• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Physics question: non-uniform circular motion

P

PCMarine

Diamond Member
A crankshaft, with a diameter of 3.00 cm, rotating at 2200 rpm comes to a halt in 1.50 s.

Q1: What is the tangential acceleration of a point on the surface of the crankshaft?

I think I have to convert 2200 rpm to m/s (velocity), to find the acceleration (change in velocity/1.50s). Then divide the acceleration by the radius. However, I don't know how to do the rpm to m/s conversion.


Q2: How many revolutions does the crankshaft make as it stops?

😕
 
Find the circumference of the shaft, which is easy. Now, you know the velocity at the surface is RPM * circumference (divide by 6000 to convert to m/s, by default it will be cm/min)

Start with that, let us know how it goes.
 
for Q2..now that you have accel...you can use a simple physics equation to calculate how far it will go before it slows to 0 m/s. Divide this by the circumference
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top