A compact disc rotates from rest up to an angular speed of 31.6 rad/s in a time

Question

A compact disc rotates from rest up to an angular speed of 31.6 rad/s in a time of 0.913 s.

(a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform?

______________ rad/s2

(b) Through what angle does the disc turn while coming up to speed?

rad

(c) If the radius of the disc is 4.45 cm, find the tangential speed of a microbe riding on the rim of the disc.

____________ m/s

(d) What is the magnitude of the tangential acceleration of the microbe at the given time?

_____________ m/s2

(Use the values from ABOVE)

(a) What are the angular speed and angular displacement of the disc 0.320 s after it begins to rotate?

? = _______ rad/s

?? = ________rad

(b) Find the tangential speed at the rim at this time.

_____________ m/s

Explanation / Answer

here,

final angular speed , w = 31.6 rad/s

time taken , t = 0.913 s

a)

let the angular accelration be alpha

w = w0 + alpha * t

31.6 = 0 + alpha * 0.913

alpha = 34.6 rad/s^2

b)

the angle travelled , theta = w0 * t + 0.5 * alpha * t^2

theta = 0 + 0.5 * 34.6 * 0.913^2 rad

theta = 14.4 rad

c)

radius , r = 4.5 cm = 0.045 m

the tangential speed , v = r * w

v = 0.045 * 31.6 = 1.42 m/s

d)

the magnitude of tangential accelration , a = r * alpha

a = 0.045 * 34.6 m/s^2

a = 1.56 m/s^2

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