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The torque vectau on a body about a given point is found to be equal to vecA and vecL where vecA is a constant vector, and vecL is the angular momentum of the body about that point. From this it follows that |
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Answer» `(dvecL)/(dt)` is perpendicular to `vecL` at all instants of time `vectau=(dvecL)/(dt)therefore(dvecL)/(dt)=vecAxxvecL` By the rule of cross-product of VECTORS, `(dvecL)/(dt)` is always perpendicular to the plane containing `VECA` and `vecL`. Hence option (a) is correct (b) For vector `vecL`, the magnitude of L is constant but `vecL` is not constant. It changes. `therefore` Let `vecL = (a cos theta)hati + (a sin theta)hatj` where a is a constant. Differentiate it to obtain `vectau`. `therforevectau=-(asintheta)hati+(acostheta)hatj` `vecL.vectau=-a^(2)sinthetacostheta+a^(2)sinthetacostheta` or `vecL.vectau=0` or `vecL` is perpendicular to `vectau` `vecA` is a constant vector and is always `bot` to `vectau` Let `vecA=Ahatk` `vecL.vecA=[(acostheta)hati+(asintheta)hatj].[Ahatk]` or `vecL.vecA=0` `therforevecL` is perpendiclar to `vecA`. `therefore` Component of `vecL` along `vecA` is ZERO. ( `because Lcos90° = 0`) `therefore` Component of `vecL` along `vecA` does not change with time. Hence option (b) is correct. By the rule of dot product of vectors, `vecL.vecL = L^(2)` Differentiate it w.r.t. time. `thereforevecL.(dvecL)/(dt)+(dvecL)/(dt).vecL=2L.(dL)/(dt)` where `L=absvecL` or `2vecL.(dvecL)/(dt)=2L(dvecL)/(dt)` Since `vecL` is perpendicular to `(dvecL)/(dt)` their dot product is zero. `therefore0=2L(dL)/(dt)` or `2L(dL)/(dt)=0` This is possible if L is a constant. `therefore` Magnitude of `vecL` = constant or Magnitude of `vecL` does not change with time. Hence option ( c) is correct. ( d) `vecL` changes with time on account of change in its direction. Magnitude of does not change with time, as shown at option (c). Hence options (a), (b) and (c) are correct. Option (d) is not correct. |
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