The roulette wheel is given in initial velocity of 20rad/s and then allowed to s

Question

The roulette wheel is given in initial velocity of 20rad/s and then allowed to slowly come to a stop while the 'pea' bounces around. Assume a constant angular deceleration of -0.75 rad/s^2 until the wheel comes to rest. The 'pea' stops 15 seconds after the wheel is started in the number "zero" a distance of 12" from the center of the wheel. Find the angular velocity of the wheel then the pea stops, the total acceleration of the pea when it has landed on the number "zero" and the number of revolutions the wheel has made when the pea lands.

Explanation / Answer

a) at t = 15 s,

w = wo + alfa*t

= 20 + (-0.75)*15

= 8.75 rad/s

b) a_rad = r*w^2

= 0.3048*8.75^2 (since 12 inch = 1 feet = 0.3048 m)

= 23.3 m/s^2

a_tan = r*alfa

= 0.3048*0.75

= 0.2286 m/s^2

a_total = sqrt(a_tan^2 + a_rad^2)

= sqrt(0.2286^2 + 23.3^2)

= 23.3 m/s^2

c) angular dispalcement of wheel, theta = wo*t + (1/2)*alfa*t^2

= 20*15 + (1/2)*(-0.75)*20^2

= 150 rad

= 150/(2*pi)

= 23.9 revolutions

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