How Do You Calculate the Acceleration of a Particle in Polar Coordinates?

AI Thread Summary
To calculate the acceleration of a particle in polar coordinates with constant radial and angular velocities, the magnitude of the velocity is given by v = √((r dot)² + (r * theta dot)²). The acceleration can be determined by taking the time derivative of this velocity equation. Since r dot is constant, its derivative is zero, simplifying the process. The acceleration formula incorporates both radial and tangential components, leading to the expression |a| = d/dt(√((r dot)² + (theta dot)² * r²)). Understanding these relationships is crucial for accurately calculating the particle's acceleration.
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Homework Statement


A particle moves in a plane with constant radial velocity (r dot) and constant angular velocity (theta dot). When the particle is at distance r from the origin, determine the magnitude of the acceleration.

r dot = 4.3 m/s
theta dot = 2.5 rad/s
r = 3.1 m
speed of the particle at r = 8.86 m/s


Homework Equations


v = (((r dot)^2)+((r*(theta dot))^2))^.5


The Attempt at a Solution


I have attempted to find the derivative of the above equation, but I am not sure if I am doing the derivate wrong.
 
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toastie said:

Homework Statement


A particle moves in a plane with constant radial velocity (r dot) and constant angular velocity (theta dot). When the particle is at distance r from the origin, determine the magnitude of the acceleration.

r dot = 4.3 m/s
theta dot = 2.5 rad/s
r = 3.1 m
speed of the particle at r = 8.86 m/s

Homework Equations


v = (((r dot)^2)+((r*(theta dot))^2))^.5
You have the right expression for the magnitude of v as a function of r. Take the derivative with respect to time. Since \dot r is constant with time, what is its derivative?

|a| = |dv/dt| = \frac{d}{dt}\sqrt{|\dot r|^2 + \omega^2r^2}

AM
 
thanks for the help
 
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