Mistake in Exercises for the Feynman Lectures?

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The discussion centers on the interpretation of the velocity components of a point on a rolling wheel, specifically focusing on the angle θ and its impact on velocity calculations. The initial assertion is that the velocity should be expressed as v = V((1+sinθ)i -(cosθ)j), based on the expected behavior of the wheel's motion. Clarification is sought on whether this expression accurately predicts the speed of point P on the rim relative to the ground contact point. The question highlights the relationship between the angle of rotation and the resulting velocity components. Ultimately, a misreading of the question is acknowledged, leading to a better understanding of the problem.
suh112
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Homework Statement
This is problem 14.1 from "Exercises for the Feynman Lectures on Physics":

14.1 A rigid wheel of radius R is rolling without slipping on a horizontal surface. The plane of the wheel is vertical, and the axis of the wheel is moving horizontally with a speed V relative to the surface. If the axis of the wheel is parallel to the z-axis, V is in the positive x-direction, and ##\theta## the angle through which the wheel has rotated since a certain point P on the rim was in contact with the ground, show that the instantaneous velocity (speed and direction) of the point P is given by
v = V ((1- cos##\theta##)i + (sin ##\theta##) j).
Relevant Equations
v = V ((1- cos##\theta##)i + (sin ##\theta##) j)
It seems to me that the answer should be v = V((1+sinθ)i -(cosθ)j) intuitively since ##V_x## should be zero at θ = −π\2 and should be greatest when the angle is 90 degrees. Similarly, the component of velocity in the y direction should be greatest when the angle ##\theta## is 180 degrees and zero when ##\theta## is 0 degrees. Am I doing something wrong?
 

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According to the statement of the question, angle θ is "the angle through which the wheel has rotated since a certain point P on the rim was in contact with the ground". If the wheel rolls without slipping what is the speed of certain point P on the rim relative to the point of contact with the ground? Does the given expression predict that? Does your expression predict that?
 
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kuruman said:
According to the statement of the question, angle θ is "the angle through which the wheel has rotated since a certain point P on the rim was in contact with the ground". If the wheel rolls without slipping what is the speed of certain point P on the rim relative to the point of contact with the ground? Does the given expression predict that? Does your expression predict that?
Oh I misread the question. This makes sense thanks.
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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