Centripetal acceleration changing radians/s into rpm's

AI Thread Summary
To achieve a centripetal acceleration of 100g for a copper sleeve with inner and outer radii of 2.10 cm and 2.20 cm, the required angular speed can be calculated using the formula ac = rω². The calculation leads to ω = √(100g / 0.021 m), which simplifies to approximately 216 radians per second. Converting this to revolutions per minute involves using the conversion factor of 1 rev per 2π radians and multiplying by 60 seconds per minute, resulting in about 2060 rpm. The user is seeking clarification on using a TI-83 calculator for these calculations and has noted a potential error in their initial approach. The discussion emphasizes the importance of showing work in detail rather than just providing an answer.
mhuffman
Messages
2
Reaction score
0

Homework Statement


Centrifugal castings- a copper sleeve of inner radius 2.10 cm and outer radius 2.20 cm is to be cast. To eliminate bubbles and give high structural integrity, the centripetal acceleration of each bit of metal should be 100g. What rate of rotation is required? State answer in revolutions per minute.


Homework Equations

r=radius, ac=centripetal acceleration, ω=angular speed, g=gravity.
What I have right now is: ac=rω2, so every particle must undergo 100g of acceleration giving ac=100g or rω2=100g reduced to;
ω=√100(9.8m/s)/2.10 x 10-2m
the conversion factor will be the answer in rads/s (1rev/2π)(60s/1min)


The Attempt at a Solution


My problem is converting this into radians per second then into rpm's. I am not looking for the answwer per say, but how to put this into my Ti-83 plus and doing the math. I also need to be able to show my work in long hand and just throwing out an answer will not work. I tried using the r from the angle setting and using both radians and degree mode, I looked in the manual but it use of no use. I am sure I am just overlooking something simple, any help would be great.

Thanks and have a great day.


 
Physics news on Phys.org
Oops, I obviously made a mental/keystroke error. I believe I got it right, I have:
ω=√100g/2.10 x 10-2m
ω=216 rads/s=rpm=216rads/s(1rev/2∏rad)(60.0s/1min)=216(30/∏) leaving rpm's
216rads/s=2.06 x 103 rev/min
If you happen to see any mistakes, feel free to comment,

Thanks and have a great day,
m
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top