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Algebra · real student question

Solve 2*pi*R*k*sin(a) - 2*pi*a*k = m*g*tan(a) / (2*pi) for k.

Question

Make kk the subject of

2πRksin⁡α−2πak=mgtan⁡α2π2\pi Rk\sin\alpha-2\pi ak=\frac{mg\tan\alpha}{2\pi}

Step-by-step solution

  1. Spot that kk appears in both terms on the left. Neither side needs expanding; the whole problem is a factorisation. Both left-hand terms contain the common factor 2πk2\pi k: 2πRksin⁡α−2πak=2πk(Rsin⁡α−a).2\pi Rk\sin\alpha-2\pi ak=2\pi k\left(R\sin\alpha-a\right). Collecting kk into a single factor is what makes it possible to isolate it in one division.

  2. Rewrite the equation with kk collected. 2πk(Rsin⁡α−a)=mgtan⁡α2π2\pi k\left(R\sin\alpha-a\right)=\frac{mg\tan\alpha}{2\pi} Everything multiplying kk now sits in one bracket, and everything else is on the right.

  3. Divide by the whole coefficient of kk. Dividing both sides by 2π(Rsin⁡α−a)2\pi\left(R\sin\alpha-a\right) gives k=mgtan⁡α2π2π(Rsin⁡α−a).k=\frac{\dfrac{mg\tan\alpha}{2\pi}}{2\pi\left(R\sin\alpha-a\right)}. Note that the 2π2\pi already in the denominator on the right does not cancel the new one - dividing by 2π2\pi a second time multiplies the denominators together.

  4. Simplify the double fraction. Multiplying the two denominators, k=mgtan⁡α4π2(Rsin⁡α−a).k=\frac{mg\tan\alpha}{4\pi^{2}\left(R\sin\alpha-a\right)}. The 4π24\pi^{2} is the step people most often get wrong, writing 2π2\pi or 2π22\pi^{2} instead.

  5. State the restriction. The division is only valid when Rsin⁡α−a≠0,R\sin\alpha-a\neq 0, i.e. sin⁡α≠a/R\sin\alpha\neq a/R. If Rsin⁡α=aR\sin\alpha=a the left side is identically zero, so the equation has no solution for kk unless mgtan⁡αmg\tan\alpha is also zero, in which case every kk works.

  6. Check the rearrangement with sample numbers. Take R=2R=2, a=0.5a=0.5, α=30∘\alpha=30^\circ, m=1m=1, g=9.8g=9.8. The formula gives k=9.8×0.5773502694π2(2×0.5−0.5)=0.286639282.k=\frac{9.8\times 0.577350269}{4\pi^{2}(2\times 0.5-0.5)}=0.286639282. Substituting back into the original left side: 2π(2)(0.286639282)(0.5)−2π(0.5)(0.286639282)=0.9005038632\pi(2)(0.286639282)(0.5)-2\pi(0.5)(0.286639282)=0.900503863, and the right side is 9.8×0.5773502692π=0.900503863\dfrac{9.8\times 0.577350269}{2\pi}=0.900503863. The two agree, confirming the algebra.

Answer

k=mgtan⁡α4π2(Rsin⁡α−a),Rsin⁡α≠ak=\frac{mg\tan\alpha}{4\pi^{2}\left(R\sin\alpha-a\right)},\qquad R\sin\alpha\neq a

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