A 10.0 kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 16.0 m/s, determine (a) the translational kinetic energy of its center of gravity.(b) the rotational kinetic energy about its center of gravity, and (c) its total kinetic energy.
KE(tr) =m•v²/2,
KE(rot) = I•ω²/2 =m•R²•v²/2•2• R²=m•v²/4.
KE= KE(tr)+ KE(rot)=
=m•v²/2 + m•v²/4= 3• m•v²/4.
To solve this problem, we need to understand the concept of translational and rotational kinetic energy.
(a) The translational kinetic energy of an object is given by the formula:
KE_translational = (1/2) * m * v^2
where:
- KE_translational is the translational kinetic energy
- m is the mass of the object
- v is the speed of the object
In this problem, the mass of the cylinder is given as 10.0 kg and its center of gravity has a speed of 16.0 m/s. Plugging these values into the formula, we get:
KE_translational = (1/2) * 10.0 kg * (16.0 m/s)^2
= 1280.0 Joules
Therefore, the translational kinetic energy of the center of gravity of the cylinder is 1280.0 Joules.
(b) The rotational kinetic energy about the center of gravity of an object is given by the formula:
KE_rotational = (1/2) * I * ω^2
where:
- KE_rotational is the rotational kinetic energy
- I is the moment of inertia of the object about its center of gravity
- ω is the angular velocity of the object
For a cylinder rolling without slipping, the moment of inertia about its center of gravity is given by:
I = (1/2) * m * r^2
where:
- r is the radius of the cylinder
In this problem, the mass of the cylinder is 10.0 kg, and we assume it is a solid cylinder. The moment of inertia can be calculated as:
I = (1/2) * 10.0 kg * (r^2)
The radius of the cylinder is not given in the problem statement, so you would need to know or calculate this value. Once you have the value of the radius, you can substitute it into the equation to find the moment of inertia.
(c) Once you have calculated both the translational and rotational kinetic energies, you can find the total kinetic energy by summing them up:
Total KE = KE_translational + KE_rotational
Substitute the values you calculated in parts (a) and (b) to find the total kinetic energy.
To solve this problem, we need to use equations that relate translational and rotational motion.
Given:
Mass of the cylinder, m = 10.0 kg
Speed of the center of gravity, v = 16.0 m/s
(a) To find the translational kinetic energy of its center of gravity, use the formula:
Translational Kinetic Energy = 0.5 * m * v^2
Substituting the values, we get:
Translational Kinetic Energy = 0.5 * 10.0 kg * (16.0 m/s)^2
Translational Kinetic Energy = 0.5 * 10.0 kg * 256.0 m^2/s^2
Translational Kinetic Energy = 1280.0 Joules
Therefore, the translational kinetic energy of its center of gravity is 1280.0 Joules.
(b) To find the rotational kinetic energy about its center of gravity, use the formula:
Rotational Kinetic Energy = 0.5 * I * ω^2
Where I is the moment of inertia and ω is the angular velocity.
For a cylinder rolling without slipping, the moment of inertia about its center of gravity is given by:
I = 0.5 * m * r^2
Where r is the radius of the cylinder.
Since the cylinder rolls without slipping, its angular velocity ω can be calculated using the formula:
ω = v / r
Substituting these values, we can calculate the rotational kinetic energy.
Let's assume the radius of the cylinder is r.
Rotational Kinetic Energy = 0.5 * (0.5 * m * r^2) * (v / r)^2
Rotational Kinetic Energy = 0.5 * 0.5 * 10.0 kg * r^2 * (16.0 m/s / r)^2
Rotational Kinetic Energy = 0.5 * 0.5 * 10.0 kg * 256.0 (m^2/s^2)
Rotational Kinetic Energy = 640.0 Joules
Therefore, the rotational kinetic energy about its center of gravity is 640.0 Joules.
(c) The total kinetic energy is the sum of the translational and rotational kinetic energies.
Total Kinetic Energy = Translational Kinetic Energy + Rotational Kinetic Energy
Total Kinetic Energy = 1280.0 Joules + 640.0 Joules
Total Kinetic Energy = 1920.0 Joules
Therefore, the total kinetic energy is 1920.0 Joules.