Question

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With right-arrow is in meters and t is in seconds). (a) Find the torque acting on the particle relative to the origin at the moment 7.46 s (b) Is the magnitude of the particle’s angular momentum relative to the origin increasing, decreasing, or unchanging?

Answer 1

Complete Question

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Answer:

a

The torque acting on the particle is  $\tau = 48t \r k$

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

      The position of the particle is   $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

       The mass of the particle is $m = 3.0 kg$

        The time is  t

   

The torque acting on  the particle is mathematically represented as

           $\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

       $\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

   $\r v$ is the velocity which is mathematically represented as

           $\r v = \frac{d \r r }{dt}$

Substituting for  $\r r$

           $\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

           $\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is  mathematically evaluated as    

          $\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

                       $= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

                      $\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

       $\r l = m (8t^2 \r k)$

Substituting for m

      $\r l = 3 * (8t^2 \r k)$

      $\r l =24t^2 \r k$

Substituting into equation for torque

       $\tau = \frac{d}{dt} [24t^2 \r k]$

       $\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

        $|\r l| = \sqrt{(24 t^2) ^2}$

        $|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also