Question

a flywheel rotates about an axis . Due to friction at the Axis ,it experiences and angular retardation proportional to its angular velocity .if its angular velocity falls to half while it makes n rotations how many more rotations will it make before coming to rest ?

Answer 1

The flywheel will make n more rotations before coming to rest.

Explanation :

let a be the angular retardation, w be its angular velocity and Ф be the angular displacement.

From the given data,

- a = kw

(-ve sign because of retardation and k is the constant of proportionality)

=> dw/dt = - k dФ/dt

=> dw = -kdФ

=> w = -kФ + c

where c is the integration constant.

From the initial condition, Ф = 0 and w = w₀ , putting these in the above equation we get,

w₀ = c

=> w = -kФ + w₀

Now when it has made n rotations,

Ф = 2πn, and w = w₀/2

putting these values in the equation we get,

w₀/2 = -k2πn + w₀

=> k = w₀/4πn

Let say before stopping it has made N Rotations, so Ф = 2πN and w = 0 at this point

So putting these values in equation we get,

0 = -(w₀/4πn) 2πN + w₀

=> (w₀/4πn) 2πN = w₀

=> 2πN = 4πn

=> N = 2n

since it has already made n revolutions in the half time , it will make n more rotation afterwards.

Answer 2

Explanation:

ANSWER

α is propotional to ω

Let α=kω (∵k is a constant)

dt

dω

=kω[also

dt

dθ

=ω⇒dt=

ω

dθ

]

∴

dθ

ωdω

=kω⇒dω=kdθ

Now ∫

ω

ω/2

dω=k∫dθ

∫

ω/2

0

dω=k∫

0

θ

dθ⇒−

2

ω

=kθ⇒−

2

ω

=kθ

1

(∵θ

1

=2πn)

∴θ=θ

1

or 2πn

1

=2πn

n

1

=n