Question

conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius r. FIGURE P8.47 shows that the string traces out the surface of a cone, hence the name.
a. Find an expression for the tension T in the string.
b. Find an expression for the ball’s angular speed v.
c. What are the tension and angular speed (in rpm) for a 500 g ball swinging in a 20-cm-radius circle at the end of a 1.0-m-long string?

Short answer

a) Expression of tension is $T=\frac{mgL}{\sqrt{(L^2-r^2)}}$

b) Expression for ball's angular velocity is $\omega =\sqrt{\frac{g}{(L^2-r^2)}}$

c) For the given mass and length of string tension is T=5.001 N and angular velocity is ω=30 rpm

Step-by-step solution

Part(a) Step 1: Given information

Mass = m

Length = L

Radius = r

Part(a) Step 2 : Explanation

First draw a free body diagram as below

From the free body diagram given, the net force F_net on the z-axis is:

$F_{net}=T_z-mg=0..................\left(1\right)$

Where:
-T_z is z-component of tension T
- m is mass of the ball
- g acceleration due to gravity
And , T_zcan be expressed as

$T_z=Tcos\left(\theta \right).....................\left(2\right)$

From equation (1) and (2) we get

$T\cos \left(\theta \right)=mg............................\left(3\right)$

We can find cosθ as below from trigonometric ratio

$cos\left(\theta \right)=\frac{\sqrt{(L^2-r^2)}.}{L}.......................\left(4\right)$

Where L= Length of string and r= radius of circle.

Now substitute the given values , we get

$$\begin{gathered} T\left(\frac{\sqrt{(L^2-r^2)}}{L}\right)=mg.............................\left(5\right) \\ T=\frac{mgL}{\sqrt{(L^2-r^2)}}.....................................\left(6\right) \end{gathered}$$

Part(b) Step 1 : Given information

Mass = m

Length = L

Circle of radius = r

Part(b) Step 2: Explanation

We know from Newton's second law of motion:

$\sum F_r=m{a}_r.................................\left(7\right)$

where m is the mass and a_r is radial acceleration.

The force in radial direction

$\sum F_r=T_r=T·sin\left(\theta \right)..........................\left(8\right)$

From equation (7) and (8) we get

$Tsin\left(\theta \right)=m{a}_r...............................\left(9\right)$

From trigonometric ratio we can write

$\sin \left(\theta \right)=\frac{r}{L}................................\left(10\right)$

From equation (9) and (10) , we get

$\frac{(T.r)}{L}=m{a}_r.........................\left(11\right)$

We know radial acceleration is given as

$a_r={\omega }^{2}r.............................\left(12\right)$

From equation (12) and (11) we get:

$\frac{(T.r)}{L}=m{\omega }^2............................\left(13\right)$

Substitute the value of T from equation(6) we get

$$\begin{gathered} \frac{g}{\sqrt{(L^2-r^2)}}={\omega }^2.............................\left(14\right) \\ \omega =\sqrt{\frac{g}{\sqrt{(L^2-r^2)}}}..............................\left(15\right) \end{gathered}$$

Part(c) Step 1: Given information

m=500gm =0.5 kg

r= 20 cm=0.2 M

L= 1 M

Part(c) Step 2: Explanation

To get Tension, substitute the values given in equation (6) , we get

$$\begin{gathered} T=\frac{mgL}{\sqrt{(L^2-r^2)}} \\ T=\frac{(.5kg)(9.8m/s^2)\left(1m\right)}{\sqrt{{\left(1m\right)}^2-{(0.2m)}^2)}} \\ T=5.001N \end{gathered}$$

To find angular speed can be calculated by substituting given values in equation (15), we get

$$\begin{gathered} \omega =\sqrt{\frac{g}{\sqrt{(L^2-r^2)}}} \\ \omega =\sqrt{\frac{9.8m/s^2}{\sqrt{{\left(1m\right)}^2-{(0.2m)}^2)}}}=3.16rad/sec \\ InRPM \\ \omega =(3.16rad/s)\left(\frac{1rev}{2rad}\right)\left(\frac{60s}{1min}\right)=30rpm \end{gathered}$$