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Dynamics problems
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- 2. MOTION UNDER THEACTION OF CENTRAL FORCES PROBLEMS
- 3. Problem 1: A Smoothstraight thin tube revolves with uniform angular velocity ‘𝜔′ in a vertical plane about one extremity which is fixed; if at zero time the tube be horizontal and a particle insides it beat a distance ‘a’ from the fixed end, and be moving with velocity V along the tube, show that the distance at time ‘t’ is a cos ℎ 𝜔𝑡 + 𝑉 𝜔 − 𝑔 2𝜔2 sin ℎ 𝜔𝑡 + 𝑔 2𝜔2 sin ℎ 𝜔𝑡
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- 5. Let at timet, P be the position of the particle of mass m on the tube OB. The forces acting at P are (i) its weight mg vertically downwards and (ii)normal reaction R perpendicular to OB Let P be (𝑟, 𝜃) Angular Velocity= 𝜃 = 𝑑𝜃 𝑑𝑡 = 𝜔 Integrating, 𝜃 = 𝜔𝑡 + 𝐴 Initially when t=0,𝜃 = 0 𝜃 = 𝜔𝑡 -------(1) Resolving along the radius vector OB m( 𝑟 − 𝑟 𝜃2) = −𝑚𝑔 cos (90 𝑜 − 𝜃) = −𝑚𝑔 sin 𝜃 𝑟 − 𝑟𝜔2 = −𝑔𝑠𝑖𝑛𝜃 = −𝑔𝑠𝑖𝑛𝜔𝑡 (using (1)) 𝐷2 − 𝜔2 𝑟 = −𝑔𝑠𝑖𝑛𝜔𝑡 ---------(2) Where D= 𝑑 𝑑𝑡 The complementary function Y is found such that (𝐷2 − 𝜔2 )𝑌 = 0
- 6. The solution ofthis differential equation is 𝑦 = 𝐴𝑒 𝜔𝑡 + 𝐵𝑒−𝜔𝑡---------(3) Where A and B are constants. The particular integral u of the equation (2) is given by 𝐷2 − 𝜔2 𝑢 = −𝑔𝑠𝑖𝑛𝜔𝑡 𝑢 = − 𝑔 𝐷2−𝜔2 𝑠𝑖𝑛𝜔𝑡 = − 𝑔 𝜔2−𝜔2 𝑠𝑖𝑛𝜔𝑡 = 𝑔 2𝜔2 𝑠𝑖𝑛𝜔𝑡 -----------(4) Hence the general of (2) is 𝑟 = 𝑌 + 𝑢 = 𝐴𝑒 𝜔𝑡 + 𝐵𝑒−𝜔𝑡+ 𝑔 2𝜔2 𝑠𝑖𝑛𝜔𝑡 ------------(5) The initial conditions are : when t=0,r=a and 𝑟 = 𝑉 Hence (5) gives A+B=a --------(6) Differentiating (5) 𝑟 = 𝐴𝜔𝑒 𝜔𝑡 − 𝐵𝜔𝑒−𝜔𝑡 + 𝑔 2𝜔 cos 𝜔𝑡---------(7)
- 7. Putting t=0 and𝑟 = 𝑉 in (7) we have 𝐴𝜔 − 𝐵𝜔 + 𝑔 2𝜔 = 𝑉 or 𝐴 − 𝐵 = 𝑉 𝜔 − 𝑔 2𝜔2 ------(8) Solving (6) and (8) 𝐴 = 1 2 (𝑎 + 𝑉 𝜔 − 𝑔 2𝜔2) and 𝐵 = 1 2 (𝑎 − 𝑉 𝜔 + 𝑔 2𝜔2) substituting these values in(5) 𝑟 = 1 2 (𝑎 + 𝑉 𝜔 − 𝑔 2𝜔2 )𝑒 𝜔𝑡 + 1 2 (𝑎 − 𝑉 𝜔 + 𝑔 2𝜔2)𝑒−𝜔𝑡+ 𝑔 2𝜔2 𝑠𝑖𝑛𝜔𝑡 = 𝑎(𝑒 𝜔𝑡+𝑒−𝜔𝑡) 2 + ( 𝑉 𝜔 − 𝑔 2𝜔2)( 𝑒 𝜔𝑡−𝑒−𝜔𝑡 2 )+ 𝑔 2𝜔2 sin 𝜔𝑡 = 𝑎𝑐𝑜𝑠ℎ𝜔𝑡 + 𝑉 𝜔 − 𝑔 2𝜔2 sinh 𝜔𝑡 + 𝑔 2 𝜔2 sin 𝜔𝑡 ∎
- 8. Problem 2: Find thelaw of force towards the pole under which the curve 𝑟 𝑛 = 𝑎 𝑛 cos 𝑛𝜃 can be described.
