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The centripetal acceleration is due to the change in the direction of tangential velocity, whereas the tangential acceleration is due to any change in the magnitude of the tangential velocity. The tangential and centripetal acceleration vectors a t and a c are always perpendicular to each other, as seen in [link] . To complete this description, we can assign a total linear acceleration    vector to a point on a rotating rigid body or a particle executing circular motion at a radius r from a fixed axis. The total linear acceleration vector a is the vector sum of the centripetal and tangential accelerations,

a = a c + a t .

The total linear acceleration vector in the case of nonuniform circular motion points at an angle between the centripetal and tangential acceleration vectors, as shown in [link] . Since a c a t , the magnitude of the total linear acceleration is

| a | = a c 2 + a t 2 .

Note that if the angular acceleration is zero, the total linear acceleration is equal to the centripetal acceleration.

Figure shows a particle executing circular motion. The vector ac is at an angle between the vectors a and at.
A particle is executing circular motion and has an angular acceleration. The total linear acceleration of the particle is the vector sum of the centripetal acceleration and tangential acceleration vectors. The total linear acceleration vector is at an angle in between the centripetal and tangential accelerations.

Relationships between rotational and translational motion

We can look at two relationships between rotational and translational motion.

  1. Generally speaking, the linear kinematic equations have their rotational counterparts. [link] lists the four linear kinematic equations and the corresponding rotational counterpart. The two sets of equations look similar to each other, but describe two different physical situations, that is, rotation and translation.
    Rotational and translational kinematic equations
    Rotational Translational
    θ f = θ 0 + ω t x = x 0 + v t
    ω f = ω 0 + α t v f = v 0 + a t
    θ f = θ 0 + ω 0 t + 1 2 α t 2 x f = x 0 + v 0 t + 1 2 a t 2
    ω f 2 = ω 0 2 + 2 α ( Δ θ ) v f 2 = v 0 2 + 2 a ( Δ x )
  2. The second correspondence has to do with relating linear and rotational variables in the special case of circular motion. This is shown in [link] , where in the third column, we have listed the connecting equation that relates the linear variable to the rotational variable. The rotational variables of angular velocity and acceleration have subscripts that indicate their definition in circular motion.
    Rotational and translational quantities: circular motion
    Rotational Translational Relationship ( r = radius )
    θ s θ = s r
    ω v t ω = v t r
    α a t α = a t r
      a c a c = v t 2 r

Linear acceleration of a centrifuge

A centrifuge has a radius of 20 cm and accelerates from a maximum rotation rate of 10,000 rpm to rest in 30 seconds under a constant angular acceleration. It is rotating counterclockwise. What is the magnitude of the total acceleration of a point at the tip of the centrifuge at t = 29.0 s? What is the direction of the total acceleration vector?

Strategy

With the information given, we can calculate the angular acceleration, which then will allow us to find the tangential acceleration. We can find the centripetal acceleration at t = 0 by calculating the tangential speed at this time. With the magnitudes of the accelerations, we can calculate the total linear acceleration. From the description of the rotation in the problem, we can sketch the direction of the total acceleration vector.

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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