The bob must have tangential speed v = √ ( g L sin θ tan θ )
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Angular Speed can be formulated as follows:
[tex]\boxed {\omega = \frac{ v }{ R }}[/tex]
ω = Angular Speed ( rad/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
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Linear Speed can be formulated as follows:
[tex]\boxed {v = \frac{ 2 \pi R }{ T }}[/tex]
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 !
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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:
[tex]\Sigma F_y = T_y - mg[/tex]
[tex]0 = T\cos \theta - mg[/tex]
[tex]T \cos \theta = mg[/tex]
[tex]\boxed {T = mg \div \cos \theta}[/tex] → Equation 1
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[tex]\Sigma F_x = ma[/tex]
[tex]T_x = m \frac{v^2}{R}[/tex]
[tex]T \sin \theta = m \frac{v^2}{R}[/tex]
[tex]( mg \div \cos \theta ) \sin \theta = m \frac{v^2}{ L \sin \theta }[/tex] ← Equation 1
[tex]g \tan \theta = \frac{v^2}{L \sin \theta}[/tex]
[tex]v^2 = gL \tan \theta \sin \theta[/tex]
[tex]\boxed {v = \sqrt { gL \tan \theta \sin \theta } }[/tex]
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Grade: High School
Subject: Physics
Chapter: Circular Motion
