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  5. Relating angular and translational

10.3 Relating angular and translational quantities  (Page 3/5)

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Solution

The angular acceleration is

α = ω − ω 0 t = 0 − ( 1.0 × 10 4 ) 2 π / 60.0 s ( rad / s ) 30.0 s = −34.9 rad / s 2 .

Therefore, the tangential acceleration is

a t = r α = 0.2 m ( −34.9 rad / s 2 ) = −7.0 m / s 2 .

The angular velocity at t = 29.0 s is

ω = ω 0 + α t = 1.0 × 10 4 ( 2 π 60.0 s ) + ( −34.9 rad / s 2 ) ( 29.0 s ) = 1047.2 rad / s − 1012.71 = 35.1 rad / s .

Thus, the tangential speed at t = 29.0 s is

v t = r ω = 0.2 m ( 35.1 rad / s ) = 7.0 m / s .

We can now calculate the centripetal acceleration at t = 29.0 s :

a c = v 2 r = ( 7.0 m / s ) 2 0.2 m = 245.0 m / s 2 .

Since the two acceleration vectors are perpendicular to each other, the magnitude of the total linear acceleration is

| a → | = a c 2 + a t 2 = ( 245.0 ) 2 + ( −7.0 ) 2 = 245.1 m / s 2 .

Since the centrifuge has a negative angular acceleration, it is slowing down. The total acceleration vector is as shown in [link] . The angle with respect to the centripetal acceleration vector is

θ = tan −1 −7.0 245.0 = −1.6 ° .

The negative sign means that the total acceleration vector is angled toward the clockwise direction.

Figure shows a particle executing circular motion in the counterclockwise direction. The vector a t is pointed clockwise. Vectors a and a c point toward the center of the circle, and the label “direction of motion” points in the opposite direction of vector a t.
The centripetal, tangential, and total acceleration vectors. The centrifuge is slowing down, so the tangential acceleration is clockwise, opposite the direction of rotation (counterclockwise).

Significance

From [link] , we see that the tangential acceleration vector is opposite the direction of rotation. The magnitude of the tangential acceleration is much smaller than the centripetal acceleration, so the total linear acceleration vector will make a very small angle with respect to the centripetal acceleration vector.

Check Your Understanding A boy jumps on a merry-go-round with a radius of 5 m that is at rest. It starts accelerating at a constant rate up to an angular velocity of 5 rad/s in 20 seconds. What is the distance travelled by the boy?

The angular acceleration is α = ( 5.0 − 0 ) rad / s 20.0 s = 0.25 rad / s 2 . Therefore, the total angle that the boy passes through is
Δ θ = ω 2 − ω 0 2 2 α = ( 5.0 ) 2 − 0 2 ( 0.25 ) = 50 rad .
Thus, we calculate
s = r θ = 5.0 m ( 50.0 rad ) = 250.0 m .

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Check out this PhET simulation to change the parameters of a rotating disk (the initial angle, angular velocity, and angular acceleration), and place bugs at different radial distances from the axis. The simulation then lets you explore how circular motion relates to the bugs’ xy -position, velocity, and acceleration using vectors or graphs.

Summary

  • The linear kinematic equations have their rotational counterparts such that there is a mapping x → θ , v → ω , a → α .
  • A system undergoing uniform circular motion has a constant angular velocity, but points at a distance r from the rotation axis have a linear centripetal acceleration.
  • A system undergoing nonuniform circular motion has an angular acceleration and therefore has both a linear centripetal and linear tangential acceleration at a point a distance r from the axis of rotation.
  • The total linear acceleration is the vector sum of the centripetal acceleration vector and the tangential acceleration vector. Since the centripetal and tangential acceleration vectors are perpendicular to each other for circular motion, the magnitude of the total linear acceleration is | a → | = a c 2 + a t 2 .

Conceptual questions

Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude.

The centripetal acceleration vector is perpendicular to the velocity vector.

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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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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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