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Question

A hoop is placed on the rough surface such that it has an angular velocity ω=4 rad/s and an angular deceleration α=5rad/s2 ,Also, its centre has a velocity of V0=5m/s and a deceleration a0=2m/s2. Determine the magnitude of acceleration of point B at this instant.
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Solution

aB=a0+aB/0
Here,aB/0 has two components at (tangential acceleration) and an (normal acceleration)
at=rα=(0.3)(5)=1.5m/s2
an=rω2=(0.3)(4)2=4.8m/s2
and a0=2m/s2
aB=(ax)2+(ay)2
=(2+4.8cos451.5cos45)2+(4.8sin45+1.5sin45)2
=6.21m/s2

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