A disk 8.04 cm in radius rotates at a constant rate of 1 220 rev/min about its central axis. (a) Determine its angular speed. rad/s (b) Determine the tangential speed at a point 3.02 cm from its center. m/s (c) Determine the radial acceleration of a point on the rim. magnitude km/s2 direction (d) Determine the total distance a point on the rim moves in 2.02 s. m

A) 128 rad/s

B) 3.87 m/s

C) 1317.17 m/s²

D) 20.79m

Explanation:

A) We are given the angular speed as 1220 rev/min. Now let's convert it to rad/s.

1220 (rev/min) x (2πrad/1rev) x (1min/60sec) = (1220 x 2π)/60 rad/s = 127.76 rad/s ≈ 128 rad/s

B) Formula for tangential speed is given as;

v = ωr

We know that ω = 128 rad/s

Also, r = 3.02cm = 0.0302m

Thus, v = 128 x 0.0302 = 3.87 m/s

C) Formula for radial acceleration is given as;

a_c = v²/r

From earlier, v = ωr

Thus, a_c = v²/r = (ωr)²/r = ω²r

On the rim, r = 8.04cm = 0.0804

a_c = 128² x 0.0804 = 1317.17 m/s²

D) We know that; distance/time = speed

Thus, distance = speed x time

D = vt

From earlier, v = ωr

Thus, D = ωrt

Plugging in the relevant values ;

D = 128 x 0.0804 x 2.02 = 20.79m

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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