A community for students.
Here's the question you clicked on:
 0 viewing
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?
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?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02200rpm=2200 revolution per minutes \[\frac{ 2200 revolution }{ \min } \times \frac{ 1\min }{ 60seconds } \times \frac{ 2\pi radians }{ 1 revolution }\]=230.283 radians/seconds

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\alpha=\frac{ v }{ t }\] Angular acceleration=velocity of circular motion divided by time. \[\alpha=\frac{ 230.283 radians/seconds }{ 1.50 seconds }\] =\[15.36 radians/seconds^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a _{t}=\alpha r\] where tangential acceleration= angular acceleration x radius. Radius would have to be in meters. Radius=diameter/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The final answer was 2.30 m/s^2. I wouldn't have figured it out without your help, thank you so much.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.