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Unformatted text preview: IPHO 2003 Solution − Theoretical Question 1 (2003/07/12) p.1/7 Solution Theoretical Question 1 A Swing with a Falling Weight Part A (a) Since the length of the string θ R s L + = is constant, its rate of change must be zero. Hence we have = + θ R s (A1)* 1 (b) Relative to O , Q moves on a circle of radius R with angular velocity θ , so t s t R v Q ˆ ˆ  = = θ (A2)* (c) Refer to Fig. A1. Relative to Q , the displacement of P in a time interval ∆ t is t t s r s t s r s r ∆ θ ∆ θ ∆ ∆ ] ˆ ) ˆ )( [( ˆ ) ( ) ˆ )( ( + = + = ′ . It follows t s r s v ˆ ˆ + = ′ θ (A3)* (d) The velocity of the particle relative to O is the sum of the two relative velocities given in Eqs. (A2) and (A3) so that r s t R t s r s v v v Q ˆ ˆ ) ˆ ˆ ( θ θ θ  = + + = + ′ = (A4)* (e) Refer to Fig. A2. The ( t ˆ )component of the velocity change v ∆ is given by t v v v t ∆ θ θ ∆ ∆ = = ⋅ ) ˆ ( . Therefore, the t ˆcomponent of the acceleration t v a ∆ ∆ / = is given by θ v a t = ⋅ ˆ ˆ . Since the speed v of the particle is θ s according to Eq. (A4), we see that the t ˆcomponent of the particle’s acceleration at P is given by 2 ) ( ˆ θ θ θ θ s s v t a = = = ⋅ (A5)* Note that, from Fig. A2, the radial component of the acceleration may also be 1 An equation marked with an asterisk contains answer to the problem. t ˆ Q r ˆ v ∆θ P θ ∆ v Figure A2 O ∆θ v ∆ v v ∆ + ∆ v v t ˆ Q r ˆ s s+ ∆ s ∆θ s ∆θ P Figure A1 r ′ ∆ s+ ∆ s IPHO 2003 Solution − Theoretical Question 1 (2003/07/12) p.2/7 obtained as dt s d dt dv r a / ) ( / ˆ θ  = = ⋅ . (f) Refer to Fig. A3. The gravitational potential energy of the particle is given by mgh U = . It may be expressed in terms of s and θ as ] sin ) cos 1 ( [ ) ( θ θ θ s R mg U + = (A6)* (g) At the lowest point of its trajectory, the particle’s gravitational potential energy U must assume its minimum value U m . If the particle’s mechanical energy ....
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This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.
 Spring '11
 NA

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