Answered

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A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while accelerating?

Sagot :

Trancescient

12.8 rad

Explanation:

The angular displacement [tex]\theta[/tex] through which the wheel turned can be determined from the equation below:

[tex]\omega^2 = \omega_0^2 + 2\alpha\theta[/tex] (1)

where

[tex]\omega_0 = 0[/tex]

[tex]\omega = 34.7\:\text{rad/s}[/tex]

[tex]\alpha = 47.0\:\text{rad/s}^2[/tex]

Using these values, we can solve for [tex]\theta[/tex] from Eqn(1) as follows:

[tex]2\alpha\theta = \omega^2 - \omega_0^2[/tex]

or

[tex]\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}[/tex]

[tex]\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}[/tex]

[tex]\:\:\:\:= 12.8\:\text{rad}[/tex]

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