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

Step-by-step explanation:

The given data is as follows.

Initial velocity, u =
`3.3 rev/s * 2 \pi rad/rev`

= 20.724 rad/sec

Final velocity, v =
`6.4 rev/s * 2 * 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 =
`((250 m)/(2 \pi r)) * 2 \pi`

=
`(250)/(0.32)`

= 781.25 rad

Also, we know that,

`v^(2) = u^(2) + 2ad`

or, a =
`((v^(2) - u^(2)))/(2d)`

=
`((40.192)^(2) - (20.724)^(2)))/(2 * 781.25)`

=
`(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)`.

Answer 2

0.76 rad/s^2

Step-by-step explanation:

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

`v_o = 3.3(rev)/(s) * (2\pi rad)/(1rev) =20.73 rad/s`

`v_f = 6.4(rev)/(s) * (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 = (D)/(r) = (250m)/(0.64m/2) = 781.25 rads`

Now, we just use the formula:

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

`\alpha =(w_f^2-w_o^2)/(2x) = ((40.21rad/s)^2-(20.73rad/s)^2)/(2*781.25rad) = 0.76 rad/s^2`