Respuesta :
Answer:
A)The rotational kinetic energy is ≈ 9.9962 J
B)The number of revolutions to achieve that angular velocity = 0.0085
C)Force to stop it = 0.15152N
Explanation:
Given :
- The system has a moment of inertia of [tex]84.4kgm^{2}[/tex]
- Radius [tex]R = 1.50m[/tex]
- Torque exerted = [tex]375Nm[/tex]
A)
Rotational kinetic energy :
[tex]E = \frac{1}{2} Iw^{2}[/tex]
where,
- [tex]E[/tex] is the rotational kinetic energy [tex](J)[/tex]
- [tex]I[/tex] is the moment of inertia [tex](Kgm^{2})[/tex]
- [tex]w[/tex] is the angular velocity [tex](rads^{-1})[/tex]
⇒[tex]I=84.4\\w=29.2rpm=\frac{29.2}{60} rps=0.4867 rads^{-1}[/tex]
⇒[tex]E = \frac{1}{2} Iw^{2}\\E=\frac{1}{2} *84.4*(0.4867)^{2}\\E=9.9962 J[/tex]
Thus the energy is ≈ 9.9962 J
B) Work done by the torque = The change in rotational kinetic energy
[tex]T\alpha = \frac{1}{2} Iw^{2} - \frac{1}{2} Iw_{o}^{2}[/tex]
where ,
- [tex]T[/tex] is the torque exerted
- [tex]\alpha[/tex] is the angle traversed
- [tex]w_o[/tex] is the initial angular velocity.
Thus,
[tex]375*\alpha =\frac{1}{2} *84.4*(0.4867)^{2}-\frac{1}{2} *84.4*0\\\alpha =0.026657 rad\\=\frac{0.026657}{\pi } rev\\=0.0085 rev[/tex]
The number of revolutions = 0.0085
C)
7 revolutions :
[tex]=7*2\pirad\\=14\pirad[/tex]
[tex]T\alpha = \frac{1}{2} Iw^{2} - \frac{1}{2} Iw_{o}^{2}[/tex]
[tex]T*14\pi =9.9962\\T= 0.2273 Nm\\F=\frac{T}{R} =\frac{0.2273}{1.5} \\F=0.15152N[/tex]
Force = 0.15152N
