Velocity of an Object in Uniform Circular Motion

AI Thread Summary
An object with a mass of 0.5 kg is in uniform circular motion with a radius of 2 meters and a force of 4 N acting on it. The attempt to calculate the velocity using the centripetal force equation revealed an algebra mistake, specifically in treating a product as a sum. The discussion emphasizes the importance of maintaining significant figures only in the final answer to avoid compounding errors. The correct approach to solving the problem involves careful algebraic manipulation and understanding of the equations involved. Overall, the thread highlights common pitfalls in physics calculations and the significance of accuracy in mathematical operations.
Argella
Messages
2
Reaction score
0

Homework Statement


An object of mass 0.5 kg is swung in uniform circular motion. The radius is 2 meters, and the force exerted is 4 N. Calculate the magnitude of the velocity.

Answer Choices:
a) 0.25 m/s
b) 1 m/s
c) 4 m/s
d) 16 m/s

Homework Equations


v = 2piR/T
F = mv^2/R

The Attempt at a Solution


Since I didn't know the variable time, I couldn't use the velocity equation. Next, I tried plugging in the variables for the centripetal force equation and solving for velocity:

4 = (0.5)v^2/2
4 = 0.5/2 + v^2/2
4 = 0.3 + v^2/2 <- 0.3 because of significant figures
(subtracted 0.3 from both sides)
4 = v^2/2 <- 4 because of significant figures
(multiplied both sides by 2)
8 = v ^2
(took the square root of both sides)
v = 3 which isn't one of my answer choices.

Maybe I'm not doing my algebra correctly or maybe I'm not using the right equation, but I couldn't find another equation using the variables I have. If I knew the time it took for one revolution I could use the velocity equation, but I don't know how to find time and I couldn't find an equation for finding it.
 
Physics news on Phys.org
Argella said:
4 = (0.5)v^2/2
4 = 0.5/2 + v^2/2
There is an algebra mistake in going from the first to the second step. You have essentially turned a product into a sum instead.

On a side note, when taking significant figures into account, do so only on the final answer. If you keep rounding off at each step, you will find that the rounding-off error grows substantially larger after each step, and the final answer you obtain might be significantly off the correct one.
 
Ohh, okay, that makes a lot more sense. I'll be sure to keep that in mind about significant figures in the future, too.

Thanks a lot!
 
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