Answer:
Centripetal acceleration, a = 0.03125 m/s²
Explanation:
Given that,
Distance from center of a rotating platter, r = 0.08 m
Tangential speed of the pickle, v = 0.05 m/s
The centripetal acceleration is given by the formula
a = v²/r
Substituting the given values in the above equation
a = (0.05 m/s)² / 0.08 m
= 0.03125 m/s²
Hence, the the pickle's centripetal acceleration a = 0.03125 m/s²
The pickle's centripetal acceleration is 0.03125 m/s²
Centripetal Acceleration can be formulated as follows:
[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]
a = Centripetal Acceleration ( m/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
[tex]\texttt{ }[/tex]
Centripetal Force can be formulated as follows:
[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]
F = Centripetal Force ( m/s² )
m = mass of Particle ( kg )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
To find the pickle's centripetal acceleration can be carried out in the following way:
Given:
Radius of Circular Motion = 0.08 m
Tangential Speed of Pickle = 0.05 m/s
Unknown:
Centripetal Acceleration of Pickle = a = ?
Solution:
[tex]a = v^2 \div R[/tex]
[tex]a = 0.05^2 \div 0.08[/tex]
[tex]a = 0.03125 \texttt{ m/s}^2[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Circular Motion
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
