Question

A hollow cylinder (hoop) is rolling on a horizontal surface at speed v = 3.0 m/s when it reaches a 15° incline. (a) How far up the incline will it go? (b) How long will it be on the incline before it arrives back at the bottom?

Short answer

(a) The hollow cylinder will go up the incline to a distance of 3.5 m.

(b) The total time for which the cylinder will be on the incline before it returns to the bottom is 4.7 s.

Step-by-step solution

Relation between linear and angular velocity

If an object is rotating about a fixed axis, the linear velocity (v) of any point on the object located at a distance r from the axis of rotation is related to the angular velocity $\left(\omega \right)$of the object by the following relation:

$v=r\omega$

In this problem, the linear velocity of the rolling cylinder is related to its angular velocity by the above relation.

Given information

The linear speed of the hollow cylinder at the bottom of the incline is$v_0=3.0\mathrm{m}/\mathrm{s}$.

The angle of inclination of the incline is$\theta ={15}^{\circ }$.

The linear speed of the hollow cylinder at the top of the incline up to the point it travels will be$v=0\mathrm{m}/\mathrm{s}$.

Let m and r be the mass and radius of the rolling cylinder. Assume that the cylinder rolls without slipping.

(a) Calculation of the distance covered by the hoop on the incline

Consider the horizontal surface as the reference level.If the cylinder covers height hduring its motion on the incline, then the change in its potential energy will be $\mathrm{\Delta }PE=mgh$.

The angular velocity of the rolling cylinder is related to its linear velocity as$\omega =\frac{v}{r}$.

The final kinetic energy of the cylinder after reaching height h on the incline will be zero. So, the change in kinetic energy will be equal to the initial kinetic energy of the cylinder at the bottom.The rolling cylinder possesses both rotational motion as well as translational motion at the bottom. Thus, the kinetic energy of the cylinder at the bottom of the incline is the sum of the rotational kinetic energy and translational kinetic energy. So,

$\begin{array}{r}c\mathrm{\Delta }KE=K{E}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{a}\mathrm{l}}+K{E}_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{a}\mathrm{l}}\\ =\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2\\ =\frac{1}{2}\left(m{r}^2\right)\frac{v^2}{r^2}+\frac{1}{2}m{v}^2\\ =m{v}^2\end{array}$

According to the law of conservation of mechanical energy, the change in the potential energy of the cylinder will be equal to the change in its total kinetic energy, i.e.,

$\begin{array}{r}c\mathrm{\Delta }PE=\mathrm{\Delta }KE\\ mgh=m{v}^2\\ h=\frac{v^2}{g}\\ =\frac{{\left(3.0\mathrm{m}/\mathrm{s}\right)}^2}{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ =0.918\mathrm{m}\end{array}$

If the vertical height covered is 0.918 m, then the distance covered $\left(x\right)$on the inclined surface is: $\begin{array}{r}cx=\frac{h}{\sin \theta }\\ =\frac{0.918\mathrm{m}}{\sin {15}^{\circ }}\\ =\frac{0.918\mathrm{m}}{0.259}\\ =3.5\mathrm{m}\end{array}$

Thus, the hollow cylinder will go up the incline to a distance of 3.5 m.

(b) Calculation of the time for which the hollow cylinder will be on the incline

Since the acceleration of the hollow cylinder is constant on the incline, the average velocity of the cylinder while going up the plane is:

$\overline{v}=\frac{v_0+v}{2}$ ... (i)

Also, the average velocity of the cylinder can be written as:

$\overline{v}=\frac{x}{t}$ ... (ii)

From (i) and (ii),

$\begin{array}{r}c\frac{x}{t}=\frac{v_0+v}{2}\\ t=\frac{2x}{v_0+v}\\ =\frac{2\left(3.5\mathrm{m}\right)}{\left(3.0\mathrm{m}/\mathrm{s}+0\mathrm{m}/\mathrm{s}\right)}\\ =2.33\mathrm{s}\end{array}$

The time required to return to the bottom is equal to the time taken to go up the incline.Thus, the total time for which the cylinder will be on the incline before it returns to the bottom is:

$\begin{array}{r}c2t=2\left(2.33\mathrm{s}\right)\\ =4.66\mathrm{s}\\ \approx 4.7\mathrm{s}\end{array}$