Question

The rear wheels of a sports car have a radius of \(46.65 \mathrm{~cm}\). The sports car goes from rest to a speed of \(29.13 \mathrm{~m} / \mathrm{s}\) in \(3.945 \mathrm{~s}\) with constant accelera. tion. What is the angular acceleration of the rear wheels? Assume that the wheels roll without slipping.

Short answer

Answer: The angular acceleration of the rear wheels of the sports car is approximately \(15.83\ \mathrm{rad/s^2}\).

Step-by-step solution

Find the linear acceleration

The sports car goes from rest (initial speed \(v_0 = 0\)) to a final speed \(v_f = 29.13\mathrm{~m/s}\) in a time \(t = 3.945\mathrm{~s}\). Using the equation for linear acceleration \(a = \frac{v_f - v_0}{t}\), we can calculate the linear acceleration. $$a = \frac{29.13\mathrm{~m/s} - 0\mathrm{~m/s}}{3.945\mathrm{~s}} = \frac{29.13}{3.945} \mathrm{ms^{-2}}$$

Calculate the value of linear acceleration

We can now calculate the numerical value for linear acceleration as follows: $$a = \frac{29.13}{3.945} \mathrm{ms^{-2}} \approx 7.39\ \mathrm{ms^{-2}}$$

Find the angular acceleration

Since the wheels roll without slipping, the linear acceleration of the car is equal to the product of the angular acceleration (\(\alpha\)) and radius (r) of the wheels. In other words, \(a = \alpha r\). Therefore we can find the angular acceleration (\(\alpha\)) using the formula: $$ \alpha = \frac{a}{r} $$

Calculate the angular acceleration

Given the radius of the wheels \(r = 46.65\ \mathrm{cm}\) (make sure to convert the radius to meters), we can calculate the angular acceleration as follows: $$\alpha = \frac{7.39 \mathrm{ms^{-2}}}{0.4665\ \mathrm{m}} \approx 15.83\ \mathrm{rad/s^2}$$

Write the final answer

The angular acceleration of the rear wheels of the sports car is approximately \(15.83\ \mathrm{rad/s^2}\).