Question

Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m, the tire's rotation has increased to 6.4 rev/s. What was the tire's angular acceleration? Give your answer in rad/s^2.

Answer 1

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

$v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s$

$v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s$

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

$D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads$

Now, we just use the formula:

$w_f^2-w_o^2=2\alpha*x$

$\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2$

Answer 2

Explanation:

The given data is as follows.

   Initial velocity, u = $3.3 rev/s \times 2 \pi rad/rev$

                           = 20.724 rad/sec

   Final velocity, v = $6.4 rev/s \times 2 \times pi rad/rev$    

                           = 40.192 rad/sec

Now, we will calculate the value of angular rotation (d) as follows.

          d = No. of revolutions × $2 \pi rad/rev$

         d = $(\frac{250 m}{2 \pi r}) \times 2 \pi$

            = $\frac{250}{0.32}$

            = 781.25 rad

Also, we know that,

                $v^{2} = u^{2} + 2ad$

or,              a = $\frac{(v^{2} - u^{2})}{2d}$

                      = $\frac{(40.192)^{2} - (20.724)^{2})}{2 \times 781.25}$

                      = $\frac{1615.39 - 429.48}{1562.5}$

                     = 0.7589 $rad/s^{2}$

Thus, we can conclude that the tire's angular acceleration is 0.7589 $rad/s^{2}$.