A Ferris wheel, rotating initially at an angular speed of 0.500 rad/s, accelerates over a 7.00-s interval at a rate of 0.0500 rad/s2. What angular displacement does the Ferris wheel undergo in this 7.00-s interval?Select one:a.2.50 radb.4.73 radc.4.48 radd.0.500 rad
Question
A Ferris wheel, rotating initially at an angular speed of 0.500 rad/s, accelerates over a 7.00-s interval at a rate of 0.0500 rad/s2. What angular displacement does the Ferris wheel undergo in this 7.00-s interval?Select one:a.2.50 radb.4.73 radc.4.48 radd.0.500 rad
Solution
To solve this problem, we can use the formula for angular displacement in uniformly accelerated rotational motion, which is similar to the formula for linear displacement in uniformly accelerated linear motion. The formula is:
θ = ω₀t + 0.5αt²
where: θ is the angular displacement, ω₀ is the initial angular speed, t is the time, and α is the angular acceleration.
Given in the problem are: ω₀ = 0.500 rad/s, t = 7.00 s, and α = 0.0500 rad/s².
Substituting these values into the formula gives:
θ = (0.500 rad/s)(7.00 s) + 0.5(0.0500 rad/s²)(7.00 s)² θ = 3.50 rad + 0.5(0.0500 rad/s²)(49 s²) θ = 3.50 rad + 0.5(0.0500 rad/s²)(49 s²) θ = 3.50 rad + 1.225 rad θ = 4.725 rad
So, the angular displacement of the Ferris wheel in this 7.00-s interval is approximately 4.73 rad. Therefore, the correct answer is (b) 4.73 rad.
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