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Solution

  1. Calculating the slope, we get
    α = ω ω 0 t t 0 = ( 0 30.0 ) rad/s ( 5.0 0 ) s = −6.0 rad/s 2 .

    We see that this is exactly [link] with a little rearranging of terms.
  2. We can find the area under the curve by calculating the area of the right triangle, as shown in [link] .
    Figure is a graph of the angular velocity in rads per second plotted versus time in seconds. Angular velocity decreases linearly with time, from 30 rads per second at zero seconds to zero at 5 seconds. The area under the curve is filled.
    The area under the curve is the area of the right triangle.

    Δ θ = area ( triangle ) ; Δ θ = 1 2 ( 30 rad/s ) ( 5 s ) = 75 rad .

    We verify the solution using [link] :
    θ f = θ 0 + ω 0 t + 1 2 α t 2 .

    Setting θ 0 = 0 , we have
    θ 0 = ( 30.0 rad / s ) ( 5.0 s ) + 1 2 ( −6.0 rad / s 2 ) ( 5.0 rad / s ) 2 = 150.0 75.0 = 75.0 rad .

    This verifies the solution found from finding the area under the curve.

Significance

We see from part (b) that there are alternative approaches to analyzing fixed-axis rotation with constant acceleration. We started with a graphical approach and verified the solution using the rotational kinematic equations. Since α = d ω d t , we could do the same graphical analysis on an angular acceleration-vs.-time curve. The area under an α -vs.- t curve gives us the change in angular velocity. Since the angular acceleration is constant in this section, this is a straightforward exercise.

Summary

  • The kinematics of rotational motion describes the relationships among rotation angle (angular position), angular velocity, angular acceleration, and time.
  • For a constant angular acceleration, the angular velocity varies linearly. Therefore, the average angular velocity is 1/2 the initial plus final angular velocity over a given time period:
    ω = ω 0 + ω f 2 .
  • We used a graphical analysis to find solutions to fixed-axis rotation with constant angular acceleration. From the relation ω = d θ d t , we found that the area under an angular velocity-vs.-time curve gives the angular displacement, θ f θ 0 = Δ θ = t 0 t ω ( t ) d t . The results of the graphical analysis were verified using the kinematic equations for constant angular acceleration. Similarly, since α = d ω d t , the area under an angular acceleration-vs.-time graph gives the change in angular velocity: ω f ω 0 = Δ ω = t 0 t α ( t ) d t .

Conceptual questions

If a rigid body has a constant angular acceleration, what is the functional form of the angular velocity in terms of the time variable?

straight line, linear in time variable

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If a rigid body has a constant angular acceleration, what is the functional form of the angular position?

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If the angular acceleration of a rigid body is zero, what is the functional form of the angular velocity?

constant

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A massless tether with a masses tied to both ends rotates about a fixed axis through the center. Can the total acceleration of the tether/mass combination be zero if the angular velocity is constant?

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Problems

A wheel has a constant angular acceleration of 5.0 rad / s 2 . Starting from rest, it turns through 300 rad. (a) What is its final angular velocity? (b) How much time elapses while it turns through the 300 radians?

a. ω = 54.8 rad / s ;
b. t = 11.0 s

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During a 6.0-s time interval, a flywheel with a constant angular acceleration turns through 500 radians that acquire an angular velocity of 100 rad/s. (a) What is the angular velocity at the beginning of the 6.0 s? (b) What is the angular acceleration of the flywheel?

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The angular velocity of a rotating rigid body increases from 500 to 1500 rev/min in 120 s. (a) What is the angular acceleration of the body? (b) Through what angle does it turn in this 120 s?

a. 0.87 rad / s 2 ;
b. θ = 66,264 rad

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A flywheel slows from 600 to 400 rev/min while rotating through 40 revolutions. (a) What is the angular acceleration of the flywheel? (b) How much time elapses during the 40 revolutions?

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A wheel 1.0 m in diameter rotates with an angular acceleration of 4.0 rad / s 2 . (a) If the wheel’s initial angular velocity is 2.0 rad/s, what is its angular velocity after 10 s? (b) Through what angle does it rotate in the 10-s interval? (c) What are the tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval?

a. ω = 42.0 rad / s ;
b. θ = 200 rad ; c. v t = 42 m / s a t = 4.0 m / s 2

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A vertical wheel with a diameter of 50 cm starts from rest and rotates with a constant angular acceleration of 5.0 rad / s 2 around a fixed axis through its center counterclockwise. (a) Where is the point that is initially at the bottom of the wheel at t = 10 s? (b) What is the point’s linear acceleration at this instant?

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A circular disk of radius 10 cm has a constant angular acceleration of 1.0 rad / s 2 ; at t = 0 its angular velocity is 2.0 rad/s. (a) Determine the disk’s angular velocity at t = 5.0 s . (b) What is the angle it has rotated through during this time? (c) What is the tangential acceleration of a point on the disk at t = 5.0 s ?

a. ω = 7.0 rad / s ;
b. θ = 22.5 rad ; c. a t = 0.1 m / s

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The angular velocity vs. time for a fan on a hovercraft is shown below. (a) What is the angle through which the fan blades rotate in the first 8 seconds? (b) Verify your result using the kinematic equations.

Figure is a graph of the angular velocity in rev per minute plotted versus time in seconds. Angular velocity is zero when the time is equal to zero and increases linearly with time.
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A rod of length 20 cm has two beads attached to its ends. The rod with beads starts rotating from rest. If the beads are to have a tangential speed of 20 m/s in 7 s, what is the angular acceleration of the rod to achieve this?

α = 28.6 rad / s 2 .

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Practice Key Terms 1

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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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