Question

Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic energy kb of particle



b. remember that both particles rotate about the y axis. express your answers in terms of m, ω, and r separated by a comma.

Answer 1

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Answer:

The kinetic energy for particle a is  $K_a  =  \frac{9}{2}  mr^2 w^2$

The kinetic energy for particle b is   $K_b  =  m w^2r^2$

Explanation:

From the question we are told that

   The mass of particle a  is  $m$  

   The mass of particle b is  $2 m$  

Generally the kinetic energy for  particle a is mathematically represented as  

       $K_a  =  \frac{1}{2} m v^2$

Here $v_a$ is the linear velocity of particle a which is mathematically represented as

         $v_b  =  w *  3r$

So

      $K  =  \frac{1}{2} m {w *  3r}^2$

=>   $K_a  =  \frac{9}{2}  mr^2 w^2$

Generally the kinetic energy for  particle a is mathematically represented as  

    $K_b  =  \frac{1}{2} 2m v_b^2$

Here $v_b$ is the linear velocity of particle a which is mathematically represented as

         $v_b  =  w *  r$

So

      $K_b  =  m (wr)^2$

=>    $K_b  =  m w^2r^2$

Answer 2

The kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is $\frac{1}{2} m \omega ^2 r^2$.

The given parameters;

• kinetic energy of the particle, K.E = ¹/₂mv²

The kinetic energy of the of the particle in terms of the angular velocity is calculated as follows;

$K_b = \frac{1}{2} m (\omega r)^2\\\\K_b = \frac{1}{2} m \omega ^2 r^2$

where;

• m is the mass of the object
• ω is the angular speed of the object
• r is the radius of the circular path

Thus, the kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is $\frac{1}{2} m \omega ^2 r^2$.

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