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A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total acceleration vector at the edge of the wheel at 1.0 s?

A. 0.18 m/s², tangential
B. 0.36 m/s², radial
C. 0.54 m/s², tangential
D. 0.72 m/s², radial

User Kevinnls
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1 Answer

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Final answer:

The magnitude and direction of the total acceleration vector at the edge of the wheel at 1.0 s can be determined using the formulas for angular acceleration and magnitude of total acceleration.

Step-by-step explanation:

The magnitude and direction of the total acceleration vector at the edge of the wheel at 1.0 s can be determined using the formula:

angular acceleration = final angular velocity - initial angular velocity / time

Given that the wheel rotates from rest to 3 rev/s in 5 s, the initial angular velocity is 0 rad/s and the final angular velocity is 6π rad/s.

Substituting the values into the formula:

angular acceleration = (6π rad/s - 0 rad/s) / 5 s = 1.2π rad/s²

The magnitude of the total acceleration vector at the edge of the wheel can be calculated using the formula:

magnitude of total acceleration = radius × angular acceleration

Given that the radius of the wheel is 0.3 m:

magnitude of total acceleration = 0.3 m × 1.2π rad/s² ≈ 0.94 m/s²

Since the wheel starts from rest, the initial direction of the total acceleration vector is tangential.

Therefore, the correct answer is option C. 0.54 m/s², tangential.

User Hernani Fernandes
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