Question

61. Consider a wheel rotating around a
fixed axis. If the rotating angle '@' varies
with time as 0 = at?, the total acceleration
of a point A on the rim of the wheel is
(v being the tangential velocity)
V
1) V1+4a’14 2)
t
3) (1+422“) 4) /(1+42°4)​

Answer 1

ANSWER

Here, angle is, φ=at

2

So, angular speed is, Ω=

dt

dφ

=2at=2×0.2t=0.4t

and, angular accelration is, α=

dt

dΩ

=2a=0.4 rad/s

2

So linear velocity is, v=ΩR⟹0.4tR=0.65

At t=2.5s,

0.4×2.5×R=0.65⟹R=0.65 m

So tangential acceleration is a

t

=αR=0.4×0.65=0.26 m/s

2

Radial acceleration is a

r

=v

2

/R=(0.65)

2

/0.65=0.65 m/s

2

Net acceleration is ω=

0.26

2

+0.65

2

=0.7

So, 10ω=7

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