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Monday, March 27, 2023

# Subtleties Missed in Friction Questions: Object Slides Down Ramp

## Downside assertion (simplified)

An object slides down a ramp at angle θ to come across degree floor. Each surfaces have kinetic friction: μ’ on the ramp, μ on the extent. The item reaches the bottom at pace u. What’s its pace when first totally on the extent?

There are a number of lacking particulars, which suggests the writer missed sure subtleties that may come up in friction issues.

For concreteness, I’ll take the thing to be a uniform rectangular block base size B, peak A, diagonal 2r, and A/B=tan(α).

I’ll assume it’s supposed that each ends of the bottom of the block ought to stay involved with the surfaces always.

Suppose the transition from the ramp to the extent floor is by way of an arc of radius R. The diagram provided implies R is vanishingly small, however I’ll begin with an easier case. In any occasion, I’ll assume u2>>g max(R,r), which makes gravity irrelevant.

There are two most important instances to think about:

1. r<<R

2. r>>R

## Case 1: r<<R

The utmost rotational KE might be small c.w. the translational KE and might be ignored.

Whereas traversing the arc at pace v=v(θ), the traditional drive is the centripetal drive, mv2/R and the frictional drive is μmv2/R,  So μv2=-v(dv/dθ), v= ue-μθ .  The fraction of KE misplaced to friction is 1-e-2μθ.

Within the unique drawback, θ=π/6, μ=0.3 for the horizontal floor, 0.2 for the ramp.  Utilizing the typical, 0.25, the fraction misplaced is 1-e-π/12, about 23%

## Case 2: r>>R

We are able to deal with this case as an impression, i.e. R=0.

To simplify the algebra, all forces and impulses might be taken to be per unit mass.

An answer posted within the thread treats it merely as an inelastic impression within the vertical. Consequently, it finds v = u cos(θ), a lack of 25% of the KE. This overlooks frictional impulse.

There are three phases within the transition to horizontal movement:

1. The forefront strikes the bottom and begins to maneuver horizontally.

2. The block continues to slip, the forefront on the bottom, and the trailing edge on the ramp.

3. The trailing edge strikes the bottom.

There are frictional losses throughout every stage.

## Stage 1

When the thing strikes the bottom, there might be a vertical impulse J from the bottom at the forefront and an impulse J’ from the ramp, regular to the ramp, on the trailing fringe of the thing.

Correspondingly, a frictional impulse μJ horizontally at the forefront and a frictional impulse μ’J’ from the ramp, up the ramp, on the trailing edge.

Two regular impulses, every with friction

Let the horizontal velocity of the forefront instantly after impression be v and the rotation charge of the thing simply after impression be ω. This means the horizontal velocity of the centre of the thing simply after impression is v-ωr sin(θ+α), and its vertical velocity is ωr cos(θ+α) (down).

Horizontal momentum conservation:

μJ+ μ’J’ cos(θ) – J’ sin(θ) = u cos(θ) -v + ωr sin(θ+α)

Vertical momentum conservation:

J+J’ cos(θ) + μ’J’ sin(θ) = u sin(θ) – ωr cos(θ+α)

For the reason that trailing edge continues parallel to the ramp,

v sin(θ) = ωB = 2ωr cos(α)

The second of inertia of the block is mr2/3.

Angular momentum conservation about block centre:

ωr2/3 = Jr cos(θ+α) – μJr sin(θ+α) – J’r cos(α) – μ’J’r sin(α)

cancelling r:

ωr/3 = J cos(θ+α) – μJ sin(θ+α) – J’ cos(α) – μ’J’ sin(α)

We now have 4 equations and 4 unknowns: J, J’, ω, v.

What to decide on for α? The unique diagram exhibits a sq. block, implying α= π/4, it specifies θ =π/6, and units μ=0.3 for the horizontal floor. Nevertheless, it seems that with these angles the again finish of the block would lose contact with the ramp when the forefront hits the bottom even with no friction.

We might compromise by utilizing α= 0.49, about 28°, which simply avoids lack of contact. This offers a lack of 37% of the KE, and that is simply the primary stage. (Even α= 0 provides a loss > 25%.)

As an alternative, I shall simplify issues by setting α= 0. That cuts the stage 1 loss to only over 25%.

## Stage 2

Sliding on each surfaces

Regardless of selecting α= 0, the equations for this dynamic case are difficult.

At first, it might appear just like Case 1, however observe the trajectory of the mass centre. Taking the centre of the rod object (as now assumed) to be at (x,y) relative to the underside of the ramp, (x+2y cot(θ))2+y2=r2, an ellipse.  I.e. it arcs downwards, requiring a downward centripetal drive.

This means lack of contact at ample preliminary velocity.

## Stage 3

Last impulse

Suppose, simply earlier than the trailing edge hits the bottom, the forefront is travelling at pace v’ horizontally. The trailing edge will need to have been travelling at v’ horizontally additionally, and v’ tan(θ) vertically down. The mass centre of the block was subsequently descending at pace v’ tan(θ)/2. On hitting the bottom full size, there’s a vertically upward impulse mv’ tan(θ)/2. and a frictional impulse μmv’ tan(θ)/2.

As well as, simply earlier than impression there was rotational KEr of (1/2)(mr2/3)ωthe place 2rω=v’ tan(θ). Therefore KEr=m(v’tan(θ))2/24.

KE simply earlier than impression = mv’2/2+mv’2tan2(θ)/8+mv’2tan2(θ)/24.

KE simply after impression = mv’2(1-μ tan(θ)/2)2/2

Fraction of preliminary KE remaining = (1- μ tan(θ)/2)2/(1+tan2(θ)/3)

With θ=π/6 and μ=0.3 that’s about 0.75, or a 25% loss.

## Different shapes

How might the issue description be adjusted to make the evaluation easier?

A technique can be to specify the form of the thing such that it doesn’t rotate. It could possibly be a triangle ABC, the place AB contacts the ramp and BC, the decrease edge, is horizontal. So ABC = π-θ.

The form of an object that gained’t rotate

With tan( ACB)=tan(θ)/2, the mass centre can be vertically above B.  This is able to tip neither whereas accelerating down the ramp nor when decelerating on the horizontal.

At impression with the bottom, there’s a vertically upward impulse mv sin(θ) and a corresponding horizontal frictional impulse μmv sin(θ). The ensuing horizontal pace is subsequently v(cos(θ) – μ sin(θ)). With θ=π/6 and μ=0.3 that’s about 0.716v, a 49% lack of KE.