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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 1: Q21E (page 295)

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s². Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship,
(a) $a_{r a d} = 蝇^{2} r$ and
$(b) a_{rad} = \frac{v^{2}}{r}$

Short Answer

Expert verified

Thus, completes the wheel its second revolution from the relationship $16.576 \frac{m}{s^{2}}$ .

Thus, Angular velocity at the end of two revolutions is $16.576 \frac{m}{s^{2}}$ .

Step by step solution

01

Step:-1 explanation

We know that the formula of angular velocity,

$蝇_{z}^{2} = 蝇_{z_{0}}^{2} + 2 伪 (胃 - 胃_{0}) 蝇_{2} = \sqrt{2 伪_{z} (胃 - 胃_{0})} = \sqrt{2 (\frac{3 r a d}{s^{2}}) (4 蟺谤补诲)} = 8.68 \frac{r a d}{s}$

02

Step:-2 concept

$a_{r a d} = \frac{v^{2}}{r}$ and we know that $v = r 蝇$ .

$a_{r a d} = \frac{{(r 蝇)}^{2}}{r} = r 蝇^{2}$

03

Step:-3 calculation

Put the value here,

$1 \frac{r a d}{s} = 9.549 \frac{m}{s^{2}} 8.68 \frac{r a d}{s} = 82.88$

Hence we get,

$a_{r a d} = (0.2 m) (82.88 \frac{m}{s^{2}}) = 16.576 \frac{m}{s^{2}}$

Completes the wheel its second revolution from the relationship $16.576 \frac{m}{s^{2}}$ .

04

Step:-4(b) explanation

We know that $a_{r a d} = \frac{v^{2}}{r}$ .

Here we know that,

Here $= v = r 蝇 a_{r a d} = \frac{{(r 蝇)}^{2}}{r}$

05

Step:-5(b) concept

We know that $a_{r a d} = \frac{{(r 蝇)}^{2}}{r}$ .

$a_{r a d} = \frac{r^{2} 蝇^{2}}{r} a_{r a d} = r 蝇^{2}$

06

Step:-6(b) Conclusion

Put the value here,

$a_{r a d} = (0.2 m) (8.68 \frac{r a d}{s})$

We know that the value of,

$1 \frac{r a d}{s} = 9.549 \frac{m}{s^{2}} 8.68 \frac{r a d}{s} = 82.88$

Hence we get,

$a_{r a d} = (0.2 m) (82.88 \frac{m}{s^{2}}) = 16.576 \frac{m}{s^{2}}$

Agular velocity at the end of two revolutions is $16.576 \frac{m}{s^{2}}$ .

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