AP EAPCET 07-Jul-2022 Shift 2 Paper

Given, ω=314‌rad∕ s at t=2 s
Let α= constant angular acceleration Then using ω1=ω0+αt We have,
ω1=αt ...[∵ω0=0]
∴314=α(2)
⇒α=

314

2

‌rad∕ s2
Now, angle covered in 20 seconds is given by,
θ1=ω0t+

1

2

αt2
⇒θ1=

1

2

αt2
So, θ1=

1

2

×

314

2

×(20)2=3.14‌rad
At t=20 s,angular‌speed‌of‌wheel,
⇒ω2=ω0+αt
⇒ω2=0+

314

2

×20
As, acceleration does not operates after t=20 s.
So, wheel now rotates freely (with constant angular speed) upto 40 s.
Angular displacement in radians covered in another 20s of freewheeling is
θ2‌=ω1×t
=314×20=628‌rad
Total angle covered by wheel, θ=(314+628)‌rad =942‌rad
∴ Number of revolutions of wheel
n=

θ

2π

=

942

2π

=

314×3

2×314

=150