Direction of friction in a yoyo rolling on an inclined plane

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Friction acts downwards on a yoyo rolling down an inclined plane to prevent slipping, as the yoyo would rotate in the opposite direction without friction. If there were no friction, the yoyo would rotate upwards, causing the contact point to slip upwards on the incline. To counteract this tendency, friction must act downwards. The difference in perimeter between the yoyo and the cylinder affects the distance traveled per revolution, necessitating friction to maintain rolling motion. The discussion also considers how the direction of the applied force affects the frictional force's direction.
dahoom102
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Homework Statement
In this problem why is the friction directed downwards not upwards? Shouldn't the velocity of the point of contact with the ground be zero and therefore mgsin = fs ? All of the problems i studied the friction was in the upward direction i just wanna know why is it different here.
Relevant Equations
F.B.D
Mgsin
fs
Screenshot_20210304-190909_Drive.jpg
 
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dahoom102 said:
In this problem why is the friction directed downwards not upwards?
If there was no friction, which way would the yoyo rotate? And, therefore, which way would the contact point of the yoyo tend to slip on the surface of the incline? Which way must friction act to prevent such slipping?
 
The perimeter of the yo-yo is bigger than the perimeter of the cylinder around which the syring is wrapped.
For one revolution, the traveled distance and the length of string are different: one must yield.

Please, see:
https://www.physicsforums.com/insights/explaining-rolling-motion/

:cool:
 
dahoom102 said:
In this problem why is the friction directed downwards not upwards?
Take moments about the centre of the yoyo. What does that tell you about F and fs?
What if the string were wound the other way around, so that F is applied above the centre?
 
Thread 'Chain falling out of a horizontal tube onto a table'

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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