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A wheel rotates with a constant angular velocity of [tex]2.00 \, \text{rad/s}[/tex]. Compute the radial acceleration of a point [tex]0.500 \, \text{m}[/tex] from the axis. The radial acceleration is [tex]2.00 \, \text{rad/s}^2[/tex]. What is the tangential speed of the point?

A. [tex]1.00 \, \text{m/s}, \, 4.00 \, \text{m/s}^2[/tex]
B. [tex]1.00 \, \text{m/s}, \, 2.00 \, \text{m/s}^2[/tex]
C. [tex]2.00 \, \text{m/s}, \, 4.00 \, \text{m/s}^2[/tex]
D. [tex]2.00 \, \text{m/s}, \, 2.00 \, \text{m/s}^2[/tex]

Answer :

Final Answer:

The radial acceleration of the point is 2.00 m/s² and the tangential speed of the point is 1.00 m/s, thus the correct option is b.

Explanation:

The radial acceleration of a point rotating with a constant angular velocity can be determined using the formula: a_r = r(α²), where a_r is the radial acceleration, r is the distance of the point from the axis of rotation, and α is the angular velocity of the wheel. In this case, a_r = (0.500 m)(2.00 rad/s²) = 2.00 m/s².

To find the tangential speed of the point, we use the formula: v = rω, where v is the tangential speed, r is the distance of the point from the axis of rotation, and ω is the angular velocity of the wheel. Therefore, v = (0.500 m)(2.00 rad/s) = 1.00 m/s.

In summary, the radial acceleration of the point is 2.00 m/s² and the tangential speed of the point is 1.00 m/s. This means that the point is experiencing an inward acceleration towards the axis of rotation, while moving tangentially at a constant speed of 1.00 m/s, thus the correct option is b.

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