A cylinder ( I = 1/2 MR2), a sphere ( I = 2/5 MR2), and a hoop ( I = MR2) roll without slipping down the same incline, beginning from rest at the same height. All three objects share the same radius and total mass. Which object has the greatest translational kinetic energy at the bottom of the incline?
To determine which object has the greatest translational kinetic energy at the bottom of the incline, we can compare the expressions for the moment of inertia and use the conservation of energy principle.
Let's compare the expressions for the moment of inertia for the cylinder, sphere, and hoop:
1. Cylinder: I = 1/2 MR^2
2. Sphere: I = 2/5 MR^2
3. Hoop: I = MR^2
Since all three objects have the same radius and total mass, we can ignore those factors in our comparison.
Now, let's consider the conservation of energy principle. At the top of the incline, the objects are at rest, so their initial kinetic energy is zero. The only form of energy they possess is gravitational potential energy.
As the objects roll down the incline without slipping, their gravitational potential energy is converted into translational kinetic energy and rotational kinetic energy.
The total kinetic energy of a rolling object is the sum of its translational kinetic energy and rotational kinetic energy.
The expression for the total kinetic energy can be written as:
KE(total) = KE(translational) + KE(rotational)
Since we are comparing the translational kinetic energy, let's check how the rotational kinetic energy differs for each object.
The rotational kinetic energy can be expressed as:
KE(rotational) = 1/2 I ω^2
where ω is the angular velocity.
Now, let's compare the rotational kinetic energy for each object:
1. Cylinder: KE(rotational) = 1/2 (1/2 MR^2) ω^2 = 1/4 MR^2 ω^2
2. Sphere: KE(rotational) = 1/2 (2/5 MR^2) ω^2 = 1/5 MR^2 ω^2
3. Hoop: KE(rotational) = 1/2 (MR^2) ω^2 = 1/2 MR^2 ω^2
We can see that the rotational kinetic energy for the cylinder is the smallest, followed by the sphere, and the hoop has the highest rotational kinetic energy.
Since the total kinetic energy is the sum of translational and rotational kinetic energy, and the rotational kinetic energy is subtracted from the total kinetic energy, the object with the least rotational kinetic energy will have the greatest translational kinetic energy.
Therefore, the cylinder will have the greatest translational kinetic energy at the bottom of the incline.
To determine which object has the greatest translational kinetic energy at the bottom of the incline, we need to compare the expressions for the kinetic energy of each object.
The kinetic energy of an object can be calculated using the formula:
KE = (1/2) mv^2
Where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.
For a rolling object without slipping, the velocity of the object is related to its rotational motion. The relationship between the velocity and angular velocity for different objects is given by:
v = ωr
Where v is the linear velocity, ω is the angular velocity, and r is the radius of the object.
Now, let's calculate the translational kinetic energy for each object.
For the cylinder:
The moment of inertia for a cylinder about its axis of rotation is given by I = (1/2) MR^2.
Using the relationship between velocity and angular velocity, we have v = ωr. Rearranging, we get ω = v/r.
Substituting the moment of inertia expression into the kinetic energy formula, we have:
KE_cylinder = (1/2) mv^2 = (1/2) m(ωr)^2 = (1/2) m(v^2/r^2) r^2 = (1/2) mv^2
So, the translational kinetic energy for the cylinder is (1/2) mv^2.
For the sphere:
The moment of inertia for a sphere about its axis of rotation is given by I = (2/5) MR^2.
Using the same approach as above, we substitute the moment of inertia expression:
KE_sphere = (1/2) mv^2 = (1/2) m(ωr)^2 = (1/2) m(v^2/r^2) r^2 = (1/2) mv^2
So, the translational kinetic energy for the sphere is also (1/2) mv^2.
For the hoop:
The moment of inertia for a hoop about its axis of rotation is given by I = MR^2.
Again, using the same approach, we substitute the moment of inertia expression:
KE_hoop = (1/2) mv^2 = (1/2) m(ωr)^2 = (1/2) m(v^2/r^2) r^2 = (1/2) mv^2
So, the translational kinetic energy for the hoop is also (1/2) mv^2.
From the calculations above, we can see that the translational kinetic energy is the same for all three objects: (1/2) mv^2.
Therefore, the cylinder, sphere, and hoop have the same translational kinetic energy at the bottom of the incline, assuming they have the same radius, mass, and starting height.