Question

A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle \theta\:=\:θ = 35.4 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?

Answer 1

Answer:

The velocity is  $v = 2.84 1 \ m/s$

Explanation:

The  diagram showing this set up is shown on the first uploaded image (reference Physics website )

From the question we are told that

    The mass is  m =  4 kg

    The  length of the string is $L = 2.0 \ m$

    The constant angle is  $\theta = 35.4 ^o$

     

Generally the vertical forces acting on the mass to keep it at equilibrium vertically is mathematically represented as

           $Tcos (\theta ) - mg = 0$

=>        $mg = Tcos (\theta )$

Now let the force acting on mass horizontally be k  so from SOHCAHTOA rule

         $sin (\theta ) = \frac{k }{T}$

=>      $k = T sin \theta$

Now this k is also equivalent to the centripetal force acting on the mass which is mathematically represented as

          $F_v = \frac{m v^2}{r}$

So

          $k = F_v$

Which

=>       $T sin \theta= \frac{ m v^2}{ r }$

     

So

        $\frac{Tsin (\theta )}{Tcos (\theta )} = \frac{mg}{ \frac{mv^2}{r} }$

=>      $Tan (\theta ) = \frac{v^2}{ r * g }$

=>      $v = \sqrt{r * g * tan (\theta )}$

Now the radius is evaluated using SOHCAHTOA rule as

       $sin (\theta) = \frac{ r}{L}$

=>    $r = L sin (\theta)$

substituting values

       $r = 2 sin ( 35.4 )$

       $r = 1.1586 \ m$

So

       $v = \sqrt{1.1586* 9.8 * tan (35.4 )}$

       $v = 2.84 1 \ m/s$