Tension & Circular Motion Question - Looking for speed

Click For Summary
The discussion revolves around a tension and circular motion problem, where the original poster seeks confirmation on their interpretation of the question and calculations. They utilize the equation a = v²/r, relating radius and speed, but encounter discrepancies in their final numerical answer. Other participants confirm that while the steps are correct, the poster's calculator was set to radians instead of degrees, leading to errors. Additionally, there is a caution about handling negative signs in calculations, emphasizing the importance of accuracy in physics problems. Overall, the conversation highlights common pitfalls in physics calculations and the importance of double-checking settings and signs.
dcmf
Messages
16
Reaction score
5
Homework Statement
A person sitting in a chair (combined mass 80 kg) is attached to a 6.0-m-long cable. The person moves in a horizontal circle. The cable angle θ is 62 degrees below the horizontal. What is the person's speed? Note: The radius of the circle is not 6.0 m.
Relevant Equations
a = v^2/r
I have attached a screenshot of my rough work. First of all, is my interpretation of the question correct? Please see the diagram in purple. To me, this makes sense because a=v^2/r is the only equation from my coursework that seems to relates radius (which you can find from the length of the cable) and speed.
1707605967159.png
 
Physics news on Phys.org
Hello @dcmf,
:welcome:

dcmf said:
is my interpretation of the question correct?
I guess so. Apparently the person is not in a wheelchair going around slowly ?

What do you do with the minus sign ?
And when I do the calculation, I get a different answer.

##\ ##
 
Welcome, @dcmf !

All the steps seem to be correct, but the final numerical calculation is incorrect.
As Tx and Ty are directly proportional to the horizontal and vertical accelerations respectively, you could have used those values directly.

##tan~28=ac/g##
 
Hi all, thank you for your replies. When you both said you're getting a different number doing the same calculation, I realized my calculator has been in radians and not degrees this entire time 🤡

Thanks for your patience and help :')
 
  • Like
Likes Lnewqban, PhDeezNutz and BvU
dcmf said:
my calculator has been in radians and not degrees
Happens often (and to all of us :wink:). Reason the more to check things, e.g. ##\sin(28^\circ)\approx 0.5## -- so you learn to smell a rat if you get ##0.27##

##\ ##
 
Screen Shot 2024-02-10 at 7.11.41 PM.png

How did you handle the negative sign under the radical? You can't simply ignore it because it shouldn't be there. Think about this because you might will get into trouble if you replace ##g## with ##-9.80~\text{m/s}^2## indiscriminately.
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
I treat this question as two cases of Doppler effect. (1) When the sound wave travels from bat to moth Speed of sound = 222 x 1.5 = 333 m/s Frequency received by moth: $$f_1=\frac{333+v}{333}\times 222$$ (2) When the sound wave is reflected from moth back to bat Frequency received by bat (moth as source and bat as observer): $$f_2=\frac{333}{333-v}\times f_1$$ $$230.3=\frac{333}{333-v}\times \frac{333+v}{333}\times 222$$ Solving this equation, I get ##v=6.1## m/s but the answer key is...