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Chapter 4: Q.71 (page 109)

Your $64 cm$ diameter car tire is rotating at $3.5 rev / s$ when suddenly you press down hard on the accelerator. After traveling $200 m$ , the tire鈥檚 rotation has increased to $6.0 rev / s$ . What was the tire鈥檚 angular acceleration? Give your answer in $rad / s^{2}$ .

Short Answer

Expert verified

The tire鈥檚 angular acceleration is $0.75 rad / s^{2}$

Step by step solution

01

Given information

Diameter of car tire is $D = 64 cm$

Car tire is rotating at $蝇_{i} = initial angular velocity = 3.5 rev / s$

After traveling $d = 200 m$ , the tire鈥檚 rotation has increased to $蝇_{f} = final angular velocity = 6.0 rev / s$ .

We have to find the tire鈥檚 angular acceleration $伪$ in $rad / s^{2}$ .

02

Formula

Perimeter of circle is $2 蟺 r$

$蝇_{f}^{2} = 蝇_{i}^{2} + 2 伪鈭� 胃$

03

Explanation

By putting values in formula

$P = 2 蟺谤$

For radius

$r = \frac{d}{2} r = \frac{64}{2} r = 32 cm$

Putting values:

$P = 2 脳 3.14 脳 32 cm = 200.96 cm$

Converting the car tire in revolutions:

$胃 = \frac{d}{P} 胃 = \frac{200 cm 脳 100}{200.96 cm} 胃 = 99.52 rev$

For angular acceleration

$6^{2} = 3 . 5^{2} + 2 伪脳 99.52 伪 = 0.1193 rev / s^{2}$

Now, convert $rev / s^{s}$ in $rad / s^{2}$

localid="1648908260302" $伪 = 0.1193 rev / s^{s} 脳 2 蟺 = 0.1193 rev / s^{s} 脳 2 脳 3.14 伪 = 0.75 rad / s^{2}$

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