Question

Supermodified race cars, such as the one shown in the image, are built with their massive engines mounted to the left side of the car between the two left wheels.

What advantage does this give the car?

The offset engine moves the race car's center of mass to the left, making it easier to turn left.

Moving the engine has no effect on the race car. It can be treated as a point mass.

Moving the engine only adds mass since it requires more mounting brackets.

The offset engine moves the race car's center of mass to the right, making it easier to turn right.

Answer 1

Center of mals it easier to consider car shifted towards left. Making turn left for calculation we as a point mass at com

Answer 2

The axis of rotation coincides with the center of curvature of the path of the race around the track, labeled in the figure as ?.$R\text{.}$ Since the dimensions of the engine are small compared to ?,$R\text{,}$ then the engine can be approximated as a point mass. The moment of inertia ?$I$ of a body with mass ?$m$ a distance ?$r$ from an axis of rotation is

?=??2$$$$

Therefore, the equation

?=??2=??2

describes fairly accurately the moment of inertia ?$I$ of the engine about the axis of rotation where ?$m$ is the mass of the engine. By shifting the engine closer to the axis by Δ?,$\mathrm{\Delta }R\text{,}$ the moment of inertia of the race car is decreased to

$$I_1=m{\left(R-\mathrm{\Delta }R\right)}^2$$

The percent difference is the difference between the two moments of inertia divided by the original value.

$$\mathrm{\Delta }\mathrm{\%}=\left|\frac{I-I_1}{I}\right|\times 100\mathrm{\%}$$

The moment of inertia of the engine in the centered position is then

?=??2=(1030 kg)(30.5 m)2=9.58×105 kg⋅m2$$\begin{array}{rl}I& =m{R}^2\\ & =\left(1030\text{ kg}\right){\left(30.5\text{ m}\right)}^2\\ & =9.58\times {10}^5\text{ kg}\cdot {\text{m}}^2\end{array}$$

The moment of inertia of the engine in the offset position is

?1=?(?−Δ?)2=(1030 kg)(30.5 m−(50.3 cm×1 m100 cm))2=9.26×105 kg⋅m2$$\begin{array}{rl}{I}_1& =m{\left(R-\mathrm{\Delta }R\right)}^2\\ & =\left(1030\text{ kg}\right){\left(30.5\text{ m}-\left(50.3\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2\\ & =9.26\times {10}^5\text{ kg}\cdot {\text{m}}^2\end{array}$$

These two moments of inertia can be used to calculate the percent difference.

Δ%=||||?−?1?||||×100%=|||||9.58×105 kg⋅m2−9.26×105 kg⋅m29.58×105 kg⋅m2|||||×100%=3.3%

Answer 3

The axis of rotation coincides with the center of curvature of the path of the race around the track, labeled in the figure as ?.$R\text{.}$ Since the dimensions of the engine are small compared to ?,$R\text{,}$ then the engine can be approximated as a point mass. The moment of inertia ?$I$ of a body with mass ?$m$ a distance ?$r$ from an axis of rotation is

$$I=m{r}^2$$

Therefore, the equation

$$I=m{r}^2=m{R}^2$$

describes fairly accurately the moment of inertia ?$I$ of the engine about the axis of rotation where ?$m$ is the mass of the engine. By shifting the engine closer to the axis by Δ?,$\mathrm{\Delta }R\text{,}$ the moment of inertia of the race car is decreased to

$$I_1=m{\left(R-\mathrm{\Delta }R\right)}^2$$

The percent difference is the difference between the two moments of inertia divided by the original value.

$$\mathrm{\Delta }\mathrm{\%}=\left|\frac{I-I_1}{I}\right|\times 100\mathrm{\%}$$

The moment of inertia of the engine in the centered position is then

?=??2=(1030 kg)(30.5 m)2=9.58×105 kg⋅m2$$\begin{array}{rl}I& =m{R}^2\\ & =\left(1030\text{ kg}\right){\left(30.5\text{ m}\right)}^2\\ & =9.58\times {10}^5\text{ kg}\cdot {\text{m}}^2\end{array}$$

The moment of inertia of the engine in the offset position is

?1=?(?−Δ?)2=(1030 kg)(30.5 m−(50.3 cm×1 m100 cm))2=9.26×105 kg⋅m2$$\begin{array}{rl}{I}_1& =m{\left(R-\mathrm{\Delta }R\right)}^2\\ & =\left(1030\text{ kg}\right){\left(30.5\text{ m}-\left(50.3\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2\\ & =9.26\times {10}^5\text{ kg}\cdot {\text{m}}^2\end{array}$$

These two moments of inertia can be used to calculate the percent difference.

Δ%=||||?−?1?||||×100%=|||||9.58×105 kg⋅m2−9.26×105 kg⋅m29.58×105 kg⋅m2|||||×100%=3.3%