Question

A solid sphere of radius R, a solid cylinder of radius R, and a hollow cylinder of radius R all have the same mass, and all three are rotating with the same angular velocity. The sphere is rotating around an axis through its center, and each cylinder is rotating around its symmetry axis. Which one has the greatest rotational kinetic energy?

(A) both cylinders have the same rotational kinetic energy
(B) the solid cylinder
(C) the solid sphere
(D) they all have the same rotational kinetic energy
(E) the hollow cylinder

Answer 1

Answer: A)

Explanation: In order to explain this problem we have to consider the kinetic energy of rotation  for rigid body which is given by:

Ekr=1/2*I*ω^2 where I and w are the inertia mometum and angular velocity, respectively.

Also we know that the radius and mass are the same for all rigid bodies. The inertia momentum for the  hollow and solid cylinder  around its symmetry axis, is equal to:

1/2 m*R^2.

The inertia momentum for the sphere, around an axis through its center, is: 2/5*m*R^2 ;

then the greatest rotational kinetic energy is for the cylinders.

Answer 2

The rotational kinetic energy of the solid cylinder is greater than the solid sphere.

What is rotational kinetic energy?

Rotational kinetic energy is a type of energy, which a body is gains due to the rotational motion. The rotational kinetic energy of a body can be found with the following formula,

$KE_{rotational}=\dfrac{1}{2}I\omega^2$

Here, (I) is the moment of inertia around the axis of rotation, and (ω) is the angular speed of the body.

A solid sphere of radius R, a solid cylinder of radius R, and a hollow cylinder of radius R all have the same mass, and all three are rotating with the same angular velocity.

The sphere is rotating around an axis through its center, and each cylinder is rotating around its symmetry axis.  Kinetic energy of the cylinder is given by,

$KE_{cylinder}=\dfrac{1}{2}I_{cylinder}\omega_{cylinder}^2\\KE_{cylinder}=\dfrac{1}{2}\times \dfrac{1}{2}mR^2\omega_{cylinder}^2\\KE_{cylinder}=\dfrac{1}{4}mR^2\omega_{cylinder}^2$

The kinetic energy of the sphere is given by,

$KE_{sphere}=\dfrac{1}{2}I_{sphere}\omega_{sphere}^2\\KE_{sphere}=\dfrac{1}{2}\times \dfrac{2}{5}mR^2\omega_{sphere}^2\\KE_{sphere}=\dfrac{1}{5}mR^2\omega_{sphere}^2$

One fourth is greater then, one fifth. Thus, the rotational kinetic energy of the solid cylinder is greater than the solid sphere.

Learn more about the kinetic energy here;

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