Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 2.0 m diameter and a mass of 260kg . Its maximum angular velocity is 1300rpm.
1) How much energy is store in the flywheel?
2) The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.0 s . What is the average power delivered to the machine?
3)How much torque does the flywheel exert on the machine?
1) How much energy is store in the flywheel?
2) The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.0 s . What is the average power delivered to the machine?
3)How much torque does the flywheel exert on the machine?
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Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 2.0 m diameter and a mass of 260kg . Its maximum angular velocity is 1300rpm.
1) How much energy is store in the flywheel?
The energy of an object moving at a constant velocity is its kinetic energy.
Rotational kinetic energy = ½ * Moment of Inertia * Angular velocity
The website below has formulas for moment of inertia of various shaped objects, and equations for rotational motion.
http://hyperphysics.phy-astr.gsu.edu/hba…
The flywheel is similar to a cylindrical disc.
Moment of inertia = ½ * mass * radius^2 = ½ * 260 * 1^2 = 130 kg*m^2
Angular velocity must be in radians per second. 1 rpm is one revolution per minute. 1 rpm is 2 * π radians per 60 seconds
1300 rpm = 2600 * π radians per 60 seconds
Angular velocity = 2600 * π ÷ 60 =
Rotational kinetic energy = ½ * 130 * (2600 * π ÷ 60)^2
2) The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.0 s. What is the average power delivered to the machine?
Half the Rotational kinetic energy = ¼ * 130 * (2600 * π ÷ 60)^2
Power = energy ÷ time
Power = [¼ * 130 * (2600 * π ÷ 60)^2] ÷ 2
Power = 16.25 * (2600 * π ÷ 60)^2
3)How much torque does the flywheel exert on the machine?
Torque = Moment of inertia * angular acceleration
Angular acceleration = (Final angular velocity – Initial angular velocity) ÷ time
Initial angular velocity = 2600 * π ÷ 60
Half the Rotational kinetic energy = ¼ * 130 * (2600 * π ÷ 60)^2
Determine the angular velocity for the flywheel with this much KE.
¼ * 130 * (2600 * π ÷ 60)^2 = ½ * 130 * ω^2
32.5 * (2600 * π ÷ 60)^2 = 65 * ω^2
Divide both sides by 65
½ * (2600 * π ÷ 60)^2 = ω^2
ω = ¼ * (2600 * π ÷ 60) = Final angular velocity
(Final angular velocity – Initial angular velocity) = ¼ * (2600 * π ÷ 60) – (2600 * π ÷ 60) = -¾ * (2600 * π ÷ 60)
Angular acceleration = -¾ * (2600 * π ÷ 60)
Torque = 130 * -¾ * (2600 * π ÷ 60)
1) How much energy is store in the flywheel?
The energy of an object moving at a constant velocity is its kinetic energy.
Rotational kinetic energy = ½ * Moment of Inertia * Angular velocity
The website below has formulas for moment of inertia of various shaped objects, and equations for rotational motion.
http://hyperphysics.phy-astr.gsu.edu/hba…
The flywheel is similar to a cylindrical disc.
Moment of inertia = ½ * mass * radius^2 = ½ * 260 * 1^2 = 130 kg*m^2
Angular velocity must be in radians per second. 1 rpm is one revolution per minute. 1 rpm is 2 * π radians per 60 seconds
1300 rpm = 2600 * π radians per 60 seconds
Angular velocity = 2600 * π ÷ 60 =
Rotational kinetic energy = ½ * 130 * (2600 * π ÷ 60)^2
2) The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.0 s. What is the average power delivered to the machine?
Half the Rotational kinetic energy = ¼ * 130 * (2600 * π ÷ 60)^2
Power = energy ÷ time
Power = [¼ * 130 * (2600 * π ÷ 60)^2] ÷ 2
Power = 16.25 * (2600 * π ÷ 60)^2
3)How much torque does the flywheel exert on the machine?
Torque = Moment of inertia * angular acceleration
Angular acceleration = (Final angular velocity – Initial angular velocity) ÷ time
Initial angular velocity = 2600 * π ÷ 60
Half the Rotational kinetic energy = ¼ * 130 * (2600 * π ÷ 60)^2
Determine the angular velocity for the flywheel with this much KE.
¼ * 130 * (2600 * π ÷ 60)^2 = ½ * 130 * ω^2
32.5 * (2600 * π ÷ 60)^2 = 65 * ω^2
Divide both sides by 65
½ * (2600 * π ÷ 60)^2 = ω^2
ω = ¼ * (2600 * π ÷ 60) = Final angular velocity
(Final angular velocity – Initial angular velocity) = ¼ * (2600 * π ÷ 60) – (2600 * π ÷ 60) = -¾ * (2600 * π ÷ 60)
Angular acceleration = -¾ * (2600 * π ÷ 60)
Torque = 130 * -¾ * (2600 * π ÷ 60)