An electric motor rotating a workshop grinding wheel at 1.00 3 102 rev/min is switched off. assume the wheel has a constant negative angular acceleration of magnitude 2.00 rad/s2. (a) how long does it take the grinding wheel to stop? (b) through how many radians has the wheel turned during the time interval found in part (a)?

(a) 5.25 second

(b) 27.5625 rad

Explanation:

fo = 1.003 x 10^2 rev / min = 1.672 rev / sec

ωo = 2 x 3.14 x fo = 2 x 3.14 x 1.672 = 10.5 rad/s

α = - 2 rad/s^2

(a) Let t be the time taken to stop.

ω = 0 rad/s

Use first equation of motion for rotational motion

ω = ωo + α t

0 = 10.5 - 2 x t

t = 5.25 second

(b) Let it rotates by an angle θ in the given time.

use third equation of motion for rotational motion

ω² = ωo² + 2 α θ

0 = 10.5 x 10.5 - 2 x 2 x θ

θ = 27.5625 rad

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

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