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A particle of mass 0.2 kg moves along a path given by the relation : `vec(r)=2cos omegat hat(i)+3 sin omegat hat(j)`. Then he torque on the particle about the origin is :A. `sqrt(13) hat(k) Nm`B. `sqrt((2)/(3)) hat (k) Nm`C. `sqrt((3)/(2)) hat(k) Nm`D. `0 hat(k)` |
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Answer» Correct Answer - D `L = m lfloorvec(v) xx vec(V)rfloor` `V = -2omega sin omega that(i) + 3omega cos omega t hat(j)` `|(hat(i),hat(j),hat(k),|),(2 cos omega t,3 sin omega t,0,= hat(R) [6 omega cos^(2) omegat + 6omegat sin^(2) omegat]),(-2 omega sin omega t,3 omega cos omega t,0,=6omega hat(k) "cons" tan t)|` |
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