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26P

Expert-verifiedFound in: Page 326

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**A small object with mass 4.00 kg moves counter-clockwise with constant angular speed 1.50 rad/s in a circle of radius 3.00 m centered at the origin. It starts at the point with position vector $3.00\hat{i}m$. It then undergoes an angular displacement of 9.00 rad. (a) What is its new position vector? Use unit-vector notation for all vector answers. (b) In what quadrant is the particle located, and what angle does its position vector make with the positive x axis? (c) What is its velocity? (d) In what direction is it moving? (e) What is its acceleration? (f) Make a sketch of its position, velocity, and acceleration vectors. (g) What total force is exerted on the object?**

The new position vector of object is $\left(-2.74\hat{i}+1.22\hat{j}\right)m$.

The mass of object is m = 4 kg

The constant angular speed of object is $\omega =1.50rad/s$

The radius of circle is R = 3m

The angular displacement of object is $\delta =9rad$

**The angular speed of an object is the rate of variation of the angular displacement of the object with time.**

The angle for the new position vector is given as:

$\alpha =\delta \left(\frac{180\xb0}{\pi}\right)-360\phantom{\rule{0ex}{0ex}}=\left(9rad\right)\left(\frac{180\xb0}{\pi rad}\right)-360\xb0\phantom{\rule{0ex}{0ex}}=156\xb0$

The new position vector of object is given as:

$\overrightarrow{r}=R\left(\mathrm{cos}\alpha \hat{i}+\mathrm{sin}\alpha \hat{j}\right)\phantom{\rule{0ex}{0ex}}=\left(3m\right)\left[\left(\mathrm{cos}156\xb0\right)\hat{i}+\left(\mathrm{sin}156\xb0\right)\hat{j}\right]\phantom{\rule{0ex}{0ex}}=\left(-2.74\hat{i}+1.22\hat{j}\right)m$

Therefore, the new position vector of object is $\left(-2.74\hat{i}+1.22\hat{j}\right)m$ .

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