Question

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see figure below). One such pendulum is constructed with a string of length
L =10.7 cm and bob of mass 0.344 kg. The string makes an angle = 5.58° with the vertical.

(a) What is the radial acceleration of the bob?
magnitude


(b) What are the horizontal and vertical components of the tension force exerted by the string on the bob? (Assume radially inward to be the positive x axis and vertically upward to be the
positive y axis. Express your answer in vector form.)
T= N

Answer 1

Answer:

a

The  radial acceleration is  $a_c = 0.9574 m/s^2$

b

The horizontal Tension is  $T_x = 0.3294 i \ N$

The vertical Tension is  $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  $L = 10.7 \ cm = 0.107 \ m$

     The mass of the bob is  $m = 0.344 \ kg$

     The angle made  by the string is  $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

         $F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

     $F = Tsin \theta$ where T is the tension on the rope  and v is the linear velocity  

     So

          $Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is  mathematically represented as

          $Tcos \theta = mg$

So

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

=>    $tan \theta = \frac{v^2}{rg}$

=>   $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      $a_c = \frac{v^2}{r}$

=>  $a_c = gtan \theta$

substituting values

     $a_c = 9.8 * tan (5.58)$

     $a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

     $T_x = Tsin \theta = ma_c$

substituting value

   $T_x = 0.344 * 0.9574$

    $T_x = 0.3294 \ N$

The vertical component of  tension is  

    $T_y = T \ cos \theta = mg$

substituting value

     $T_ y = 0.344 * 9.8$

      $T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       $T = T_x i + T_y j$

substituting value  

      $T = [(0.3294) i + (3.3712)j ] \ N$

         

Answer 2

The radical acceleration of the bob is 0.9575 m/s². The horizontal and vertical components of the tension force exerted by the string on the bob are 0.329 N and 3.37 N respectively.

Taking the vertical component of the tension where the weight mass is balanced, then:

T sin θ = mg

$\mathbf{T = \dfrac{mg}{sin \theta}}$

However, the centripetal force of the system is given by the horizontal component of the tension which can be expressed as:

T cos θ = m$\mathbf{a_r}$

Making $\mathbf{a_r}$ the subject, we have:

$\mathbf{a_r = \dfrac{Tcos \theta }{m}}$

replacing the value of tension (T), we have:

$\mathbf{a_r = \dfrac{ \dfrac{mg}{sin \theta}cos \theta }{m}}$

$\mathbf{a_r=g tan \theta}$

where;

• angle  θ = 5.58°

$\mathbf{a_r=9.8 m/s^2 \times tan 5.58}$

$\mathbf{a_r=9.8 m/s^2 \times 0.0977}$

$\mathbf{a_r=0.9575 \ m/s^2}$

Thus, the radical acceleration of the bob is 0.9577 m/s²

On the positive x-axis, the horizontal component of the tension force is:

$\mathbf{T_x =Tcos \theta}$

$\mathbf{T_x =ma_r}$

$\mathbf{T_x =0.344 \ kg \times 0.9575 \ m/s^2}$

$\mathbf{T_x =0.329 \ N}$

On the positive y-axis, the vertical component of the tension force is:

$\mathbf{T_y =Tsin \theta}$

$\mathbf{T_y =mg}$

$\mathbf{T_y=0.344 \ kg \times 9.8 \ m/s^2}$

$\mathbf{T_x =3.37 \ N}$

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