A Ferris wheel, rotating initially at an angular speed of 0.50 rad/s, accelerates over a 4.0-s interval at a rate of 0.10 rad/s2. What angular displacement does the Ferris wheel undergo in this 7.00-s interval?Select one:a.2.40 radb.2.80 radc.0.80 radd.2.20 rade.3.60 rad
Question
A Ferris wheel, rotating initially at an angular speed of 0.50 rad/s, accelerates over a 4.0-s interval at a rate of 0.10 rad/s2. What angular displacement does the Ferris wheel undergo in this 7.00-s interval?Select one:a.2.40 radb.2.80 radc.0.80 radd.2.20 rade.3.60 rad
Solution
To solve this problem, we need to use the equations of motion.
The equation for angular displacement is given by:
θ = ωit + 0.5α*t^2
where: θ is the angular displacement, ωi is the initial angular speed, t is the time, and α is the angular acceleration.
Given in the problem, we have:
ωi = 0.50 rad/s, t = 7.00 s, and α = 0.10 rad/s^2.
Substituting these values into the equation, we get:
θ = (0.50 rad/s * 7.00 s) + 0.5 * (0.10 rad/s^2 * (7.00 s)^2) θ = 3.50 rad + 0.5 * 0.10 rad/s^2 * 49 s^2 θ = 3.50 rad + 2.45 rad θ = 5.95 rad
However, the acceleration only occurs over a 4.0-s interval. So, we need to subtract the displacement that occurs during the last 3.0 s (7.0 s - 4.0 s) of uniform motion (α = 0).
The displacement during uniform motion is given by:
θ_uniform = ω*t θ_uniform = 0.50 rad/s * 3.0 s θ_uniform = 1.50 rad
Subtracting this from the total displacement, we get:
θ_total = θ - θ_uniform θ_total = 5.95 rad - 1.50 rad θ_total = 4.45 rad
However, this answer is not in the options. There might be a mistake in the problem or the options.
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