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!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top