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lakkis [162]
3 years ago
6

A wheel decelerates from 13.5 rad s−1 to 6.0 rad s−1 in 7 s. Calculate the angular displacement​

Physics
1 answer:
Igoryamba3 years ago
3 0

Answer:

<em>Angular displacement=68.25 rad</em>

Explanation:

<u>Circular Motion</u>

If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

\displaystyle \alpha=\frac{\omega_f-\omega_o}{t}

The angular displacement is given by:

\displaystyle \theta=\omega_o.t+\frac{\alpha.t^2}{2}

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:

\displaystyle \alpha=\frac{6-13.5}{7}

\displaystyle \alpha=\frac{-7.5}{7}

\displaystyle \alpha=-1.071 \ rad/s^2

Thus, the angular displacement is:

\displaystyle \theta=13.5*7+\frac{-1.071*7^2}{2}

\displaystyle \theta=94.5-26.25

\boxed{\displaystyle \theta=68.25\ rad}

Angular displacement=68.25 rad

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Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

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From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

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Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

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Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

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<h2>Answer</h2>

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<h2>Explanation</h2>

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