An electric motor's rotor makes 10 rotations around the centre per second. The rotor weighs 0.05kg and has a radius of 1cm. If the electric motor is hooked up to a 1.5V DC energy source, then how much current flows through the motor?
Circumference = 2πr
C = 2 (3.14)(1cm)
C = 6.28 centimetres
Displacement = Circumference x 10
d = 6.28cm x 10
d = 62.8cm
Velocity x Time = Displacement
V x 1s = 62.8cm
V = 62.8cm/s = 0.628m/s
Kinetic Energy = 1/2 (Mass) (Velocity)^2
Ek = 1/2(0.05kg)(0.628m/s)^2
Ek = 1/2(0.05kg)(0.394384m/s)
Ek = 0.0098596 Joules
Joules = Coulomb x Volt
0.0098596J = C x 1.5V
0.0098596J/1.5V = A x 1s
0.00657306666 = Amperes
7 milliamperes flow through the motor.
Circumference = 2πr
C = 2 (3.14)(1cm)
C = 6.28 centimetres
Displacement = Circumference x 10
d = 6.28cm x 10
d = 62.8cm
Velocity x Time = Displacement
V x 1s = 62.8cm
V = 62.8cm/s = 0.628m/s
Kinetic Energy = 1/2 (Mass) (Velocity)^2
Ek = 1/2(0.05kg)(0.628m/s)^2
Ek = 1/2(0.05kg)(0.394384m/s)
Ek = 0.0098596 Joules
Joules = Coulomb x Volt
0.0098596J = C x 1.5V
0.0098596J/1.5V = A x 1s
0.00657306666 = Amperes
7 milliamperes flow through the motor.