A pulley of radius 2m is rotates about its axis by s force F = (20t - st²) N ( where , t is measured in seconds ) applied tangentially . if the moment of inertia of the pulley about its axis of rotation is 10kg-m² , then the number of rotation made by the pulley before its direction of motion is reserved is ? [AIEEE 2011]
Answers
Answered by
2
Answer:
to reverse the direction
∫τdθ=0
τ=(20t−5t
2
)×2=40t−10t
2
α=
I
τ
=
10
40t−10t
2
=4t−t
2
ω=∫
0
t
αdt=2t
2
−
3
t
3
at ω=0⟹2t
2
−
3
t
3
=0⟹t
3
=6t
2
⟹t=6
θ=∫
0
6
ωdt=∫
0
6
(2t
2
−
3
t
3
)dt=[
3
2t
3
−
12
t
4
]
0
6
=36rad=
2π
36
<6
Explanation:
Answered by
3
Explanation:- Given force , F = 20t -5t² , R = 2m inertia of the pulley about its axis
- α = FR/I = (20t- 5t²) 2/10 = 4t-t²
- ⟹ dω/dt = 4t - t²
- ⟹ ₀∫ʷdω = ₀∫ᵗ(4t-t²)dt
- ⟹ ω = 2t²- t³/3
- ⟹ 0= 6t²-t³
- ⟹ t = 0,6
When direction is reserved , ω = 0 .,t = 0 ,6s
Now, dθ = ωdt
⟹ ₀∫ᵗʰᵉᵗᵃdθ
⟹ ₀∫⁶(2t²- t³/3)dt
⟹ θ = [2t³/3- t²/12]₀⁶
⟹θ = 144 - 108 = 36 rad
⟹∴ number of rotation ,n = θ/2π = 36/2π <6(∴π =3.14)
since answer ⟹number of rotation more than 3 but less then 6
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