Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

No problem!