Angular velocity of a rolling object

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ZD_zainy
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Homework Statement
The question is in the attachment. I chose the 1st graph. But according to the scheme the correct one is the 3rd graph. I do not know what causes the sudden increase in the 3rd graph. Plz explain it
Relevant Equations
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Handwritten physics question: a cylinder at rest on a rough incline above a floor, with four labelled graphs 1 to 4 of angular velocity omega against time t, each showing a different curve shape.
 
Attachments
  • Handwritten physics problem: a cylinder resting at the top of a rough inclined surface above a floor, with four candidate graphs of angular velocity ω against time t labelled 1 to 4.
    5529EEF2-DBDF-4835-91E9-8F24B0157634.jpeg
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Hello @ZD_zainy , :welcome: !

Please read the PF guidelines: we are not allowed to help if you don't show your own attempt. What does one see in the graphs ?
 
It can help to consider extreme cases. What if the cylinder were already at the end of the slope when released?
 
BvU said:
Hello @ZD_zainy , :welcome: !

Please read the PF guidelines: we are not allowed to help if you don't show your own attempt. What does one see in the graphs ?
How can I show my attempt which I made to find a graph?
 
haruspex said:
It can help to consider extreme cases. What if the cylinder were already at the end of the slope when released?
Oh sorry . The question’s diagram itself shows how the cylinder is in the slope
 
ZD_zainy said:
How can I show my attempt which I made to find a graph?
You write up what leads you to pick 1) and discuss the differences between 1) and 3). Where does it occur ?

Personally, I think the increase in ##\omega## is exaggerated (it looks almost like a jump)
 
For the angular velocity to exhibit a sudden increase at the very end when it flies off the incline, there must be an increased torque acting on it. Remember that a rolling object rotates about the instantaneous point of contact. What happens when it reaches the edge just before contact is lost?
 
ZD_zainy said:
Oh sorry . The question’s diagram itself shows how the cylinder is in the slope
You miss my point.
You are puzzled that the correct answer shows a sudden acceleration just as the cylinder reaches the end of the slope. The reason may become clear if you consider the cylinder released from rest when it is already touching the end of the slope.
Draw the diagram for that.
 
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