A roller in a sheet-metal processing plant has a diameter of 1.77 m and a mass of 5871 kg. Assume the roller is a solid cylinder.
a) What is the moment of inertia of the roller?
b) What is the kinetic energy of the roller when it spins at 4.28 rad/s?
Please provide all work and equations. Please include the actual answer and the steps that were taken to come up wih that answer. The equation with numbers filled in does not completely help me. Thanks.
a) I=mR²/2
http://cnx.org/content/m14292/latest/
b) KE= Iω²/2
http://www.dummies.com/how-to/content/how-to-calculate-rotational-kinetic-energy.html
To find the moment of inertia of the roller, we can use the formula for the moment of inertia of a solid cylinder:
I = (1/2) * m * r^2
where I is the moment of inertia, m is the mass of the roller, and r is the radius of the roller.
a) Substitute the given values into the formula:
I = (1/2) * 5871 kg * (1.77 m/2)^2
I = (1/2) * 5871 kg * (0.885 m)^2
I = (1/2) * 5871 kg * 0.782 m^2
I = 2271.339 kg * m^2
Therefore, the moment of inertia of the roller is 2271.339 kg * m^2.
To find the kinetic energy of the roller, we can use the formula for the kinetic energy of a rotating object:
KE = (1/2) * I * w^2
where KE is the kinetic energy, I is the moment of inertia of the roller, and w is the angular velocity of the roller.
b) Substitute the given values into the formula:
KE = (1/2) * (2271.339 kg * m^2) * (4.28 rad/s)^2
KE = (1/2) * (2271.339 kg * m^2) * 18.3584 rad^2/s^2
KE = 20643.348 kg * m^2 * rad^2/s^2
Therefore, the kinetic energy of the roller when it spins at 4.28 rad/s is 20643.348 kg * m^2 * rad^2/s^2.
To find the moment of inertia of the roller, you can use the formula for the moment of inertia of a solid cylinder:
I = (1/2) * m * r^2
where I is the moment of inertia, m is the mass of the roller, and r is the radius of the roller.
In this case, we are given the diameter of the roller, so we need to calculate the radius first. The radius can be calculated by dividing the diameter by 2:
r = diameter / 2
= 1.77 m / 2
= 0.885 m
Now we can plug in the values into the formula:
I = (1/2) * m * r^2
= (1/2) * 5871 kg * (0.885 m)^2
= 1157.69 kg * m^2
Therefore, the moment of inertia of the roller is 1157.69 kg * m^2.
To find the kinetic energy of the roller, we need to use the formula for rotational kinetic energy:
KE = (1/2) * I * ω^2
where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
In this case, we are given the angular velocity, so we can directly plug in the values:
KE = (1/2) * (1157.69 kg * m^2) * (4.28 rad/s)^2
= (1/2) * (1157.69) * (18.3184)
= 10583.215 J
Therefore, the kinetic energy of the roller when it spins at 4.28 rad/s is 10583.215 Joules.