question_answer

A spot light S rotates in a horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light P moves along the wall at a distance 3 m. What is the velocity of the spot P when\[\text{ }\!\!\theta\!\!\text{ }=\text{ 45}{}^\circ \]?

Answer:

                Here, \[\omega =0.1\,rad/s\], SO = 3 m, \[\text{ }\!\!\theta\!\!\text{ }=\text{ 45}{}^\circ \], vel. of spot, \[\upsilon =?\]At any time, let the spot of light P be at a distance x from O, Fig. 5(HT). 10. Let \[\angle OSP=\phi \] and\[\angle OPS=\theta \]. Clearly, \[\upsilon =\frac{dx}{dt}\] and \[\omega =\frac{d\phi }{dt}=0.1rad/s\] In \[\Delta OSP\], \[\tan \phi =\frac{OP}{OS}=\frac{x}{3}\] \[x=3\tan \phi \] Differentiating, we get, \[\frac{dx}{dt}=3{{\sec }^{2}}\phi \frac{d\phi }{dt}\] \[\upsilon =3{{\sec }^{2}}\phi (\omega )\]                                                            .... (i) As \[\phi ={{90}^{\text{o}}}-\text{ }\!\!\theta\!\!\text{ =9}{{\text{0}}^{\text{o}}}-{{45}^{\text{o}}}={{45}^{\text{o}}}\], \[\therefore \]  \[\sec \phi =\sec {{45}^{\text{o}}}=\sqrt{2}\] From(i), \[\upsilon =3{{(\sqrt{2})}^{2}}\times 0.1=0.6m/s\]