Tangential Acceleration of Object moving in cylindrical wall

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Homework Help Overview

The problem involves an object sliding along the base of a vertical circular wall, analyzing the forces acting on it, particularly focusing on tangential acceleration and the role of friction from both the wall and the ground.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the object, particularly the frictional forces from the wall and the ground. Questions arise about the direction of these forces and their contributions to tangential acceleration.

Discussion Status

Participants are exploring the relationship between the forces and the resulting tangential acceleration. Some guidance has been provided regarding the direction of frictional forces and their impact on the motion of the object.

Contextual Notes

There is an ongoing examination of assumptions regarding the direction of frictional forces and their roles in the context of circular motion. The problem setup includes specific coefficients of friction and gravitational effects that are under discussion.

RATKING
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Homework Statement


An object slides along the ground at speed v at the base of a circular wall of radius r. The object is in contact with both the wall and the ground, and friction acts at both contacts. The wall is vertical and provides no force in the vertical direction. (i) Show that the tangential acceleration of the object is given by dv/dt = −µGg− µWv2/r , where µG and µW are the coefficients of kinetic friction between the object and the ground and wall respectively, and g is the acceleration due to gravity.

Homework Equations


Centripetal Force=(mv2)/r
Frictional Force when moving = coefficient of kinetic friction * Normal force

The Attempt at a Solution


Frictional Force Due to Wall =FW= µW * NW
Normal due to wall is in same direction to centripetal force, these must be equal as the ball doesn't move in this plane and there are no other forces acting here.
Therefore Fc=NW
We know:
Fc=mv2/r
Therefore: mv2/r=NW
Substituting back into equation for Frictional force due to wall:
FWW *mv2/r

The frictional force is in the direction tangent to circle.
My problem is that I do not see any other force to be acting tangential, so do not see where a second term comes from.

I say:

Tangential Force =- Frictional Force due to Wall=- µW *mv2/r (as it acts in opposite direction to velocity)

Therefore dv/dt=Tangential Acceleration =- µW *v2/r (Dividing through by m)
 
Last edited:
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Hello RATKING. Welcome to PF!

Your analysis of the friction from the wall looks good.

What is the direction of the force of friction from the ground?
 
Hi TSny!

I have consulted with some friends and I think I see the problem. I have assumed that the friction due to the ground must always be radially directed due to the simple uniform case of circular motion. However, here the centripetal force keeping the body in circular motion is provided by the normal force from the wall on the body. I also know think all friction opposes motion and so as there is no motion in radial direction and no other direction except tangentially, all the frictional force acts tangential. Therefore the frictional force due to ground also contributes to tangential force causing the extra term in the solution. do you think this is the correct reasoning?
 
Yes. Good. The motion is in the tangential direction, and the sliding friction (from the wall and the ground) acts opposite to the direction of motion in this problem.
 

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