# A high-speed sander has a disk 4.00 cm in radius that rotates about its axis at aconstant rate of 1265 rev/min.Determine (a) the angular speed of the

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A high-speed sander has a disk 4.00 cm in radius that rotates about its axis at aconstant rate of 1265 rev/min.Determine
(a) the angular speed of the disk in radians persecond,
(b) the linear speed of a point 2.2 cmfrom the disk's center,
m/s
(c) the centripetal acceleration of a point on the rim, and
$$\displaystyle\frac{{m}}{{s}^{{{2}}}}$$
(d) the total distance traveled by a point on the rim in1.96 s.
m

2021-05-10
r = 0.04m
1265 rev/min
a.)$$\displaystyle{\left({\frac{{{1265}{r}{e}{v}}}{{\min{u}{t}{e}}}}\right)}{\left({\frac{{{1}\min}}{{{60}{\sec}}}}\right)}{\left({\frac{{{2}\pi{r}{a}{d}}}{{{r}{e}{v}}}}\right)}={132.47}{\frac{{{r}{a}{d}}}{{{\sec}}}}$$
$$\displaystyle\omega={132.47}{r}{a}\frac{{d}}{{\sec{}}}$$
b.) The linear speed is called the tangential velocityalso
$$\displaystyle{V}_{{{\tan}}}={r}\omega$$
$$\displaystyle{V}_{{{\tan}}}={\left({0.022}\right)}{\left({132.47}\right)}$$
$$\displaystyle{V}_{{{\tan}}}={2.914}\frac{{m}}{{s}}$$
c.)$$\displaystyle{a}_{{{c}{e}{n}{t}}}={r}\omega^{{{2}}}$$
$$\displaystyle{a}_{{{c}{e}{n}{t}}}={\left({0.04}\right)}{\left({132.47}\right)}^{{{2}}}$$
$$\displaystyle{a}_{{{c}{e}{n}{t}}}={701.9}\frac{{m}}{{s}^{{{2}}}}$$
d.)$$\displaystyle{V}_{{{\tan}}}={r}\omega$$
$$\displaystyle{V}_{{{\tan}}}={\left({0.04}\right)}{\left({132.47}\right)}$$
$$\displaystyle{V}_{{{\tan}}}={5.299}\frac{{m}}{{s}}$$
D=VT
D= (5.299)(1.96)
D= 10.386 m