Question

A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its side at the top of a 3.00-m-long incline that is at$\mathbf{25}\mathbf{°}$to the horizontal and is then released to roll straight down. It reaches the bottom of the incline after 1.50 s. (a) Assuming mechanical energy conservation, calculate the moment of inertia of the can. (b) Which pieces of data, if any, are unnecessary for calculating the solution? (c) Why can’t the moment of inertia be calculated from$\mathit{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{r}}^{\mathbf{2}}$ for the cylindrical can?

Short answer

1. The moment of inertia of the can,$I=1.21\times {10}^{-4}\mathrm{kg}·{\mathrm{m}}^2$
2. Height of the can
3. We know that the density of the metal can is larger than that of the soup, so the mass is not uniformly distributed, which is why we cannot calculate the moment of inertia using$I=\frac{1}{2}m{r}^2$.

Step-by-step solution

Step 1:To find the moment of inertia of the cylinder

(a)We have a cylinder with a mass of m = 215 g, the height of H = 10.8 cm, and a diameter D=6.38cm. It is placed at the top of an incline with a length of d = 3.00 m and an angle of$\theta =25.0°$to the horizontal. Then it is released to roll straight down. It is given that it reaches the bottom of the incline within t= 1.50 s. First, we need to find the moment of inertia of the cylinder, the system of the cylinder and earth is isolated, so the energy is conserved, that is,

$\mathrm{\Delta }K+\mathrm{\Delta }U=0$

Initially, the cylinder has a potential energy of “mgh” where $h=d\sin \theta$, at the bottom of the incline the potential energy is zero, the cylinder has only transitional kinetic energy and rotational kinetic energy, so we can write,

$\left(\frac{1}{2}m{v}_{\mathrm{CM}}^2+\frac{1}{2}I{\omega }^2-0\right)+(0-mgh)=0$

Solve for the moment of inertia as:

$\begin{array}{c}I=\frac{2mgh-m{v}_{\mathrm{CM}}^2}{{\omega }^2}\\ =\left(2mgh-m{v}_{\mathrm{CM}}^2\right)\left(\frac{r^2}{v_{\mathrm{CM}}^2}\right)\\ =m{r}^2\left(\frac{2gh}{v_{CM}^2}-1\right)\end{array}$

Where, $v_{\mathrm{CM}}$is the velocity of the center of mass of the cylinder. Since the cylinder is under a constant acceleration model, we can write,

$\begin{array}{c}{v}_{\mathrm{CM},\mathrm{avg}}=\frac{0+v_{\mathrm{CM}}}{2}\\ v_{\mathrm{CM}}=2v_{\mathrm{CM},\mathrm{avg}}\\ =\frac{2d}{t}\end{array}$

Substitute into (!) we get,

$\begin{array}{c}I=m{r}^2\left(\frac{2gh{t}^2}{4d^2}-1\right)\\ =m{r}^2\left(\frac{2g\left(d\sin \theta \right)t^2}{4d^2}-1\right)\\ =m{r}^2\left(\frac{g\left(\sin \theta \right)t^2}{2d}-1\right)\end{array}$

To find the I value

The radius is calculated as,

$\begin{array}{c}r=\frac{D}{2}\\ =\frac{6.38 \mathrm{cm}}{2}\\ =3.19 \mathrm{cm}\end{array}$

Substitute with the givens to get,

$\begin{array}{c}I=(0.215 \mathrm{kg})(0.0319{)}^2\left(\frac{\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(\sin 25.0°\right){(1.50 \mathrm{s})}^2}{2(3.00 \mathrm{m})}-1\right)\\ =1.21\times {10}^{-4} \mathrm{kg}·{\mathrm{m}}^2\\ =1.21\times {10}^{-4} \mathrm{kg}·{\mathrm{m}}^2\end{array}$

Unnecessary data

(b)

We can see the height of the can is unnecessary data.

(c)

We know that the density of the metal can is larger than that of the soup, so the mass is not uniformly distributed, which is why we cannot calculate the moment of inertia using$I=\frac{1}{2}m{r}^2$.