A solid uniform cylinder of mass 4.6 kg and radius 0.051 m rolls without slipping at a speed of 0.78 m/s. What is the cylinder’s total kinetic energy? please show some work so that I can learn from it.
To find the total kinetic energy of the rolling cylinder, we need to consider both its translational kinetic energy (due to its linear motion) and its rotational kinetic energy (due to its rotational motion). The formula for the translational kinetic energy is given by:
KE(translational) = (1/2) * mass * velocity^2
where mass is the mass of the cylinder and velocity is its linear velocity.
In this case, the mass of the cylinder is given as 4.6 kg, and the linear velocity is given as 0.78 m/s. Plugging these values into the formula, we get:
KE(translational) = (1/2) * 4.6 kg * (0.78 m/s)^2 = 0.89304 J (rounded to 5 decimal places)
Now, let's calculate the rotational kinetic energy. The formula for rotational kinetic energy is given by:
KE(rotational) = (1/2) * moment of inertia * angular velocity^2
where moment of inertia is a property that depends on the mass distribution of the object and angular velocity is its rotational velocity.
For a solid cylinder rolling without slipping, the moment of inertia is given by:
moment of inertia = (1/2) * mass * radius^2
Here, the mass of the cylinder is given as 4.6 kg, and the radius is given as 0.051 m. Plugging these values into the formula, we get:
moment of inertia = (1/2) * 4.6 kg * (0.051 m)^2 = 0.0058581 kg⋅m^2 (rounded to 7 decimal places)
Since the cylinder is rolling without slipping, its angular velocity is related to its linear velocity by the equation:
angular velocity = velocity / radius
Plugging in the given values, we have:
angular velocity = 0.78 m/s / 0.051 m = 15.2941 rad/s (rounded to 4 decimal places)
Now we can calculate the rotational kinetic energy:
KE(rotational) = (1/2) * 0.0058581 kg⋅m^2 * (15.2941 rad/s)^2 = 0.84070 J (rounded to 5 decimal places)
Finally, to find the total kinetic energy of the rolling cylinder, we add the translational and rotational kinetic energies:
Total kinetic energy = KE(translational) + KE(rotational)
= 0.89304 J + 0.84070 J
= 1.73374 J (rounded to 5 decimal places)
Therefore, the total kinetic energy of the cylinder is approximately 1.73374 J.