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

(a). Angular speed
`\omega = 130.83 (rad)/(s)`

(b). The tangential speed at a point 3.06 cm from its center is V = 4
`(m)/(s)`

(c). The radial acceleration of a point on the rim is
`a_(r) =` 1.37
`(km)/(s^(2) )`

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

Step-by-step explanation:

Given data

Radius = 8.02 cm = 0.0802 m

N = 1250 RPM

(a). Angular speed

`\omega = (2 \pi N)/(60)`

`\omega = (2 \pi 1250)/(60)`

`\omega = 130.83 (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
`(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
`(m)/(s^(2) )`

`a_(r) =` 1.37
`(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