Hard kinematics problem: Block pushed onto a moving conveyor belt

AI Thread Summary
The discussion revolves around solving a kinematics problem involving a block on a moving conveyor belt, focusing on the effects of friction and velocity components. Participants emphasize the importance of using the coefficient of friction, denoted as μ, even when not explicitly provided, to analyze the motion. They explore the relationship between the block's velocity and the forces acting on it, particularly in the belt's frame of reference. The conversation includes attempts to derive equations of motion and clarify the signs of velocity and acceleration components. Overall, the key takeaway is the need to carefully consider the frame of reference and the effects of friction on the block's motion.
  • #51
PeroK said:
##\frac 2 {\sqrt 5} m/s## is the answer I get.

In general the minimum speed is ##v_b \cos \theta'##, where ##v_b## is the speed of the belt and ##\theta'## is the angle of motion relative to the belt.
can you please explain further will be a great help
 
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  • #52
PeroK said:
##\frac 2 {\sqrt 5} m/s## is the answer I get.

In general the minimum speed is ##v_b \cos \theta'##, where ##v_b## is the speed of the belt and ##\theta'## is the angle of motion relative to the belt.
is my approach right just want to confirm??
 
  • #53
IamVector said:
can you please explain further will be a great help
That's what comes out of the algebra and calculus. I suspect the geometric approach could be generalised.
 
  • #54
PeroK said:
That's what comes out of the algebra and calculus. I suspect the geometric approach could be generalised.
is my approach correct?? otherwise I will try it with different approach cause sometimes answer is correct approach is wrong
 
  • #55
IamVector said:
is my approach correct?? otherwise I will try it with different approach cause sometimes answer is correct approach is wrong
I haven't seen your approach.
 
  • #56
PeroK said:
I haven't seen your approach.
ok , so i got this equation (-1+μ gt/(5^1/2))i + (2-2μ gt/(5^1/2))j so to convert it into lab frame or ground frame I added 1m/s ie velocity of belt then I took the mod of the resultant vector ie (-1+μ gt/(5^1/2))i + (2-2μ gt/(5^1/2))j which in return formed a quadratic in μgt now we were asked to find the minimum vector so it will be minimum when the value of quadratic will be minimum so by differentiating i found the min val to be 4/5 thus on sqrt gives 2/(5^1/2)
 
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  • #57
PeroK said:
I haven't seen your approach.
I am preparing for Olympiads a word of advice will be appreciated and are the level of difficulty of questions I am doing sufficient??
 
  • #60
IamVector said:
Is it g/2? I am not sure
That's a quite a hard problem I think. You can't just guess that the answer is ##g/2## or ##3g/13## or whatever. You have to work it out.
 
  • #61
PeroK said:
That's a quite a hard problem I think. You can't just guess that the answer is ##g/2## or ##3g/13## or whatever. You have to work it out.
Well I took the component of mg acting on rod but I doubt it's a right approach.
 
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