Starting from rest at t = 0 s, a wheel undergoes a constant angular acceleration. When t = 3.8 s, the angular velocity of the wheel is 4.8 rad/s. The acceleration continues until t = 20 s, when the acceleration abruptly changes to 0 rad/s2. Through what angle does the wheel rotate in the interval t = 0 s to t = 55 s?

Question

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Respuesta :

aristeus aristeus · Answer 1 · 2019-11-13T06:46:06+01:00

Answer:

Total displacement will be 1136.79 rad

Explanation:

We have given that wheel starts from the rest so initial angular velocity [tex]\omega _0=0rad/sec[/tex]

Angular velocity after 3.8 sec is given as 4.8 rad/sec

So [tex]\omega _f=4.8rad/sec[/tex]

From second equation of motion we know that

[tex]\omega _f=\omega _0+\alpha t[/tex]

[tex]\alpha =\frac{\omega _f-\omega _i}{t}=\frac{4.8-0}{3.8}=1.2631rad/sec^2[/tex]

This acceleration is constant until 20 sec

So displacement in 20 sec

[tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 20+\frac{1}{2}\times 1.2631\times 20^2=252.62rad[/tex]

Total time is given as t = 55 sec

So left time = 55 - 20 =35 sec

After 20 sec angular acceleration is 0

So displacement in 35 sec

[tex]Theta =\omega _0t[/tex]

Angular velocity after 20 sec [tex]\omega _0=\alpha t=1.2631\times 20=25.262rad/sec[/tex]

So displacement [tex]\Theta =35\times 25.262=884.17rad[/tex]

So total displacement in 55 sec = 252.62 +884.17 = 1136.79 rad