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A rotating wheel requires 2.92 seconds to rotate through 37.0 revolutions. Its angular speed at the end of the 2.92-second interval is 97.8 rad/s. What is the constant angular acceleration of the wheel?

Answer :

Answer:

Therefore the constant angular acceleration of the wheel 12.43 rad /s².

Explanation:

[tex]\triangle \theta =\omega_{av}t=\frac{\omega_f+\omega_i}{2}t[/tex]

[tex]\Rightarrow \triangle \theta =\frac{\omega_f+\omega_i}{2}t[/tex]

[tex]\Rightarrow \frac{2\triangle \theta}{t} ={\omega_f+\omega_i}[/tex]

[tex]\Rightarrow {\omega_i= \frac{2\triangle \theta}{t} -\omega_f}[/tex]

and

[tex]\alpha =\frac{\omega_f-\omega_i}{t}[/tex]

[tex]\omega_i[/tex] = initial angular velocity

[tex]\omega_f[/tex] = final angular velocity

[tex]\theta[/tex] = displacement

[tex]\alpha[/tex] = angular acceleration

t = time

Here [tex]\triangle \theta[/tex] = 37.0 revolution [tex]=37 \times(2\pi)[/tex] rad [ since one revolution = [tex]2 \pi[/tex]]

t=2.92 s

Final angular velocity = [tex]\omega_f[/tex] = 97.8 rad/s

To find the angular velocity, first we need to find out the initial angular velocity [tex](\omega_i)[/tex].

[tex]{\omega_i= \frac{2\triangle \theta}{t} -\omega_f}[/tex]

[tex]=\frac{2\times2\pi \times 37.0}{2.92}-97.8[/tex]

= 61.50 rad /s

Then the angular velocity is

[tex]\alpha =\frac{\omega_f-\omega_i}{t}[/tex]

[tex]=\frac{97.8-61.50}{2.92}[/tex] rad /s²

=12.43 rad /s²

Therefore the constant angular acceleration of the wheel 12.43 rad /s².

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Rewritten by : Jeany

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