First slide

Circular motion

Question

A particle moves in xy plane. The position vector at any time / is $\vec{r} = \{(2 t) \hat{i} + (2 t^{2}) \hat{j}\} m$ . The rate of change of $θ$ at time t = 2 second (where $θ$ is the angle which its velocity vector makes with positive x-axis) is

Moderate

A

$\frac{2}{17} rad / s$
B

$\frac{1}{14} rad / s$
C

$\frac{4}{7} rad / s$
D

$\frac{6}{5} rad / s$

Solution

$\begin{array}{l} x = 2 t \Rightarrow v_{x} = \frac{d x}{d t} = 2 \\ Y = 2 t^{2} \Rightarrow v_{y} = \frac{d y}{d t} = 4 t \\ \tan θ = \frac{v_{y}}{v_{x}} = \frac{4 t}{2} = 2 t \end{array}$
$∴$ Differentiating with respect to time we get,
$\begin{array}{l} (\sec^{2} θ) \frac{d θ}{2} = 2 \\ or (1 + \tan^{2} θ) \frac{d θ}{2} = 2; or (1 + 4 t^{2}) \frac{d θ}{d t} = 2 \\ or \frac{d θ}{d t} = \frac{2}{1 + 4 t^{2}}; \frac{d θ}{d t} at t = 2 is \\ \frac{d θ}{d t} = \frac{2}{1 + 4 (2)^{2}} = \frac{2}{17} rad / s \end{array}$

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