Flywheels are large, massive wheels used to store energy. They canbe spun up slo

Question

Flywheels are large, massive wheels used to store energy. They canbe spun up slowly, then the wheel's energy can be released quicklyto accomplish a task that demands high power. An industrialflywheel has a 1.2 m diameter and a massof 225 kg. Its maximum angular velocity is1150 rpm. (a) A motor spins up the flywheel with aconstant torque of 45 N · m. Howlong does it take the flywheel to reach top speed?
1 s

(b) How much energy is stored in the flywheel?
2 J

(c) The flywheel is disconnected from the motor and connected to amachine to which it will deliver energy. Half the energy stored inthe flywheel is delivered in 2 seconds. What is the average powerdelivered to the machine?
3 kW

(d) How much torque does the flywheel exert on the machine?
4 N · m
(a) A motor spins up the flywheel with aconstant torque of 45 N · m. Howlong does it take the flywheel to reach top speed?
1 s

(b) How much energy is stored in the flywheel?
2 J

(c) The flywheel is disconnected from the motor and connected to amachine to which it will deliver energy. Half the energy stored inthe flywheel is delivered in 2 seconds. What is the average powerdelivered to the machine?
3 kW

(d) How much torque does the flywheel exert on the machine?
4 N · m

Explanation / Answer

   Moment of inertia offlywheel      I      =   (1/2)* m * R2   =   0.5 * 225 *1.22   =   162.0   kg-m2    a.   Angularacceleration         =   torque/ I   =   45 /162.0   =   0.278   rad/s2          Initialangularspeed   0    =   0,   finalangularspeed      =   1150   rpm   =   1150* 2 /60   =   120.37   rad/s             =   0   +   * t       =>   time toattain maximum angularspeed   t   =   (120.37   -   0)/0.278   =   432.97   s    b.   Energystored   E   =   (1/2)* I * 2   =   0.5 *162.0 * 120.372   =   1.17* 106   J    c.   Powerdelivered   P   =   Energysupplied / time   =   ( 1.17 *106 / 2) / 2   =   2.93 *105   W          P   =   2.93* 102   kW    d.   Power   P   =   Torque* angular speed          2.93 *105   =    *120.37       torqueapplied      =   2.93* 105 / 120.37   =   2.44* 103   N-m          Initialangularspeed   0    =   0,   finalangularspeed      =   1150   rpm   =   1150* 2 /60   =   120.37   rad/s             =   0   +   * t       =>   time toattain maximum angularspeed   t   =   (120.37   -   0)/0.278   =   432.97   s    b.   Energystored   E   =   (1/2)* I * 2   =   0.5 *162.0 * 120.372   =   1.17* 106   J    c.   Powerdelivered   P   =   Energysupplied / time   =   ( 1.17 *106 / 2) / 2   =   2.93 *105   W          P   =   2.93* 102   kW    d.   Power   P   =   Torque* angular speed          2.93 *105   =    *120.37       torqueapplied      =   2.93* 105 / 120.37   =   2.44* 103   N-m

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.