Drawback 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?
(Unique 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 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, top 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 predominant instances to contemplate:
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 conventional 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 unique downside, 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 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:
The forefront strikes the bottom and begins to maneuver horizontally.
The block continues to slip, the vanguard on the bottom, and the trailing edge on the ramp.
The trailing edge strikes the bottom.
There are frictional losses throughout every stage.
When the thing strikes the bottom, there might 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 thing.
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.
Let the horizontal velocity of the vanguard instantly after impression be v and the rotation price 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)
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 reveals 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 through the use of 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%.
Regardless of selecting α= 0, the equations for this dynamic case are difficult.
At first, it could 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 pressure.
This means lack of contact at enough preliminary velocity.
Suppose, simply earlier than the trailing edge hits the bottom, the vanguard 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)ω2 the 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.
How may the issue description be adjusted to make the evaluation less complicated?
A method could 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.
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 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.