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 ?
Answers
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.
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