Rotational Kinematics

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A wheel starting from rest reaches an angular velocity of 20 rad/s in 4 seconds while subject to a constant angular acceleration. Through what total angle does the wheel turn during this interval?

RankGuru · Accepted Answer

Rotational kinematics under constant angular acceleration mirrors linear kinematics, with angle, angular velocity, and angular acceleration replacing displacement, velocity, and acceleration respectively. First find the angular acceleration from $\alpha = \frac{\omega - \omega_0}{t} = \frac{20 - 0}{4} = 5	ext{ rad/s}^2$. Then the angle swept from rest is $	heta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(5)(4)^2 = \frac{1}{2}(5)(16) = 40	ext{ rad}$. The value 80 rad wrongly uses the final speed for the whole interval as if the motion were uniform. The value 20 rad mistakenly multiplies an average speed by half the time. The value 10 rad applies an incorrect averaging that halves the answer twice over. This applies the NCERT rotational kinematic equations directly, exactly paralleling the equations of uniformly accelerated linear motion. As a cross-check, the average angular velocity over the interval is $\frac{0 + 20}{2} = 10	ext{ rad/s}$, and over 4 s that gives $10 	imes 4 = 40	ext{ rad}$, matching the result from the kinematic equation.

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