PHALIN THAKKAR / PORTFOLIO

PHALIN THAKKAR

Energy gives speed. Contact decides the path.

P8 · Advanced extension. Combine work and energy with the condition that a track can push, but cannot pull.

Prerequisites: work, energy, circular motion and radial force balance. Review the foundation force lesson.

A small particle of mass m is released from rest on a smooth ramp, at height H above a horizontal track. It then crosses a rough horizontal section of length L = 1.00 m with μk = 0.20 and enters the bottom of a smooth vertical circular loop of radius R = 0.50 m. It moves on the inside of the loop.

(a) Find the minimum H needed to complete the loop while maintaining contact.

(b) At that limiting H, find the normal reaction at the bottom.

(c) If H = 1.20 m, find the angle at which contact is lost and the speed then. Measure θ from the downward radius, so θ = 0 at the bottom.

Then select every correct statement:

A. The minimum release height is 1.45 m.

B. At that limiting height, the bottom reaction is 6mg.

C. At H = 1.20 m, the particle reverses along the track before losing contact.

D. At H = 1.20 m, contact is lost when cosθ = −2/3.

Ramp, rough section and inside loopA ramp releases the particle from height H onto a rough horizontal section of length 1 metre, followed by a smooth loop of radius 0.5 metres. Theta is measured from the downward radius. At theta 132 degrees the inward normal is down and left toward the centre; weight is vertically down.HRough: L = 1 mμₖ = 0.20R = 0.50 mθSmooth loopN inwardmg ↓Bottom: θ = 0
Schematic, not to force scale. N points toward the centre while contact persists; it becomes zero at detachment. The particle moves on the inside of the loop.

1. Account for the energy loss

12mvb2=mgH−μkmgL\tfrac12mv_b^2=mgH-\mu_kmgL

The rough section removes energy before entry. Inside the loop the track is smooth, so the change in speed is due to the height gained, not friction.

2. Check the radial equation

N−mgcos⁡θ=mv2R,N≥0N-mg\cos\theta=\frac{mv^2}{R},\qquad N\ge0

At the top the minimum allowable speed is √(gR). Reaching the top with zero speed would satisfy a height calculation but fail the contact condition. Once N would turn negative, the circular model must stop: the particle follows a free flight trajectory.

Full solution and limiting checks for P8

Crossing the rough section dissipates μkmgL. The loop entry speed satisfies:

vb2=2g(H−μkL).v_b^2=2g(H-\mu_kL).

At the top, energy gives vt² = vb² − 4gR. Contact requires vt² ≥ gR. Therefore:

Hmin⁡=μkL+52R=0.20+1.25=1.45 m.H_{\min}=\mu_kL+\frac52R=0.20+1.25=1.45\ {\rm m}.

At that height vb² = 5gR. The bottom reaction is:

Nb=m(vb2/R+g)=6mg.N_b=m(v_b^2/R+g)=6mg.

For a general angle θ from the bottom:

v2=2g[H−μkL−R(1−cos⁡θ)],v^2=2g[H-\mu_kL-R(1-\cos\theta)],
N=m(v2R+gcos⁡θ)=mg[2(H−μkL)R−2+3cos⁡θ]N=m\left(\frac{v^2}{R}+g\cos\theta\right)=mg\left[\frac{2(H-\mu_k L)}{R}-2+3\cos\theta\right]

For H = 1.20 m, set N = 0:

cos⁡θ=−23,θ≈131.8103∘.\cos\theta=-\frac23,\qquad \theta\approx131.8103^\circ.

The speed is still positive:

v2=−gRcos⁡θ=103,v≈1.8257 m/s.v^2=-gR\cos\theta=\frac{10}{3},\qquad v\approx1.8257\ {\rm m/s}.

It therefore loses contact before reversing on the track.

Correct statements: A, B and D.

Check: energy determines the available speed; the normal reaction determines whether the prescribed circular path is physically possible.

Attempt P8 on the practice sheet · P8 worked solution

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