Two problems involving angular velocity:

AI Thread Summary
The discussion revolves around calculating the angular velocity of the Sun as it orbits the center of the Milky Way galaxy. The initial calculation used the formula 2π/T but resulted in an incorrect distance due to unit conversion errors. After clarification, it was confirmed that the correct approach involves converting the period from years to seconds and ensuring consistent units throughout the calculation. The final correct answer was achieved by converting centimeters to kilometers. The importance of unit conversion in physics problems is emphasized in this discussion.
Bob Loblaw
Messages
67
Reaction score
0
Two problems involving angular velocity:

Edit: Problem one solved! Hooray!

2.The Milky Way galaxy rotates about its center with a period of about 192 million years. The Sun is 2x10^20 m from the center of the galaxy. How fast is the Sun moving with respect to the center of the galaxy?

For this problem I took 2pi/T which gave me 2pi/192000000 and multiplied the result by 2x10^20m. I then expressed the answer in kilometers which gave me 6544984695km. This was not correct. Any suggestions?

Thanks for your generous help!
 
Last edited:
Physics news on Phys.org
The number you have is in units of km/year... are these the units you need?
 
Yes - it was indeed a problem of units. I took 2pi/(195 million years)(second in a year) * 2.0x10^20 and came up with the correct answer once I converted centimeters into kilometers.

Thanks!
 
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