A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s2. The diameter of the wheel is 40 cm. What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s2

v = 0.42m/s

Explanation:

In order o calculate the linear speed of the point at the border of the wheel, you first take into account that the total acceleration of such a point is given by:

      (1)

atotal: total acceleration = 1.2m/s^2

ar: radial acceleration of the wheel

at: tangential acceleration

The tangential acceleration is also given by:

    (2)

r: radius of the wheel = (40cm/2 )= 20cm = 0.2m

α: angular acceleration = 4.0rad/s^2

You replace the expression (2) into the expression (1) and solve for the radial acceleration:

Next, you use the following formula for the radial acceleration and solve for the linear speed:

The linear speed of the point at the border of the wheel is 0.42m/s

the answer is 12 as i may remember

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