- 9. Solution Given curve, 𝑟𝑛 = 𝑎 𝑛 cos 𝑛𝜃 Since 𝑟 = 1 𝑢 , the equations is un 𝑎 𝑛 cos 𝑛𝜃 = 1 ------(1) Taking log on both sides 𝑛 log 𝑢 + 𝑛 log 𝑎 + log cos 𝑛𝜃 = 0-----(2) Differentiating (2) with respect to 𝜃 , 𝑛. 1 𝑢 𝑑𝑢 𝑑𝜃 − 𝑛𝑠𝑖𝑛 𝑛𝜃 cos 𝑛𝜃 = 0 (ie) 𝑑𝑢 𝑑𝜃 = 𝑢 tan 𝑛𝜃-------(3) Differentiating (3) with respect to 𝜃𝑑2 𝑢 𝑑𝜃2 = 𝑢 𝑛 sec2 𝑛𝜃 + tan 𝑛𝜃 . 𝑑𝑢 𝑑𝜃 𝑢 + 𝑑2 𝑢 𝑑𝜃2 = 𝑢 + 𝑛𝑢 sec2 𝑛𝜃 + 𝑢 tan2 𝑛𝜃 = 𝑛𝑢 sec2 𝑛𝜃 + 𝑢(1 + tan2 𝑛𝜃)
- 10. = 𝑛𝑢 𝑠𝑒𝑐2𝑛𝜃 + 𝑢 𝑠𝑒𝑐2 𝑛𝜃 = 𝑛 + 1 𝑢 𝑠𝑒𝑐2 𝑛𝜃 = 𝑛 + 1 𝑢. 𝑢2𝑛 𝑎2𝑛 using (1) to substitute for sec2 𝑛𝜃 = 𝑛 + 1 𝑎2𝑛 𝑢2𝑛+1 P∝ 1 𝑟2𝑛+3 which means that the central acceleration varies inversely as the(2n+3) rd power of the distance. ∎
- 11. Problem 3: Find thelaw of force to an internal point under which a body will describe a circle.
- 12. Solution : From thepedal equation of the circle for general position of the pole is 𝑐2 = 𝑟2 + 𝑎2 − 2𝑎𝑝 -----(1) Differentiating with respect to r, 0 = 2𝑟 − 2𝑎 𝑑𝑝 𝑑𝑟 (i.e) 𝑑𝑝 𝑑𝑟 = 𝑟 𝑎 Now the central acceleration 𝑃 = ℎ2 𝑝3 𝑑𝑝 𝑑𝑟 substituting for p from (1) ∎
- 13. Problem 4: A particlemoves in an ellipse under a force which is always directed towards its focus. Find the law of force, the velocity at any point of the path and its periodic time.
