Relative velocity between lift and screw

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Homework Help Overview

The discussion revolves around the relative velocity between a lift and a screw, specifically addressing the initial velocities of both objects and the implications of gravity on their motion. Participants explore the concepts of free fall, initial velocity, and the effects of acceleration in a vertical motion context.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the reasoning behind using the lift's velocity as the initial velocity for the screw and discuss the implications of dropping the screw from the lift. There are inquiries about the conditions under which the screw's velocity should be considered equal to that of the lift, as well as the effects of gravity on both objects.

Discussion Status

The discussion is active, with participants exploring various interpretations of the problem. Some have offered insights into the relationship between the screw's motion and the lift's acceleration, while others are clarifying the conditions under which initial velocities are relevant. There is no explicit consensus, but productive dialogue is occurring.

Contextual Notes

Participants are navigating assumptions about reference frames, particularly the Earth as a reference for velocity. The problem involves complexities related to free fall and relative motion, with some participants suggesting the need for visual aids like velocity graphs to enhance understanding.

robax25
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Homework Statement
A screw inside a lift falls down when the lift moves upward at an acceleration of 4m/s². find out the time is taken by screw to fall down. It is 0.659s by calculation.Now the lift has initial velocity is 16m/s. Find out the impact velocity.
Relevant Equations
v=v0 +at (for lift)
v=v0-at(for screw)
v=16m/s+4*.659s=18.636m/s (for lift)
v0=16m/s-9.81*.659s=9.53m/s
my question is that why 16m/s is considered for screw velocity ? it is given for lift .what does it mean 9.53m/s? If you drop something , it gets acceleration and I calculate it 6.45m for screw while falling and why should I deduct from v0 to get result impact velocity for screw?
 
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robax25 said:
v=16m/s+4*.659s=18.636m/s (for lift)
v0=16m/s-9.81*.659s=9.53m/s
my question is that why 16m/s is considered for screw velocity ? it is given for lift .what does it mean 9.53m/s? If you drop something , it gets acceleration and I calculate it 6.45m for screw while falling and why should I deduct from v0 to get result impact velocity for screw?
Is the screw inside the lift when it is dropped?
 
yes
 
robax25 said:
yes
So it has the same initial velocity as the lift?
 
why?
 
robax25 said:
why?
If you are in a car moving along the road at 30m/s, say, then you are moving at 30m/s. Not just the car!
 
To answer your question "why subtract". When you drop something, the acceleration is downward. The way I read it: v0 was upward. The elevator (lift) continues to accelerate upward so v0 and the lifts acceleration are in the same direction (up)
 
PeroK said:
If you are in a car moving along the road at 30m/s, say, then you are moving at 30m/s. Not just the car!
yes, I get it. But, if screw is in free fall and lift has impact with screw and let's say, lift has no any roof. Screw has free fall and lift is moving upward, in this case, screw is not inside the lift. Now what is the impact velocity for screw? In this case, Screw don't have initial velocity. Right?
 
robax25 said:
yes, I get it. But, if screw is in free fall and lift has impact with screw and let's say, lift has no any roof. Screw has free fall and lift is moving upward, in this case, screw is not inside the lift. Now what is the impact velocity for screw? In this case, Screw don't have initial velocity. Right?
If someone drops an object down the lift shaft that is different from someone inside the lift, and moving up with the lift, dropping something.

In the first case, the dropped object has zero initial velocity, relative to the Earth.

In the second case, the dropped object has a non-zero initial velocity relative to the Earth. This is the velocity of the lift at the time it is dropped.
 
  • #10
Ps if you are standing on the ground looking at the screw. You are not looking at anything else. Focus on the screw. Is it moving or not before it is dropped? Do you see the screw moving upwards?
 
  • #11
no, screw move downward.my question is that when don't you need to consider screw velocity is equal to lift velocity?
 
  • #12
robax25 said:
no, screw move downward.my question is that when don't you need to consider screw velocity is equal to lift velocity?

If the screw isn't moving upwards how did it get off the ground?

Note that in this problem the screw is always moving upwards.

I think you are making a fundamental mistake by assuming that when the screw is dropped it immediately loses its upward velocity and immediately starts moving downwards relative to the Earth.

The screw in this case is like a projectile. Its journey in the lift has given it an initial upward velocity relative to the Earth. After being dropped it will continue moving upwards relative to the Earth, but gravity will slow it down and after a time it will reach its highest point and then start to fall back down.

If you use your logic, then it would be impossible to throw a ball upward. By your logic, as soon as the ball leaves your hand it just falls back to Earth without going any higher than your hand.

Relative to the lift the screw falls to the floor. But relative to the Earth the screw continues moving upwards in projectile motion.
 
  • #13
Actually, you do not understand my point. I mean which case or which system can be that screw don't need to consider initial velocity?
 
  • #14
robax25 said:
Actually, you do not understand my point. I mean which case or which system can be that screw don't need to consider initial velocity?

The solution to this problem implicitly assumes we are using the Earth as a reference frame. In that reference frame the screw has an initial velocity.

If you really understand what I'm saying then you wouldn't be in any doubt over the solution to this problem.
 
  • #15
your solution and your explanation are ok. I do not have any doubt. Let's consider system that lift has initial velocity and acceleration (upward) and screw or an object falls down from a certain height, I mean from any height for example, the roof of a building. A person throws a screw on a lift's roof. In this case, lift has intital velocity and acceleration and screw has only downward acceleration.so they collide once they meet together.Now the equation for screw's impact velocity , v= -at and for lift, v=v0+at is it right? I hope I am clear what I mean?
 
  • #16
robax25 said:
your solution and your explanation are ok. I do not have any doubt. Let's consider system that lift has initial velocity and acceleration (upward) and screw or an object falls down from a certain height, I mean from any height for example, the roof of a building. A person throws a screw on a lift's roof. In this case, lift has intital velocity and acceleration and screw has only downward acceleration.so they collide once they meet together.Now the equation for screw's impact velocity , v= -at and for lift, v=v0+at is it right? I hope I am clear what I mean?

In any case the impact velocity of two objects will be the difference between their velocities at impact.

In your example, you calculate the velocity of the falling object and the velocity of the lift at time of impact and subtract one from the other.

In general, impact velocity is just a special case of the relative velocity between two objects.
 
  • #17
get it
 
  • #18
Perhaps it may help you to understand if you sketch out the velocity graphs of the screw (relative to the ground) and the lift. In making these, the graph must stay continuous (no breaks). A discontinuity would mean infinite acceleration at that point, with is impossible for an object having mass.
 

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