Home Physics Subtleties Neglected in Friction Questions: Object Slides Down Ramp

## Downside assertion (simplified)

An object slides down a ramp at angle θ to come across stage floor. Each surfaces have kinetic friction: μ’ on the ramp, μ on the extent. The article 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 neglected sure subtleties that may come up in friction issues.

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

I’ll assume it’s meant 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 equipped implies R is vanishingly small, however I’ll begin with a less complicated case. In any occasion, I’ll assume u2>>g max(R,r), which makes gravity irrelevant.

There are two important circumstances to contemplate:

1. r<<R

2. r>>R

## Case 1: r<<R

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

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

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

## Case 2: r>>R

We will deal with this case as an influence, i.e. R=0.

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

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

There are three levels within the transition to horizontal movement:

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

2. The block continues to slip, the vanguard 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 item strikes the bottom, there shall be a vertical impulse J from the bottom at the vanguard and an impulse J’ from the ramp, regular to the ramp, on the trailing fringe of the item.

Correspondingly, a frictional impulse μJ horizontally at the vanguard 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 vanguard instantly after influence be v and the rotation charge of the item simply after influence be ω. This suggests the horizontal velocity of the centre of the item simply after influence 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. Nonetheless, it seems that with these angles the again finish of the block would lose contact with the ramp when the vanguard hits the bottom even with no friction.

We may 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 a substitute, 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 could appear much like Case 1, however be aware 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 pressure.

This suggests lack of contact at enough preliminary velocity.

## Stage 3

Ultimate impulse

Suppose, simply earlier than the trailing edge hits the bottom, the vanguard is travelling at pace v’ horizontally. The trailing edge should have been travelling at v’ horizontally additionally, and v’ tan(θ) vertically down. The mass centre of the block was due to this fact 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 influence 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 influence = mv’2/2+mv’2tan2(θ)/8+mv’2tan2(θ)/24.

KE simply after influence = 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 may the issue description be adjusted to make the evaluation less complicated?

A method could be to specify the form of the item 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 received’t rotate

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

At influence 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 due to this fact v(cos(θ) – μ sin(θ)). With θ=π/6 and μ=0.3 that’s about 0.716v, a 49% lack of KE.