Accelerating Turntable

Classical Mechanics Level 2

A turntable is spun from rest with a constant angular acceleration of $\frac{\pi}{2} \text{ rad}/\text{s}^2$ . After completing six full revolutions, what is its angular velocity in $\text{rad}/\text{s}$ ?

12\pi^2

2\pi\sqrt{3}

3\pi

\pi\sqrt{3}

2 solutions

Matt DeCross
Feb 2, 2016

Note that the total angular displacement over six full revolutions is $12\pi$ . The angular velocity is then found from the equation:

$\omega^2 = 2\alpha \theta = (\pi \text{ rad}/\text{s}^2)(12\pi) = 12\pi^2 \text{ rad}/\text{s}^2.$

Taking square roots yields the answer $\omega = \sqrt{12\pi^2} \text{ rad}/\text{s} = 2\pi \sqrt{3} \text{ rad}/\text{s}$ .

It's a good explanation, although in the task the author should mention the value of initial angular velocity in order to make things clearer, because it makes the beginners feel a bit confused, the precision I think is the key to a better comprehension, because it makes things clearer because the reader knows that it would be difficult to solve the equation if a different formula was used than the given one.

Weronika Kieca - 2 years, 6 months ago

The problem does tell you the value of initial angular velocity - the turntable is spun from rest .

Matt DeCross - 2 years, 5 months ago

Gandoff Tan
Mar 25, 2019

$\theta =6(2\pi)=12\pi \quad \alpha =\frac { π }{ 2 } \Rightarrow ω=\frac { π }{ 2 } t\Rightarrow t=\frac { 2 }{ π } ω$

$\begin{aligned} \theta & = & \frac { 1 }{ 2 } t\omega \\ 12π & = & \frac { 1 }{ 2 } (\frac { 2 }{ π } ω)ω \\ 12π & = & \frac { 1 }{ π } ω^{ 2 } \\ ω^{ 2 } & = & 12{ \pi }^{ 2 } \\ ω & = & \boxed { 2π\sqrt { 3 } } \end{aligned}$

Accelerating Turntable

2 solutions

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