Centripetal Acceleration and Tangential Acceleration?

Distinguish between the tangential acceleration of motion along the circumference of a circle and the centripetal acceleration. How is each of these quantities related to the angular acceleration? Any answers to this will be greatly appreciated! Thanks!

3 Answers

  • 1 decade ago
    Favorite Answer

    If the particle is moving in a straight line. There can be only tangential acceleration only. Of course the tangential acceleration may be zero as well, in which case the particle will move with constant velocity.

    If the motion of particle is not in straight line, there may be tangential as well as normal acceleration. If both are nonzero then both the speed or direction or both may change. The resultant motion may be circular or may not be. If it is circular motion and tangential acceleration is zero then the sped of the particle will not change. As angular velocity multiplied by radius of curvature gives you instantaneous linear velocity, angular acceleration multiplied by radius of curvature will give you linear acceleration.

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  • Anonymous
    1 decade ago

    The tangential acc is the vector component of the angular vector that is paralel to the circles part where the moving ovject is.

    The tangential vector is at a 90 degree to the centripetal vector and together they are the another vector I think... I didn't study this in english, but since you said ANY answers I thought I'd pitch in...

    The two vectors together make the angular acceleration. If an object is also having tangential acceleration in addition to the centripetal acc the object starts to move inside and will reach the center of the circle... Since their common vector (the angular vector) will be facing inwards, so the object will I think move inside the circle until it reaches the center.

    Then it might just rotate... I think... But don't listen to me, I'm not sure what I'm talking about.

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  • 4 years ago

    Let w = angular speed, rad/sec Then, at some time t = T, w^2*r = r*(dw/dt), r = radius of bit So, w^2 = 2dw/dt Let a = angular accel. = constant, rad/s^2 w = aT Then, (aT)^2 = 2*(dw/dt) = 2a a = 2/T^2 Let total angle traveled = theta = (a*T^2)/2 Then theta = (a*T^2)/2 = (T^2/2)*(2/T^2) = 1 radian Answer

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