Solve the following :

Question:

A uniform rod of mass $\mathrm{m}$ and length $\mathrm{I}$ is struck at an end by a force $F$ perpendicular to the rod for short time interval t. Calculate

(a) the speed of the centre of mass, (b) the angular speed of the rod about the centre of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the centre of mass after the force has stopped to act. Assume that $t$ is so small that the rod does not appreciably change its direction while the force acts.

Solution:

(a) $F=m a$

acceleration, $\mathrm{a}=\frac{F}{m} ; \mathrm{u}=0 ; \mathrm{t}$

$v=u+a t$

$\mathrm{v}=\frac{F}{m} t$

(b) $\mathrm{T}=1 \mathrm{a}$

$F\left(\frac{E}{6}\right)=\frac{M L^{2}}{12} \cdot a$

$\stackrel{L}{\mathrm{~L}}\left({ }^{2}\right)=\frac{M L^{2}}{12}$

$\frac{6 F}{M L} \cdot \omega_{0}=0$

$\alpha=\omega_{0}+\mathrm{t}=\mathrm{t}$

$\omega=\frac{6 F t}{M L}$

$\alpha=$

(c) $\mathrm{KE}=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$

$=\frac{1}{2} m\left(\frac{F t}{m}\right)^{2}+\frac{1}{2}\left(\frac{M L^{2}}{12}\right)\left(\frac{6 F t}{m L}\right)^{2}$

$=\frac{2 F^{2} t^{3}}{2 m}$

(d) $L=1^{\omega}$

$=\left(\frac{M L^{2}}{12}\right)\left(\frac{6 F t}{M L}\right)$

$\mathrm{L}=2 \mathrm{~m}$

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