Rotational Mechanics#0

Classical Mechanics Level 2

NOte- Enter the correct option number as the answer


The answer is 2.

Aakhyat Singh by Aakhyat Singh , 20, India

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1 solution

Steven Chase
Feb 19, 2018

Balance equations:

T c o s θ = m g T s i n θ = m v 2 l s i n θ T \, cos \theta = m g \\ T \, sin \theta = \frac{m v^2}{l \, sin \theta}

Re-arranging:

g c o s θ s i n θ = v 2 l s i n θ v = g l s i n 2 θ c o s θ \frac{g}{cos \theta} \, sin \theta = \frac{v^2}{l \, sin \theta} \\ v = \sqrt{\frac{g l \, sin^2 \theta}{cos \theta}}

Angular momentum:

L = r m v = l s i n θ m g l s i n 2 θ c o s θ = m 2 g l 3 s i n 4 θ c o s θ L = r m v = l \, sin \theta \, m \ \sqrt{\frac{g l \, sin^2 \theta}{cos \theta}} \\ = \sqrt{\frac{m^2 \,g \, l^3 \, sin^4 \theta}{cos \theta}}

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Just solved this problem. Here is my approach.

The position vector of the bob is:

r ⃗ = [ L sin ⁡ θ cos ⁡ ϕ L sin ⁡ θ sin ⁡ ϕ − L cos ⁡ θ ] \vec{r} = \left[\begin{matrix} L\sin{\theta}\cos{\phi}\\ L\sin{\theta}\sin{\phi} \\ -L\cos{\theta} \end{matrix} \right]

Linear momentum is:

p ⃗ = [ − m L ϕ ˙ sin ⁡ θ sin ⁡ ϕ m L ϕ ˙ sin ⁡ θ cos ⁡ ϕ 0 ] \vec{p} = \left[\begin{matrix} -mL\dot{\phi}\sin{\theta}\sin{\phi}\\ mL\dot{\phi}\sin{\theta}\cos{\phi} \\ 0 \end{matrix} \right]

Here θ \theta is constant.

Now, according to the force balance:

T sin ⁡ θ = m ϕ ˙ 2 L sin ⁡ θ T \sin{\theta} = m \dot{\phi}^2 L \sin{\theta} T cos ⁡ θ = m g T \cos{\theta} = mg

⟹ g tan ⁡ θ L sin ⁡ θ = ϕ ˙ 2 \implies \frac{g\tan{\theta}}{L\sin{\theta}} = \dot{\phi}^2

The vectors r ⃗ \vec{r} and p ⃗ \vec{p} are perpendicular to each other. Therefore, the magnitude of the angular momentum is:

∣ r ⃗ × p ⃗ ∣ = ∣ r ⃗ ∣ ∣ p ⃗ ∣ sin ⁡ ( π 2 ) \lvert \vec{r} \times \vec{p} \rvert = \lvert \vec{r} \rvert \lvert \vec{p} \rvert \sin\left(\frac{\pi}{2}\right) ∣ r ⃗ × p ⃗ ∣ = L ( m L sin ⁡ θ ϕ ˙ ) sin ⁡ ( π 2 ) \lvert \vec{r} \times \vec{p} \rvert = L (mL\sin{\theta} \dot{\phi}) \sin\left(\frac{\pi}{2}\right) ∣ r ⃗ × p ⃗ ∣ = m L 2 sin ⁡ θ g tan ⁡ θ L sin ⁡ θ \lvert \vec{r} \times \vec{p} \rvert = mL^2\sin{\theta} \sqrt{ \frac{g\tan{\theta}}{L\sin{\theta}} } ∣ r ⃗ × p ⃗ ∣ = m 2 L 4 sin ⁡ 2 θ ( g tan ⁡ θ L sin ⁡ θ ) \lvert \vec{r} \times \vec{p} \rvert = \sqrt{ m^2L^4\sin^2{\theta} \left(\frac{g\tan{\theta}}{L\sin{\theta}} \right)} ∣ r ⃗ × p ⃗ ∣ = m 2 g L 3 sin ⁡ 2 θ cos ⁡ θ \lvert \vec{r} \times \vec{p} \rvert = \sqrt{ \frac{m^2gL^3\sin^2{\theta}}{\cos{\theta}} }

I cannot find my mistake here. Any comments would be helpful.

Karan Chatrath - 3 months, 3 weeks ago

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@Steven Chase What are your thoughts on this?

Karan Chatrath - 3 months, 3 weeks ago

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For the angular momentum calculation, we want r ⃗ \vec{r} to be defined with respect to the center of the circle of rotation, rather than with respect to the origin. The difference is a factor of sin ⁡ θ \sin \theta , which I think is why your answer is off by just a factor of sin ⁡ 2 θ \sin^2 \theta under the radical.

Steven Chase - 3 months, 3 weeks ago

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@Steven Chase – Yes, I was calculating the angular momentum about the point of suspension. I did not read the question completely. Now my answer is the same as yours. Thank you for the feedback.

Karan Chatrath - 3 months, 3 weeks ago

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