Question

A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy, what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?

Answer 1

Answer:

1

Step-by-step explanation:

Hello!

Remember that kinetic energy is one that is produced by the speed of a body that has mass. The equation is as follows.

$E=\frac{MV^2}{2}$

Where

E=kinetic energy

m=mass

V=speed

To solve this problem we must raise the equation for the kinetic energy of the sphere and the kinetic energy of the cylinder.

for cilinder

$Ec=\frac{McVc^2}{2}$

for spherical shell

$Es=\frac{MsVs^2}{2}$

as the problem indicates the kinetic energy is the same, so we can match the previous equations

$\frac{MsVs^2}{2}=\frac{McVc^2}{2}$

${MsVs^2}={McVc^2}$

according to the problem the masses are equal Ms = Mc

$Vs^2=Vc^2$

we apply square root on both sides of the equation

$Vs=Vc$

$\frac{Vs}{Vc} =1$

In conclusion, if two bodies, regardless of their shape, have the same mass and the same kinetic energy, the speed of the two bodies will be the same.