UAM equations are,

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{+}}{\mathit{a}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{+}}{\frac{1}{2}}{\mathit{a}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{+}}{\mathbf{2}}{\mathit{a}}{\mathbf{\u2206}}{\mathit{x}}}$

Moment of inertia of a circular disk,

$\overline{){\mathbf{I}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{r}}}^{{\mathbf{2}}}}$

Velocity, v, and angular velocity, ω,

$\overline{){\mathbf{v}}{\mathbf{=}}{\mathbf{r}}{\mathbf{\omega}}}$

Gravitational potential energy,

$\overline{){\mathbf{U}}{\mathbf{=}}{\mathbf{m}}{\mathbf{g}}{\mathbf{h}}}$

Kinetic energy,

$\overline{){\mathit{K}}{\mathit{E}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{m}}{{\mathit{v}}}^{{\mathbf{2}}}}$

**Time taken by the box:**

The component of weight acting on the objects downslope is, **mg sin θ **

From the dimensions of the incline, sin θ = **h/s**

Where, s is the length of the inclined plane and h is the height of the inclined plane.

Thus, **s = h / sin θ**

For the box,

ΣF = ma = mg sin θ.

Thus, a = g sin θ.

Time taken by the box downslope:

Using the third UAM equation,

- t = ?
- Δx = s = h / sin θ
- a = g sin θ
- v
_{0}= 0 - v
_{f}= ?

$\begin{array}{rcl}\frac{\mathbf{h}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\theta}}& \mathbf{=}& \mathbf{0}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{g}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\theta}\mathbf{)}{\mathbf{t}}^{\mathbf{2}}\\ \frac{\mathbf{2}\mathbf{h}}{\mathbf{g}\mathbf{}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{}\mathbf{\theta}}& \mathbf{=}& {\mathbf{t}}^{\mathbf{2}}\\ \frac{\mathbf{1}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\theta}}\sqrt{\frac{\mathbf{2}\mathbf{h}}{\mathbf{g}}}& \mathbf{=}& \mathbf{t}\end{array}$

t_{box} = **(1/ sin θ)(2h/g)**^{0.5}

This applet shows the results of releasing a frictionless block and a rolling disk with equal masses from the top of identical inclined planes.

This applet shows the same situation, but it also shows, through bar graphs that change with time, the way that the energy is transformed as the box and the disk go down the inclined plane.

Assume that the box and disk each have mass *m*, the top of the incline is at height *h*, and the angle between the incline and the ground is *θ* (i.e., the incline is at an angle *θ* above the horizontal). Also, let the radius of the disk be *R.*

How much sooner does the box reach the bottom of the incline than the disk? Express your answer in terms of some or all of the variables *m*, *h*, *θ*, and *R*, as well as the acceleration due to gravity *g.*

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Conservation of Energy in Rolling Motion concept. You can view video lessons to learn Conservation of Energy in Rolling Motion. Or if you need more Conservation of Energy in Rolling Motion practice, you can also practice Conservation of Energy in Rolling Motion practice problems.