Question

the drive propeller of a ship starts from rest and accelerates at 2.09x10^-3 rad/s² for 2.58x10³ s. for the next 1.90x10³ s the propeller rotates at a constant angular speed. then it decelerates at 2.75x10^-3 rad/s² until it slows (without reversing direction) to an angular speed of 2.28 rad/s. find the total angular displacement of the propeller.

Explanation:

Step1: Find angular speed after acceleration

Use the formula \(\omega=\omega_0+\alpha t\). Here \(\omega_0 = 0\) (starts from rest), \(\alpha=2.09\times 10^{-3}\text{ rad/s}^2\), \(t = 2.58\times 10^{3}\text{ s}\).
\(\omega=(0)+(2.09\times 10^{-3})\times(2.58\times 10^{3})\)
\(\omega = 5.3922\text{ rad/s}\)

Step2: Find angular displacement during acceleration

Use the formula \(\theta_1=\omega_0t+\frac{1}{2}\alpha t^{2}\). Since \(\omega_0 = 0\), \(\theta_1=\frac{1}{2}\times(2.09\times 10^{-3})\times(2.58\times 10^{3})^{2}\)

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LATEXBLOCK0
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Step3: Find angular displacement during constant - speed

Use the formula \(\theta_2=\omega t\). Here \(\omega = 5.3922\text{ rad/s}\), \(t = 1.90\times 10^{3}\text{ s}\)
\(\theta_2=(5.3922)\times(1.90\times 10^{3})=10245.18\text{ rad}\)

Step4: Find angular displacement during deceleration

Use the formula \(\omega_f^{2}=\omega^{2}+2\alpha\theta_3\). Here \(\omega_f = 2.28\text{ rad/s}\), \(\omega = 5.3922\text{ rad/s}\), \(\alpha=- 2.75\times 10^{-3}\text{ rad/s}^2\)

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LATEXBLOCK1
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Step5: Find total angular displacement

\(\theta=\theta_1+\theta_2+\theta_3\)
\(\theta=6942.018 + 10245.18+4341.345\)
\(\theta = 21528.543\text{ rad}\approx2.15\times 10^{4}\text{ rad}\)

Answer:

\(2.15\times 10^{4}\text{ rad}\)