A horizontal disc of radius 2 m rotates anticlockwise at 2 rad/s. Let V be its velocity field and V= V . Which are correct? Cylindrical gradient, divergence, curl and Laplacian may be used.

GATE 2023 PH – Question 57

Mathematical Physics · linear differential equations: simple applications of first and second order linear differential equations and solutions · 2 marks · Multiple select

A horizontal disc of radius 2 m rotates anticlockwise at 2 rad/s. Let $\mathbf V$ be its velocity field and $V=|\mathbf V|$. Which are correct? Cylindrical gradient, divergence, curl and Laplacian may be used.

$\nabla V=2\hat r$
$\nabla\cdot\mathbf V=2$
$\nabla\times\mathbf V=4\hat z$
$\nabla^2V=4/3$ at r=1.5 m

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Show answer and explanation

Correct answer: (A) $\nabla V=2\hat r$; (C) $\nabla\times\mathbf V=4\hat z$; (D) $\nabla^2V=4/3$ at r=1.5 m

Explanation

A horizontal disc of radius 2 m rotating anticlockwise at $\omega=2$ rad/s has the velocity field
$$\mathbf V=\boldsymbol\omega\times\mathbf r=\omega r\,\hat\theta=2r\,\hat\theta .$$

Its magnitude is $V=2r$.

**A. $\nabla V=2\hat r$.** In cylindrical coordinates, $\nabla V=\dfrac{\partial V}{\partial r}\hat r=2\hat r$. ✓
**B. $\nabla\cdot\mathbf V=2$.** $\nabla\cdot\mathbf V=\dfrac1r\dfrac{\partial}{\partial\theta}V_\theta=0$ (the velocity does not depend on $\theta$). ✗
**C. $\nabla\times\mathbf V=4\hat z$.** $(\nabla\times\mathbf V)_z=\dfrac1r\dfrac{\partial(rV_\theta)}{\partial r}=\dfrac1r\dfrac{\partial(2r^2)}{\partial r}=4$, that is $2\omega$ ✓
**D. $\nabla^2V=4/3$ at $r=1.5$ m.** $\nabla^2V=\dfrac1r\dfrac{d}{dr}\left(r\dfrac{dV}{dr}\right)=\dfrac1r\dfrac{d}{dr}(2r)=\dfrac2r$, and at $r=1.5$ this is $\dfrac43$. ✓

Answer **A, C and D**.