Question

A cylinder of radius 2.09 cm and a spherical shell of radius 6.72 cm are rolling without slipping along the same floor. The two objects have the same mass. If they are to have the same total kinetic energy, what should the ratio of the cylinder's angular speed to the spherical shell's angular speed be?

wcyl/wshl

Answer 1

Here,

for wcycl and wshl

for the same kinetic energy

total kinetic energy is same

0.50 * Icycl * wcycl^2 + 0.50 m * vcycl^2 = 0.50 * Ishl * wshl^2 + 0.50 m * vshl^2

as v = r * w

putting the values for I and w

0.50 * 0.50 * m * 2.09^2 * wcycl^2 + 0.50 * m * (2.09 * wcycl)^2 = 0.50 * 2/3 * m * 6.72^2 * wsph^2 + 0.50 * m * (6.72 * wsph)^2

= 0.50 * 0.50 * 2.09^2 * wcycl^2 + 0.50 * (2.09 * wcycl)^2 = 0.50 * 2/3 * 6.72^2 * wsph^2 + 0.50 * (6.72 * wsph)^2

solving

wcyl/wshl = 3.39

the ratio of angular speed is 3.39