9.4 Rotational kinetic energy: work and energy revisited  (Page 5/9)

Calculating the speed of a cylinder rolling down an incline

Calculate the final speed of a solid cylinder that rolls down a 2.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.

Strategy

We can solve for the final velocity using conservation of energy, but we must first express rotational quantities in terms of translational quantities to end up with $v$ as the only unknown.

Solution

Conservation of energy for this situation is written as described above:

Before we can solve for $v$ , we must get an expression for $I$ from [link] . Because $v$ and $\omega$ are related (note here that the cylinder is rolling without slipping), we must also substitute the relationship $\omega =v/R$ into the expression. These substitutions yield

Interestingly, the cylinder’s radius $R$ and mass $m$ cancel, yielding

Solving algebraically, the equation for the final velocity $v$ gives

Substituting known values into the resulting expression yields

Discussion

Because $m$ and $R$ cancel, the result $v={\left(\frac{4}{3}\text{gh}\right)}^{1/2}$ is valid for any solid cylinder, implying that all solid cylinders will roll down an incline at the same rate independent of their masses and sizes. (Rolling cylinders down inclines is what Galileo actually did to show that objects fall at the same rate independent of mass.) Note that if the cylinder slid without friction down the incline without rolling, then the entire gravitational potential energy would go into translational kinetic energy. Thus, $\frac{1}{2}{\text{mv}}^2=\text{mgh}$ and $v=(2\text{gh}{)}^{1/2}$ , which is 22% greater than $(4\text{gh}/3{)}^{1/2}$ . That is, the cylinder would go faster at the bottom.