Theoretical Mechanics: Fall-2021
Practice Acceleration in Polar Coordinates (SOLUTION): Due Day 16

1. Acceleration in Polar Coordinates S1 4099S

   The acceleration vector written in polar coordinates is:

   \[\vec{a} = \big[\ddot{s}-s\dot{\phi}^2\big]\hat{s}+\big[2\dot{s}\dot{\phi}+s\ddot{\phi}\big]\hat{\phi}\]

   Below are motion diagrams, where each dot represents the location of a particle at equal time intervals.

   For each motion diagram, indicate the direction of the acceleration vector at the middle dot (point 3) by sketching velocity vectors for points 2&4 and subtracting them.

   Break the acceleration vector into polar components and match the components to terms in the acceleration equation above.

   

   1. The picture looks like:
      
      The particle is moving directly away from the origin, so \(\dot{\phi}=0\) and \(\ddot{\phi}=0\). The equation for the acceleration is: \[\vec{a} = \ddot{s}\hat{s}\] The acceleration only has a radial component.
   2. The picture looks like:
      The particle is moving in a circle with constant angular speed, so \(\dot{s}=0\), \(\ddot{s}=0\), \(\dot{\phi}=\)constant, and \(\ddot{\phi}=0\). The equation for the acceleration is: \[\vec{a} = s\dot{\phi}^2\hat{s}\] The acceleration only has a radial (centripetal) component.
   3. The picture looks like: The particle is moving in a circle with an increasing angular speed, so \(\dot{s}=0\), and \(\ddot{s}=0\). The equation for the acceleration is: \[\vec{a} = s\dot{\phi}^2\hat{s}+s\ddot{\phi}\hat{\phi}\] The acceleration only both a radial (centripetal) component and a tangential component corresponding to the increasing speed of the particle.
   4. The picture looks like:
      The particle is moving around but away from the origin so \(\dot{s}\neq0\). The radius of the motion increases the same amount in each time interval, so \(\ddot{s}=0\). The angular displacement is \(\pi/4\) for each time interval so the angular velocity is constant, so \(\ddot{\phi}=0\). The equation for the acceleration is: \[\vec{a} = s\dot{\phi}^2\hat{s}+2\dot{s}\dot{\phi}\hat{\phi}\] The acceleration only both a radial (centripetal) component and a tangential Coriolis component corresponding to the changing direction of radial component of the velocity vector as the particle moves around the origin.