A solid cylinder of mass 20 kg rotate about its axis of rotation with angular speed 100m/s. The radius of the cylinder is 0.25m. What is the K.E. associated with the rotation of the cylinder what is the magnitude of the angular momentum of the cylinder about its axis
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To calculate the kinetic energy (K.E.) associated with the rotation of the cylinder, use the formula:
K.E. = (1/2) * I * ω^2
where K.E. is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid cylinder rotating about its axis can be calculated using the formula:
I = (1/2) * m * r^2
where m is the mass and r is the radius of the cylinder.
Substituting the given values:
m = 20 kg (mass of the cylinder)
r = 0.25 m (radius of the cylinder)
ω = 100 m/s (angular speed)
Calculating the moment of inertia:
I = (1/2) * 20 kg * (0.25 m)^2
I = 0.625 kg * m^2
Now, calculating the kinetic energy:
K.E. = (1/2) * 0.625 kg * m^2 * (100 m/s)^2
K.E. = 3125 J
Therefore, the kinetic energy associated with the rotation of the cylinder is 3125 Joules.
To calculate the magnitude of the angular momentum of the cylinder about its axis, use the formula:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Substituting the given values:
I = 0.625 kg * m^2 (moment of inertia)
ω = 100 m/s (angular speed)
Calculating the angular momentum:
L = 0.625 kg * m^2 * 100 m/s
L = 62.5 kg * m^2/s
Therefore, the magnitude of the angular momentum of the cylinder about its axis is 62.5 kilogram-meter squared per second (kg * m^2/s).