(B) The velocity $v$ of a particle moving along a curved path can be resolved into radial $(v_r = dr/dt)$ and transverse $(v_{\theta} = r(d\theta/dt))

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(B) The velocity $v$ of a particle moving along a curved path can be resolved into radial $(v_r = dr/dt)$ and transverse $(v_{\theta} = r(d\theta/dt))$ velocity components.The angle $\alpha$ between the resultant velocity $v$ and the radial velocity $v_r$ is given by $\tan \alpha = v_{\theta} / v_r$.Substituting the expressions for $v_{\theta}$ and $v_r$:$\tan \alpha = \frac{r(d\theta/dt)}{dr/dt} = r \cdot \frac{d\theta}{dr} = \frac{r}{dr/d\theta}$Given that $dr/d\theta = r$,we have:$\tan \alpha = \frac{r}{r} = 1$Therefore,$\alpha = \tan^{-1}(1) = 45^{\circ}$.

Class 11 Physics 2 Motion in a Plane Mix Examples-Motion in Plane

$A$ particle at a distance of $1 \ m$ from the origin starts moving,such that $dr/d\theta = r$,where $r$ and $\theta$ are polar coordinates. Then,the angle between the resultant velocity and the tangential velocity is

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A
$30^{\circ}$
B
$45^{\circ}$
C
$60^{\circ}$
D
dependent on where the particle is

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2 Motion in a Plane

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Mix Examples-Motion in Plane

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$A$ stone of mass $m$ is tied to a string of length $l$ and rotated in a circle with a constant speed $v$. If the string is released,the stone flies:

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$A$ particle moves along a circle with a constant angular speed $\omega$. Its displacement,with respect to the initial position of the particle at time $t = 0$,is plotted against time. The graph would look like:

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$A$ projectile is fired at an angle of $30^{\circ}$ to the horizontal such that the vertical component of its initial velocity is $80\,m/s$. Its time of flight is $T$. Its velocity at $t=\frac{T}{4}$ has a magnitude of nearly $........\,m/s$.

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Particles $A$ and $B$ are moving with constant velocities along the $x$ and $y$ axes respectively. The graph of the separation $S$ between them with time $t$ is:

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The figure shows the velocity and the acceleration of a point-like body at the initial moment of its motion. The direction and the absolute value of the acceleration remain constant. Find the time when the speed becomes minimum. (Given: $a = 4\, m/s^2$,$v_0 = 40\, m/s$,$\phi = 143^o$)

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$A$ particle at a distance of $1 \ m$ from the origin starts moving,such that $dr/d\theta = r$,where $r$ and $\theta$ are polar coordinates. Then,the angle between the resultant velocity and the tangential velocity is

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$A$ stone of mass $m$ is tied to a string of length $l$ and rotated in a circle with a constant speed $v$. If the string is released,the stone flies:

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