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Direction of acceleration

  1. Jan 18, 2015 #1
    My friend and i were having a conversation on circular motion and were confused with the direction of acceleration along that circular path.what will be the direction of acceleration of an object which is in circular motion?
  2. jcsd
  3. Jan 18, 2015 #2
    Newton's second law states that the rate of change of momentum of an object is proportional to the force acting on the object. This means that the acceleration will always be in the same direction as the resultant force. Can you tell me where the centripetal force acts?
  4. Jan 18, 2015 #3
    It pulls the object towards the centre of circular path..
  5. Jan 18, 2015 #4
    Precisely. Centripetal acceleration always acts towards the center of the circle.
  6. Jan 18, 2015 #5
  7. Jan 18, 2015 #6


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    Well, the most general circular motion can be described by an angle ##\phi(t)##. Let the circle be in the origin of the ##xy## plane. Then the trajectory is given by
    $$\vec{x}(t)=R \begin{pmatrix} \cos[\phi(t)] \\ \sin [\phi(t)] \end{pmatrix}.$$
    Now you have to take the 1st and 2nd time derivatives to get velocity and acceleration:
    $$\vec{v}(t)=\dot{\vec{x}}(t)=R \dot{\phi}(t) \begin{pmatrix} -\sin[\phi(t)] \\ \cos[\phi(t)] \end{pmatrix},$$
    $$\vec{a}(t)=\dot{\vec{v}}(t)=\ddot{\vec{x}}(t) = R \ddot{\phi}(t) \begin{pmatrix} -\sin[\phi(t)] \\ \cos[\phi(t)] \end{pmatrix}-R \dot{\phi}^2(t) \begin{pmatrix} \cos[\phi(t)] \\ \sin [\phi(t)] \end{pmatrix}.$$
    As you see, the velocity is (as for any motion) always pointing along the tangent of the trajectory. The acceleration splits into two parts: The tangential acceleration of magnitude (and sign wrt. the direction of the tangent vector) ##a_{\parallel}=R \ddot{\phi}## and one perpendicular, i.e., along the position vector. The component is ##a_{\perp}=-R \dot{\phi}^2 \leq 0##, which means it's always negative, i.e., directed towards the center. The prependicular component is called centripetal acceleration.

    According to Newton's Law to maintain this motion you need the total force
    $$\vec{F}=m \vec{a}.$$
    The part in direction perpendicular to the trajectory is called centripetal force.
    Last edited: Jan 18, 2015
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