A grindstone radius 4.0 m is initially spinning with an angular speed of 8.0 rad

Question

A grindstone radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s The angular speed is then to 10 rad/s over the next 4.0 that acceleration is constant. What is the average angular speed of the grindstone? 0.5 rad/s 2.0 rad/s 4.5 rad/s 9.00 rad/s 18 rad/s What is the magnitude of the angular acceleration of the grindstone 0.50 rad/s^2 4.5 rad/s^2 18 rad/s^2 9.0 rad/s^2 Through how many revolutions does the grindstone turn during the 4.0 second interval? 0.64 3.8 4.0 36 5.7 A grindstone, initially at rest, is given a constant angular acceleration so that it makes 20.0 rev in the first 8.00 s. What is its angular acceleration? 0.313 rad/s^2 2.50 rad/s^2 3.93 rad/s^2 0.625 rad/s^2 1.97 rad/s^2 A wheel, originally rotating at 126 rad/s undergoes a constant angular deceleration of 5.00 rad/s^2 What is its angular speed after it has turned through an angle of 628 radians? 15 rad/s 19 rad/s 98 rad/s 121 rads 150 rad/s

Explanation / Answer

(1) angular acc is constant

hecne average angular veocity = (wf + wi) / 2

= (8 + 10) /2

= 9 rad /s Ans(d)

(2) wf = wi + alpha t

10 = 8 + 4(alpha)

alpha = 0.50 rad/s^2 Ans(a)

(3) wf^2 - wi^2 = 2 alpha theta

10^2 - 8^2 = 2(0.50) (theta)

theta = 36 rad

revolutions = 36 / 2pi = 5.7 Ans(d)

(4) theta = 20 x 2pi = 125.66 rad

theta = wi t + alpha t^2 / 2

125.66 = 0 + (alpha x 8^2) / 2

alpha = 3.93 rad/s^2

Ans(e)

(5) wf^2 - wi^2 = 2 alpha theta

w^2 - 126^2 = 2(-5)(628)

w = 98 rad/s

Ans(c)

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