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

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:

[tex]a_{total}^2=a_r^2+a_t^2[/tex] (1)

atotal: total acceleration = 1.2m/s^2

ar: radial acceleration of the wheel

at: tangential acceleration

The tangential acceleration is also given by:

[tex]a_t=r\alpha[/tex] (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:

[tex]a_{total}^2=a_r^2+(r\alpha)^2\\\\a_r=\sqrt{(a_{total})^2-(r\alpha)^2}\\\\a_r=\sqrt{(1.2m/s^2)^2-((0.2m)(4.0rad/s^2))^2}=0.894\frac{m}{s^2}[/tex]

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

[tex]a_r=\frac{v^2}{r}\\\\v=\sqrt{ra_r}=\sqrt{(0.2m)(0.894m/s^2)}=0.42\frac{m}{s}[/tex]

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

Lanuel Lanuel · Answer 2 · 2021-11-30T16:32:46+01:00

The linear speed of a point on the rim of this wheel is equal to 0.423 m/s

Given the following data:

Angular acceleration = 4.0 [tex]m/s^2[/tex]

Diameter = 40 cm to meter = 0.4 m.

Total linear acceleration = [tex]1.2 \;m/s^2[/tex]

Conversion:

Radius = [tex]\frac{Diameter}{2} =\frac{0.4}{2} =0.2\;m[/tex]

To determine the linear speed of a point on the rim of this wheel:

First of all, we would determine the tangential acceleration of the wheel.

[tex]Tangential \;acceleration = radius \times angular \;acceleration\\\\Tangential \;acceleration =0.2 \times 4.0[/tex]

Tangential acceleration = 0.8 [tex]m/s^2[/tex]

Mathematically, the total linear acceleration is given by the formula:

[tex]a_{t}^2 = a_r^2 + \alpha _{t}^2[/tex]

Where:

[tex]a_r[/tex] is the radial acceleration.

[tex]\alpha _{t}[/tex] is the tangential acceleration.

[tex]a_t[/tex] is the total linear acceleration.

Making the subject of formula, we have:

[tex]a_{r} = \sqrt{a_t^2 - \alpha _{t}^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]a_{r} = \sqrt{1.2^2 - 0.8^2}\\\\a_{r} = \sqrt{1.44 - 0.64}\\\\a_{r} = \sqrt{0.8}\\\\a_r = 0.894 \;m/s^2[/tex]

Now, we can determine the linear speed of a point on the rim of this wheel:

[tex]a_r = \frac{v^2}{r} \\\\V= \sqrt{ra_r} \\\\V=\sqrt{0.2 \times 0.894} \\\\V=\sqrt{0.1788}[/tex]

V = 0.423 m/s

Read more: https://brainly.com/question/8898885