Subtleties Neglected in Friction Questions: Object Slides Down Ramp
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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 thing reaches the bottom at pace u. What’s its pace when first absolutely on the extent?
(Authentic is at https://www.physicsforums.com/threads/distance-a-block-slides-along-a-surface-with-friction-given-with-an-initial-velocity.1047556/.)
There are a number of lacking particulars, which suggests the creator neglected sure subtleties that may come up in friction issues.
For concreteness, I’ll take the article 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 in touch with the surfaces always.
Suppose the transition from the ramp to the extent floor is through an arc of radius R. The diagram equipped 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 important circumstances to think about:
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r<<R
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r>>R
Case 1: r<<R
The utmost rotational KE will probably be small c.w. the translational KE and will probably be ignored.
Whereas traversing the arc at pace v=v(θ), the traditional power is the centripetal power, mv2/R and the frictional power 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 will probably 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 phases within the transition to horizontal movement:
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The vanguard strikes the bottom and begins to maneuver horizontally.
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The block continues to slip, the forefront on the bottom, and the trailing edge on the ramp.
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The trailing edge strikes the bottom.
There are frictional losses throughout every stage.
Stage 1
When the article strikes the bottom, there will probably 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 article.
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.
Let the horizontal velocity of the forefront instantly after influence be v and the rotation charge of the article simply after influence be ω. This suggests the horizontal velocity of the centre of the article 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. 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 offers a loss > 25%.)
As a substitute, I shall simplify issues by setting α= 0. That cuts the stage 1 loss to simply over 25%.
Stage 2
Regardless of selecting α= 0, the equations for this dynamic case are difficult.
At first, it could appear just like Case 1, however word 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 power.
This suggests lack of contact at adequate preliminary velocity.
Stage 3
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 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)ω2 the 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 might the issue description be adjusted to make the evaluation easier?
A technique can be to specify the form of the article such that it doesn’t rotate. It might be a triangle ABC, the place AB contacts the ramp and BC, the decrease edge, is horizontal. So ∠ABC = π-θ.
With tan(∠ ACB)=tan(θ)/2, the mass centre can be vertically above B. This may 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.
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