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Question

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A motorcyclist is traveling along a road and accelerates for 4.50 s to pass another cyclist. The angular acceleration of each wheel is +6.70 rad/s^2, and, just after passing, the angular velocity of each is +74.5 rad/s, where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time? (Chapter 8)

Answer 1

More angular kinematics

v = vo + at

v = final angular velocity (74.5 r/s)
vo = initial angular velocity (unknown)
a = angular acceleration (6.7 r/s^2)
t = time (4.5 s)

Solve for vo

Now

x - xo = vo t + 1/2 a t^2

x = final angular position
xo = initial angular position (usually zero)
vo = initial angular velocity (found in previous part)
a = angular acceleration (6.7 r/s^2)
t = time (4.5 s)

Solve for x

Answer 2

All the equations for rotary motion look like their analogues for linear motion, so for instance KE= 1/2 I w^2 where I= moment of inertia and w= angular velocity just as KE=1/2 Mv^2 for linear motion. Soooo, in the linear case, the formula you'd use would be v^2 = u^2 +2as. In this case v=74.5, u=74.5-(4.5x6.7)=44.35, a=6.7 and s= angular displacement, so 5550.25=1966.9225+2x6.7xs, so s=267.4125 radians.