## anonymous one year ago A crankshaft with a diameter of 3.0 cm, rotating at 2200 rpm comes to a halt in 1.50 s . What is the tangential acceleration of a point on the surface of the crankshaft?

1. anonymous

2200rpm=2200 revolution per minutes $\frac{ 2200 revolution }{ \min } \times \frac{ 1\min }{ 60seconds } \times \frac{ 2\pi radians }{ 1 revolution }$=230.283 radians/seconds

2. anonymous

$\alpha=\frac{ v }{ t }$ Angular acceleration=velocity of circular motion divided by time. $\alpha=\frac{ 230.283 radians/seconds }{ 1.50 seconds }$ =$15.36 radians/seconds^2$

3. anonymous

$a _{t}=\alpha r$ where tangential acceleration= angular acceleration x radius. Radius would have to be in meters. Radius=diameter/2

4. anonymous

The final answer was -2.30 m/s^2. I wouldn't have figured it out without your help, thank you so much.

5. anonymous

No problem!