Respuesta :
Answer:
The value of torque on the rod is = 3.94 N - m
Explanation:
mass = 50 kg
Radius = 0.5 m
Moment of inertia ( I ) = [tex]\frac{MR^{2} }{2}[/tex]
[tex]I = 50 \frac{0.5^{2} }{2}[/tex]
I = 6.25 kg- [tex]m^{2}[/tex]
Angular velocity [tex]\omega[/tex] = 120 [tex]\frac{Rev}{min}[/tex] = 12.6 [tex]\frac{rad}{sec}[/tex]
Δ[tex]\theta[/tex] = No. of rev × 2[tex]\pi[/tex]
Δ[tex]\theta[/tex] = 20 × 2[tex]\pi[/tex]
Δ[tex]\theta[/tex] = 126 rad
From work energy theorem
Work done is equal to change in kinetic energy.
[tex]W_{AB} = K_{B} - K_{A}[/tex] -------- (1)
Where A & B represents the initial & final states.
Since [tex]K_{A}[/tex] = 0
[tex]W_{AB} = K_{B}[/tex]
[tex]W_{AB} = \frac{1}{2} I \omega^{2}[/tex]
[tex]W_{AB} = \frac{1}{2} 6.25 (12.6)^{2}[/tex]
[tex]W_{AB} =[/tex] 496 J
We know that
[tex]W_{AB} =[/tex] T Δ[tex]\theta[/tex]
Where [tex]W_{AB} =[/tex] work done
T = torque applied on the rod
Δ[tex]\theta[/tex] = Angular displacement of the rod.
496 = T × 126
T = 3.94 N - m
Therefore the value of torque on the rod is = 3.94 N - m
