Question

A disk 8.02 cm in radius rotates at a constant rate of 1 250 rev/min about its central axis. (a) Determine its angular speed. rad/s (b) Determine the tangential speed at a point 3.06 cm from its center. m/s (c) Determine the radial acceleration of a point on the rim. magnitude km/s2 direction Correct: Your answer is correct. (d) Determine the total distance a point on the rim moves in 1.96 s. m

Answer 1

Answer:

(a). Angular speed $\omega = 130.83 \frac{rad}{s}$

(b). The tangential speed at a point 3.06 cm from its center is V = 4 $\frac{m}{s}$

(c). The radial acceleration of a point on the rim is $a_{r} =$ 1.37 $\frac{km}{s^{2} }$

(d). The total distance a point on the rim moves in 1.96 sec is D = 20.56 m

Explanation:

Given data

Radius = 8.02 cm = 0.0802 m

N = 1250 RPM

(a). Angular speed

$\omega = \frac{2 \pi N}{60}$

$\omega = \frac{2 \pi 1250}{60}$

$\omega = 130.83 \frac{rad}{s}$

(b). The tangential speed at a point 3.06 cm from its center is

V = r $\omega$

V = 0.0306 × 130.83

V = 4 $\frac{m}{s}$

(c). The radial acceleration of a point on the rim is

$a_{r} = r \omega^{2}$

$a_{r} = 0.0802$ × $130.83^{2}$

$a_{r} =$ 1369.32 $\frac{m}{s^{2} }$

$a_{r} =$ 1.37 $\frac{km}{s^{2} }$

(d). The total distance a point on the rim moves in 1.96 sec is

D = r $\omega t$

D = 0.0802 × 130.83 × 1.96

D = 20.56 m