Question

Your car tire is rotating at 3.3rev/s when suddenly you press down hard on the accelerator. After traveling 220m , the tire's rotation has increased to 6.1rev/s . The radius of the tire is 32 cm. What was the tire's angular acceleration? Give your answer in rad/s2.

Answer 1

Well, you first find the time (t) through this kinematic equation. You need to consider that the velocity (v) is equal to the agular velocity (w) times the radius (r).\[x _{f}=x _{i}+\frac{ v _{i}+v _{f} }{ 2 }t \rightarrow t=2\frac{ \Delta x }{ r \Delta w }\]The symbol Delta means change (i.e. final-initial).

Then we proceed to figure out the angular acceleration:\[\alpha =\frac{ \Delta \omega }{ t }=\frac{ \Delta \omega }{ 2 \frac{ \Delta x }{ r \Delta \omega } }=\frac{ r (\Delta w)^{2}}{ 2 \Delta x}=\frac{ 0.32m((6.1-3.3)\frac{ rev }{ s }\frac{ 2 \pi rad }{ 1 rev })^{2} }{ 2 \times 220m }\]