Question

What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle θ from the vertical? express your answer in terms of some or all of the variables m, l, and θ, as well as the acceleration due to gravity g?

Answer 1

The bob must have tangential speed v = √ ( g L sin θ tan θ )

$\texttt{ }$

Further explanation

Angular Speed can be formulated as follows:

$\boxed {\omega = \frac{ v }{ R }}$

ω = Angular Speed ( rad/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

$\texttt{ }$

Linear Speed can be formulated as follows:

$\boxed {v = \frac{ 2 \pi R }{ T }}$

T = Period of Circular Motion ( s )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

$\texttt{ }$

Given:

length of string = L

mass of bob = m

acceleration due to gravity = g

angle of string from the vertical = θ

Asked:

tangential speed = v = ?

Solution:

$\Sigma F_y = T_y - mg$

$0 = T\cos \theta - mg$

$T \cos \theta = mg$

$\boxed {T = mg \div \cos \theta}$ → Equation 1

$\texttt{ }$

$\Sigma F_x = ma$

$T_x = m \frac{v^2}{R}$

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

$( mg \div \cos \theta ) \sin \theta = m \frac{v^2}{ L \sin \theta }$ ← Equation 1

$g \tan \theta = \frac{v^2}{L \sin \theta}$

$v^2 = gL \tan \theta \sin \theta$

$\boxed {v = \sqrt { gL \tan \theta \sin \theta } }$

$\texttt{ }$

Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

Answer 2

Refer to the diagram shown below.

v = the tangential speed.
r = the radius of the horizontal circle.
T = tension in the string.
θ =  the angle that the string makes with the vertical
m =  Bob's mass  (mg = the weight)
F =  centripetal force
l = the length of the string

From geometry,
r = l sin θ

The centripetal acceleration is
$a= \frac{v^{2}}{r} = \frac{v^{2}}{l \,sin \theta}$
The centripetal force is
$F = \frac{m v^{2}}{l \, sin \theta}$

For vertical force balance,
T cosθ = mg                              (1)
For horizontal force balance,
$Tsin \theta = F = \frac{m v^{2}}{l sin \theta}$         (2)
Divide (2) by (1).
$tan \theta = \frac{v^{2}}{gl sin \theta} \\\\ v^{2} = gl\, sin \theta \,tan \theta \\\\ v= \sqrt{gl \, sin \theta \, tan \theta}$

Answer:   $v = \sqrt{gl \, sin \theta \, tan \theta}$