Relating the Linear and Angular Variables — Question 1

Question 1

Tangential and radial accelerations point in different directions

A point on a fixed-axis disk is instantaneously at $(0.40,0)\,\mathrm m$. The disk has $\omega=+3\,\mathrm{rad/s}$ and $\alpha=+5\,\mathrm{rad/s^2}$ about $+z$.

1. Find its velocity and acceleration vectors.

2. Find acceleration magnitude and explain why $r\alpha$ alone is insufficient.

Question 1 - Solution

Tangential and radial accelerations point in different directions

(a) Derivation and calculation

For fixed-axis rotation, use the perpendicular tangential and radial contributions: $$\vec v=\vec\omega\times\vec r=3\hat z\times0.40\hat x =\boxed{1.20\hat y\ \mathrm{m/s}}.$$ $$\vec a_t=\vec\alpha\times\vec r=2.00\hat y,\qquad \vec a_r=\vec\omega\times(\vec\omega\times\vec r)=-3.60\hat x.$$ $$\boxed{\vec a=(-3.60,2.00)\ \mathrm{m/s^2}}.$$

(b) Result and interpretation

Since the two components are perpendicular, $$\boxed{|\vec a|=\sqrt{(r\omega^2)^2+(r\alpha)^2} =\sqrt{3.60^2+2.00^2}=4.118\ \mathrm{m/s^2}}.$$ The term $r\alpha$ accounts for speed change; the inward term accounts for direction change.

Physical check

The velocity is tangent to the circle while the radial acceleration points toward the axis.