Question

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

Answer 1

Answer:

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:

$a_{total}^2=a_r^2+a_t^2$       (1)

atotal: total acceleration = 1.2m/s^2

ar: radial acceleration of the wheel

at: tangential acceleration

The tangential acceleration is also given by:

$a_t=r\alpha$     (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:

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

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

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

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