Relationship Between Linear And Angular Motion MCQ

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A thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$ . Its centre of mass rises to a maximum height of

$\frac{1}{3}\frac{l^{2}\omega ^{2}}{g}$

$\frac{1}{6}\frac{l\omega }{g}\;$

$\frac{1}{2}\frac{l^{2}\omega ^{2}}{g}$

$\; \frac{1}{6}\frac{l^{2}\omega^{2}}{g}$

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A particle of mass $m$ moves along line PC with velocity $\nu$ as shown. What is the angular momentum of the particle about P?

$m\nu L$

$m\nu l$

$m\nu r$

zero

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Concepts Covered - 1

Equations of Linear Motion and Rotational Motion.

	Linear Motion	Rotational Motion
I	If linear acceleration =a=0 Then u = constant and s = u t.	If angular acceleration= $\alpha = 0$ Then $\omega = constant$ and $\theta = \omega.t$
II	If linear acceleration= a = constant $a = \frac{v-u}{t}$ $v=u+at$ $s= ut +\frac{1}{2}at^{2}$ $s=\frac{v+u}{2}*t$ $v^{2}-u^{2}=2as$ $S_{n}= u+\frac{a}{2}(2n-1)$	If angular acceleration= $\alpha = \frac{\omega_f - \omega_i}{t}$ $\omega_f=\omega_i+\alpha.t$ $\theta = \omega_i.t+\frac{1}{2}.\alpha.t^2$ $\theta =\frac{\omega _f+\omega _i}{2}*t$ $\omega _f^2-\omega _i^2=2\alpha \theta$ $\theta_n = \omega_i+\frac{\alpha}{2}(2n-1)$
III	If linear acceleration= a $\neq$ constant $v = \frac{dx}{dt}$ $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$ $v.dv = a.ds$	If angular acceleration= $\alpha \neq constant$ $\omega = \frac{d\theta}{dt}$ $\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$ $\omega.d\omega = \alpha.d\theta$