A thin-walled sphere rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy through its center of mass?
RE=1/2 *I w^2=.5*v^2/r^2
But I=2/5 mr^2
so TE/Re= (1/2 mv^2)/(Mr^2*v^2/5)=2.5
check
To find the ratio of translational kinetic energy to rotational kinetic energy for a thin-walled sphere rolling along the floor, we need to consider the formulae for both types of kinetic energy.
The translational kinetic energy for an object is given by the formula:
Kt = (1/2) * m * v^2
where Kt represents translational kinetic energy, m represents the mass of the object, and v represents the velocity of the object.
The rotational kinetic energy for an object rotating about its center of mass is given by the formula:
Kr = (1/2) * I * ω^2
where Kr represents rotational kinetic energy, I represents the moment of inertia of the object, and ω represents the angular velocity of the object.
In the case of a thin-walled sphere rolling along the floor, the translational and rotational motions are related. The rotational motion of the sphere is due to its rolling motion, and the translational motion is the linear motion of the center of mass.
For a thin-walled sphere, the moment of inertia can be given as:
I = (2/5) * m * r^2
where r represents the radius of the sphere.
Since the sphere is rolling without slipping, the linear velocity is related to the angular velocity by the equation:
v = ω * r
Substituting the expression for rotational kinetic energy (Kr) and translational kinetic energy (Kt) into the ratio, we get:
Kt / Kr = [(1/2) * m * v^2] / [(1/2) * I * ω^2]
= (m * v^2) / (I * ω^2)
Substituting the expressions for moment of inertia (I) and linear velocity (v) in terms of the sphere's properties, we have:
Kt / Kr = (m * v^2) / (I * ω^2)
= (m * v^2) / [(2/5) * m * r^2 * ω^2]
= (5/2) * (v^2 / (r^2 * ω^2))
= (5/2)
Therefore, the ratio of translational kinetic energy to rotational kinetic energy for a thin-walled sphere rolling along the floor through its center of mass is 5/2 or 2.5.