S5.4 Constant acceleration equations
[A] Rotation with constant acceleration
Consider the motion of a body rotating about a fixed axis, with constant acceleration.- Since the rotation axis is fixed, displacement, velocity and acceleration vectors all point in the same direction. The motion is, in a sense, one dimensional.
The relationships between displacement, velocity and accleration (Key Point 5.4 and Key Point 5.6) are
- By analogy with the relations describing one dimensional linear motion
(Key Point 1.1
and Key Point 1.2),
we can immediately write down the constant
acceleration equations (c.f. Key Point 1.4)
Key Point 5.11
Constant Acceleration Equations:
Worked example
Reveal
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![\[ \alpha=\frac{\Delta \omega}{\Delta T}=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{33.3\times 2\pi/60 }{3}=1.16\;} rad\; s^{-2} \]](mastermathpng-2.png)

Worked example
A record player turntable (radius 0.152m) starts from rest and turns with constant angular acceleration to reach a final angular speed of 33.3 rev/min after 3s. Find the components of the linear acceleration at the rim, 2s after starting.
Solution
The angular acceleration is
![\[ \alpha=\frac{\Delta \omega}{\Delta T}=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{33.3\times 2\pi/60 }{3}=1.16\;} rad\; s^{-2} \]](mastermathpng-2.png)
From KP 6.11, the angular velocity after 2s is
ω(t=2s)=0+1.16×2=2.32 rad s-1
and thus
![\[ a_r=\frac{v^2}{r}=r\omega^2=0.818 \; m\;s^{-2} \]](mastermathpng-3.png)
at=rα=0.152×1.16=0.176 m s-2
Learning Resources
![]() | HRW Chapter 10.4 |
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