Question

2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗

Answer 1

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

   The tension is  $T = 48.255 \ N$

b

   The time taken is  $t = 0.226 \ s$

c

   The position for maximum velocity is  

        S = 0  

d

     The maximum velocity is  $V_{max} =0.384 \ m/s$

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

    The mass of the bob  is $m_b = 5 \ kg$

     The angle is  $\theta = 10^o$

     The length of the string is  $L = 0.5 \ m$

The tension on the string is mathematically represented as

              $T = mg cos \theta$

substituting values

             $T = 5 * 9.8 cos(10)$

             $T = 48.255 \ N$

The motion of the bob is mathematically represented as

             $S = A sin (w t + \frac{\pi }{2} )$

     =>   $S = A sin (wt)$

Where  $w$ is the angular speed

and  $\frac{\pi}{2}$ is the phase change

At initial position S =  0

   So   $wt = cos^{-1} (0)$

          $wt = 1$

Generally $w$ can be mathematically represented as

          $w = \frac{2 \pi }{T}$

Where T is the period of oscillation which i mathematically represented as

          $T = 2 \pi \sqrt{\frac{L}{g} }$

So  

      $t = \frac{1}{w}$

       $t = \frac{T}{2 \pi}$

       $t = \sqrt{\frac{L}{g} }$

substituting values

       $t = \sqrt{\frac{0.5}{9.8} }$

       $t = 0.226 \ s$

Looking at the equation

         $wt = 1$

We see that maximum velocity of the bob will be at  S = 0  

i. e     $w = \frac{1}{t}$

The maximum velocity is mathematically represented as

          $V_{max} = w A$

Where A is the amplitude which is mathematically represented as

         $A = L sin \theta$

So

      $V_{max} = \frac{2 \pi }{T } L sin \theta$

      $V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} } L sin \theta$  Recall   $T = 2 \pi \sqrt{\frac{L}{g} }$

     $V_{max} = \sqrt{gL} sin \theta$

substituting values        

       $V_{max} = \sqrt{9.8 * 0.5 } sin (10)$

       $V_{max} =0.384 \ m/s$