- 14. Solution The polar equationto the ellipse is 𝑙 𝑟 = 1 + 𝑒𝑐𝑜𝑠𝜃 -----(1) where e is the eccentricity and 𝑙 is the semi latus rectum. From (1) 𝑢 = 1 𝑟 = 1+𝑒 cos 𝜃 𝑙 Hence 𝑑𝑢 𝑑𝜃 = − 𝑒𝑠𝑖𝑛𝜃 𝑙 and 𝑑2 𝑢 𝑑𝜃2 = − 𝑒𝑐𝑜𝑠𝜃 𝑙 𝑢 + 𝑑2 𝑢 𝑑𝜃2 = 1 + 𝑒 cos 𝜃 𝑙 − 𝑒 cos 𝜃 𝑙 = 1 𝑙 We know that 𝑃 ℎ2 𝑢2 = 𝑢 + 𝑑2 𝑢 𝑑𝜃2 = 1 𝑙 i.e The force varies inversely as the square of the distance from the pole. 1 𝑝2 = 𝑢2 + 𝑑𝑢 𝑑𝜃 2 = 1 + 𝑒𝑐𝑜𝑠𝜃 𝑙 2 + 𝑒𝑠𝑖𝑛𝜃 𝑙 = 1 + 2𝑒𝑐𝑜𝑠𝜃 + 𝑒2 𝑙2 Hence 𝑣2 = ℎ2 𝑝2 = ℎ2(!+2𝑒𝑐𝑜𝑠𝜃+𝑒2) 𝑙2
- 15. = 𝜇𝑙 𝑙2 (1 +𝑒2 + 2 𝑙 𝑟 − 1 ) substituting for 𝑒𝑐𝑜𝑠𝜃 from (1) = 𝜇 𝑙 𝑒2 + 2𝑙 𝑟 − 1 = 𝜇 𝑙 ( 2𝑙 𝑟 − 1 − e2 ) = 𝜇[ 2 𝑟 − 1−𝑒2 𝑙 )------(2) Now if a and b are the semi axes of ellipse. We know that 𝑙 = 𝑏2 𝑎 = 𝑎2(1 − 𝑒2) 𝑎 = 𝑎(1 − 𝑒2 ) 𝑣2 = 𝜇 2 𝑟 − 1 𝑎 , giving the velocity v. Areal velocity in the orbit = 1 2 ℎ and this is constant. The total area of the ellipse = 𝜋𝑎𝑏. Periodic Time T= 𝜋𝑎𝑏 ( 1 2 ℎ) = 2𝜋𝑎𝑏 ℎ = 2𝜋𝑎𝑏 √𝜇𝑙 where 𝜇 = ℎ2 𝑙 = 2𝜋𝑎𝑏 𝜇.𝑏 . √𝑎, since 𝑙 = 𝑏2 𝑎 = 2𝜋 √𝜇 . 𝑎3/2 ∎
- 16. Problem 5: A particlemoves in a curve under a central attraction so that its velocity at any point is equal to that in a circle at the same distance and under the same attraction. Show that the path is an equiangular spiral and that the law of force is that of the inverse cube.
- 17. Solution Let the centralacceleration be P. If v is the velocity in a circle at a distance r under the normal acceleration P, then 𝑣2 𝑟 = 𝑃 i.e 𝑣2 = 𝑃𝑟 -------(1) Since v is also the velocity in the central orbit, ℎ = 𝑝𝑣 or 𝑣 = ℎ 𝑝 putting this is (1), ℎ2 𝑝2 = 𝑃𝑟 ------(2) We know that, 𝑃 = ℎ2 𝑃3 . 𝑑𝑝 𝑑𝑟 ----------(3) Substituting (3) in (2) ℎ2 𝑝2 = ℎ2 𝑝3 . 𝑑𝑝 𝑑𝑟 . 𝑟 (i.e) 𝑑𝑝 𝑝 = 𝐴 Substituting this in (3), 𝑃 = ℎ2 𝑝3 . 𝐴 = 𝐴ℎ2 𝐴3 𝑟3 using (4) = ℎ2 𝐴2 ( 1 𝑟3) (ie)𝑃 ∝ 1 𝑟3 ∎
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