Chapter 8: Q. 47 (page 201)

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

Expert verified

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

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

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:

$F_{n e t} = T_{z} - m g = 0 . . . . . . . . . . . . . . . . . . (1)$

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

$T_{z} = T c o s (θ) . . . . . . . . . . . . . . . . . . . . . (2)$

From equation (1) and (2) we get

$T \cos (θ) = m g . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)$

We can find cosθ as below from trigonometric ratio

$c o s (θ) = \frac{\sqrt{(L^{2} - r^{2})} .}{L} . . . . . . . . . . . . . . . . . . . . . . . (4)$

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

Now substitute the given values , we get

$T (\frac{\sqrt{(L^{2} - r^{2})}}{L}) = m g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5) T = \frac{m g L}{\sqrt{(L^{2} - r^{2})}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (6)$

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:

$\sum F_{r} = m a_{r} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (7)$

where m is the mass and a_r is radial acceleration.

The force in radial direction

$\sum F_{r} = T_{r} = T \cdot s i n (θ) . . . . . . . . . . . . . . . . . . . . . . . . . . (8)$

From equation (7) and (8) we get

$T s i n (θ) = m a_{r} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (9)$

From trigonometric ratio we can write

$\sin (θ) = \frac{r}{L} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (10)$

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

$\frac{(T . r)}{L} = m a_{r} . . . . . . . . . . . . . . . . . . . . . . . . . (11)$

We know radial acceleration is given as

$a_{r} = ω^{2} r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (12)$

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

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

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

$\frac{g}{\sqrt{(L^{2} - r^{2})}} = ω^{2} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (14) ω = \sqrt{\frac{g}{\sqrt{(L^{2} - r^{2})}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (15)$

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

$T = \frac{m g L}{\sqrt{(L^{2} - r^{2})}} T = \frac{(. 5 k g) (9.8 m / s^{2}) (1 m)}{\sqrt{{(1 m)}^{2} - {(0.2 m)}^{2})}} T = 5.001 N$

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

$ω = \sqrt{\frac{g}{\sqrt{(L^{2} - r^{2})}}} ω = \sqrt{\frac{9.8 m / s^{2}}{\sqrt{{(1 m)}^{2} - {(0.2 m)}^{2})}}} = 3.16 r a d / s e c I n R P M ω = (3.16 r a d / s) (\frac{1 r e v}{2 r a d}) (\frac{60 s}{1 m i n}) = 30 r p m$