AP EAPCET 23-Aug-2021 Shift 1 Paper

Given, length of rod AB,l=1m
Displacement of centre of mass, s=5√2m
Time taken, t=1s
Acceleration due to gravity, g=10ms−2 Let vertical displacement in 1s=y
Angular velocity =ω
Moment of inertia =I
I (about centre of rod) =

ml2

12

∴y=ut+

1

2

gt2
∴y=

1

2

×10×12=5m
and =√x2+y2
∴(5√2)2=x2+y2
⇒(5√2)2=x2+52
Displacement along X-axis in 1s
x2=50−25
⇒x=√25=5m
∴vx=

x

t

=5m∕s
Since,
angular impulse = change in angular momentum
J.

l

2

=lω
⇒

mvl

2

=

ml2

12

ω
⇒ω=6v=6×5=30‌rad‌s−1