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a hoop is placed on the rough surface such that it has an angular velocity `omega=4rad//s` and an angular deceleration `alpha=5rad//s^(2)` also its centre has a velocity of `v_(0)=5m//s` and a decoleration `a_(0)=2m//s^(2)` determine the magnitude of acceleration of point B at this instant. |
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Answer» `a_(B)=a_(0)+a_(B//0)` here `a_(B//0)` has two components `a_(t)` (tangential acceleration) and `a_(n)` (normal acceleration) `a_(t)=ralpha=(0.3)(5)=1.5m//s^(2)` `a_(n)=romega^(2)=(0.3)(4)^(2)=4.8m//s^(2)` and `a_(0)=2m//s^(2)` `thereforea_(B)=sqrt((suma_(x))^(2)+(suma_(y))^(2))` sqrt((2+4.8cos45^(@)-1.5cos45^(@))^(2)+(4.8sin45^(@)+1.5sin45^(@))^(2))` `=6.21m//s^(2)` |
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