Explanation:

From the question we are told that

    The radius of each  plate is  [tex]r = 30 \ cm = \frac{30}{100} = 0.3 \ m[/tex]

    The mass of each  plate is  [tex]m_p = 1 \ kg[/tex]

    The angular speed of the spinning plate is [tex]w = 0.7 \ rev \ per \ sec = 0.7 * 2 \pi = 4.3988 \ rad/sec[/tex]

  From the law of conservation of momentum

         [tex]L_i = L_f[/tex]

Where [tex]L_i[/tex] is the initial angular momentum of the system (The spinning and stationary plate ) which is mathematically represented as

          [tex]L_i = I_1 w + 0[/tex]

here [tex]I_1[/tex] is the moment of inertia of the spinning plate which mathematically represented as

              [tex]I_1 = \frac{m_pr^2}{2}[/tex]

      and the zero signify that the stationary plate do not have an angular momentum as it is at rest at the initial state

         [tex]L_f[/tex] is the final angular momentum of the system (The spinning and stationary plate) , which is mathematically represented as

              [tex]L_f = (I_1 + I_2 ) w_f[/tex]

Where  

           [tex]I_2[/tex] is the moment of inertia of the second plate (This was stationary before but now it spinning due to the first pate )   and is equal to  [tex]I_1[/tex]

and  [tex]w_f[/tex] is the final angular speed

So we have  

             [tex]I_1 w = (I_1 + I_2)w_f[/tex]

             [tex]\frac{m_p r^2}{2} * w = 2 * \frac{m_p r^2}{2} * w_f[/tex]

             [tex]w = 2 * w_f[/tex]

substituting values

             [tex]4.3988 = 2 * w_f[/tex]

            [tex]w_f = \frac{4.3988 }{2}[/tex]

            [tex]w_f = 2.1994 rad/sec[/tex]

The the rotational impulse-momentum theorem can be mathematially represented as

        [tex]\tau * \Delta t = 0.09891[/tex]

Where [tex]\tau[/tex] is the torque  and  [tex]\Delta t[/tex] is the change in time  

So at  [tex]\Delta t = 2 \ sec[/tex]

        [tex]\tau = \frac{0.09891}{2}[/tex]

        [tex]\tau = 0.0995 \ Nm[/tex]

now the angular acceleation is mathematically represented as

         [tex]\alpha = 2 * \frac{\tau}{m_p * r^2 }[/tex]

substittuting values

           [tex]\alpha = 2 * \frac{0.0995}{1 * 0.3^2}[/tex]

           [tex]\alpha = 1.099 \ rad/sec^2[/tex]

       

     

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