A different rotation

Classical Mechanics Level 4

A solid body rotates with deceleration about a stationary axis with an angular deceleration β ∝ ω \beta \propto \sqrt{\omega} , where ω \omega is its angular velocity. Find the mean angular velocity of the body averaged over the whole time of rotation if at the initial moment of time its angular velocity was equal to ω 0 = 18 rad/s \omega_0 = 18 \text{ rad/s} .

Problem is not original.


The answer is 6.

Rajdeep Dhingra by Rajdeep Dhingra , 20, India

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2 solutions

Kishore S. Shenoy
Aug 19, 2015

Let β = − (some) k ω \displaystyle \beta = -\text{(some)} k \sqrt{\omega}

We can say β = ω ⋅ d ω d θ \displaystyle \beta = \omega \cdot \frac{\mathrm{d}\omega}{\mathrm{d}\theta}

⇒ ω ⋅ d ω d θ = − k ω \displaystyle \Rightarrow \omega \cdot \frac{\mathrm{d}\omega}{\mathrm{d}\theta} = -k\sqrt{\omega}

⇒ ∫ 0 θ k ⋅ d θ = ∫ 0 ω 0 ω d ω = 2 3 ω 0 ω 0 \displaystyle \begin{aligned}\Rightarrow \int\limits_0^{\theta} k\cdot \mathrm{d}\theta &= \int\limits_0^{\omega_0}\sqrt{\omega} \mathrm{d}\omega\\ &= \frac{2}{3} \omega_0\sqrt{\omega_0}\end{aligned}

Now β = d ω d t = − k ω \displaystyle\beta = \frac{\mathrm{d}\omega}{\mathrm{d}t} = -k\sqrt{\omega}

⇒ ∫ 0 t k d t = ∫ 0 ω 0 d ω ω = 2 ω 0 \begin{aligned}\Rightarrow \displaystyle \int\limits_0^t k\mathrm{d}t &= \int\limits_0^{\omega_0} \frac{\mathrm{d}\omega}{\sqrt{\omega}}\\&=2\sqrt{\omega_0}\end{aligned}

∴ ⟨ ω ⟩ = k θ k t = ω 0 3 = 6 \displaystyle ∴ \langle\omega\rangle = \frac{k\theta}{kt} = \frac{\omega_0}{3} = \boxed{6}

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Moderator note:

Your work would benefit from some discussion of what you're doing. For instance, what is the integral you set up in the first part, and why is it important?

β = k × ω \beta = k \times \sqrt{\omega}

β = − d ω d t = k × ω \beta = \frac{-d\omega}{dt} = k \times \sqrt{\omega}

⇒ d ω 2 ω = − k d t 2 \Rightarrow \frac{d\omega}{2\sqrt{\omega}} = \frac{-kdt}{2}

Integrating we get,

ω − ω 0 = − k t 2 \sqrt{\omega} - \sqrt{\omega_0} = -k\frac{t}{2}

Without loss of generality, k = 1 k=1

The law of motion is : ω = t 2 4 + 18 − 3 2 t \omega = \frac{t^2}{4} + 18 - 3\sqrt{2}t and it stops at t 0 = 6 2 s e c t_0 = 6\sqrt{2} sec

< ω > = ∫ 0 t 0 ω d t ∫ 0 t 0 d t <\omega> = \dfrac{\displaystyle \int_{0}^{t_0} \omega dt}{\displaystyle \int_{0}^{t_0} dt}

We get < ω > = 6 <\omega> = \boxed{6}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Even if you dont assume k=1, it will cancel out eventually, creating no effect.

Pankaj Joshi - 6 years, 1 month ago

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Yeah. But why do it with someother k k when you can take 1. :-P

Vishwak Srinivasan - 6 years, 1 month ago

For a neater solution, k k is taken WLOG (without loss of generality) to be = 1 = 1 :)

Jake Lai - 6 years, 1 month ago

There are various Typos in Ur Sol

Rajdeep Dhingra - 6 years, 1 month ago

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I had forgotten the square root in the first two steps and forgot to divide by 2 in the third step. It is rectified. Check now.

Vishwak Srinivasan - 6 years, 1 month ago

You can use β = ω × d ω d θ \beta = \omega \times \frac{\mathrm{d} \omega} {\mathrm {d} \theta}

Kishore S. Shenoy - 5 years, 9 months ago

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Can you please post your solution using the formula above ?

Vishwak Srinivasan - 5 years, 9 months ago

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