Calculating Kinetic Energy of a Segment: Where Did I Go Wrong?

AI Thread Summary
The discussion revolves around calculating the kinetic energy of a segment, where the original poster struggles with their solution and seeks help. Key errors identified include the incorrect use of 1 + sin(θ) instead of the correct expression and miscalculating the integral of sin(θ) as zero. An alternative approach using angular speed and distance from the point of contact is suggested, which simplifies the problem. The poster successfully corrects their calculations and inquires about the limits of integration, which are clarified to be from 0 to π. Overall, the conversation emphasizes the importance of accurate trigonometric representation and integration limits in solving kinetic energy problems.
Saitama
Messages
4,244
Reaction score
93

Homework Statement


(see attachment 1)


Homework Equations





The Attempt at a Solution


I have attached my working, the correct answer is (A) but I can't see where I went wrong.

Any help is appreciated!
 

Attachments

  • ring.jpg
    ring.jpg
    18.2 KB · Views: 459
  • attempt.jpg
    attempt.jpg
    16.9 KB · Views: 444
Physics news on Phys.org
Your first line is wrong. How did you get 1+sin(θ)? You're measuring it down fron horizontal, so should be minus.
And in the last step, you've evaluated the integral of the sin(θ) part to zero. It won't be.

Another approach is to say the angular speed is v/R and work out the distance from the element to the point of contact.
 
haruspex said:
Your first line is wrong. How did you get 1+sin(θ)? You're measuring it down fron horizontal, so should be minus.
And in the last step, you've evaluated the integral of the sin(θ) part to zero. It won't be.
Oh yes, the integral sin(θ) won't be zero. I evaluated it correctly during the examination but did it wrong while posting up my attempt. :-p
I did the necessary corrections as you said and got the right answer. One more question, would the limits for integration change from 0 to -pi?

haruspex said:
Another approach is to say the angular speed is v/R and work out the distance from the element to the point of contact.
This approach is much easier. I am not much familiar with the instantaneous axis of rotation but this worked out pretty well. Thanks for the alternate method. :smile:
 
Pranav-Arora said:
One more question, would the limits for integration change from 0 to -pi?
No, you're measuring theta anticlockwise in the diagram from the "9 o'clock" position, so it will be 0 to pi.
 
haruspex said:
No, you're measuring theta anticlockwise in the diagram from the "9 o'clock" position, so it will be 0 to pi.

Got it, thanks haruspex!
 
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