Question

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.

Answer 1

`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)`