Classical Mechanics and Special Relativity (ACM 20050) Assignment 2

Transcription

1 Classical Mechanics and Special Relativity ( Assignment 2 Dr Lennon Ó Náraigh Tuesday 8th September Instructions: This is a graded assignment. Put your name and student number on your homework. The assignment is to be handed in on or before Monday September 24th before 4pm. Hand in your assignment by putting it into the homework box marked outside the school office (G03, Science North. The assignment will be discussed subsequently in the tutorial on Monday 24th September at 4pm.

2 . A bead moves outward with constant speed u along the spoke of the wheel. It starts from the centre at t = 0. The angular position of the spoke is given by θ = ωt, where ω is a constant. Find the velocity and acceleration. Hint: We are given that ṙ = u and θ = ω. Answer clue: v = uˆr + utω ˆθ, a = utω 2 ˆr + 2uω ˆθ. Use the standard formulae from class notes: v = ṙ ˆr + r θ ˆθ, a = ( r r θ 2 ˆr + (r θ + 2ṙ θ ˆθ. Fill in for ṙ = u = const. and θ = ω = const.. Hence, r = ut and θ = ωt, hence v = uˆr + utω ˆθ. For the acceleration, r = 0 and θ = 0, hence a = utω 2 ˆr + 2uω ˆθ. 2. A particle moves with θ = ω = const. and r = r 0 e βt, where r 0 and β are constants. Show that Deduce that a r = 0 when β = ±ω. a = (β 2 ω 2 r 0 e β 0t ˆr + 2βr 0 ωe βt ˆθ. Fill in: Also, r = r 0 e βt, ṙ = r 0 βe βt, r = r 0 β 2 e βt. θ = ω, θ = 0. Hence, a = ( r r θ 2 ˆr + (r θ + 2ṙ θ ˆθ, = (β 2 ω 2 r 0 e βt ˆr + 2βr 0 ωe βt ˆθ. Clearly, a r = 0 when β = ±ω. 2

3 3. A block of mass m slides on a frictionless table. The mass is contrained to move along the inside of a ring of radius l which is fixed to the table. At time t = 0, the block is moving along the inside of the ring (i.e. in the tangential direction with velocity v 0. The coefficient of friction between the block and the ring is µ. (a Show that the equations of motion are ml θ = µn, ml θ 2 = N, where N is the positive reaction force exerted by the ring on the particle. (b Hence, show that the tangential velocity (= l θ of the block at later times is v 0 v(t = + (µtv 0 /l. Hint: You will need to solve a simple ODE to arrive at this result. If you are stuck you can use Wolfram Alpha to try to solve the ODE, only show your workings clearly. Start with the unconstrained equations of motion, with no external forces: m ( r r θ 2 = 0, ( m r θ + 2ṙ θ = 0, Now put in the constraints into the equation of motion. In the Normal direction there is the normal/constraining force which constrains the particle to move in the circle, such that ṙ = 0 and r = l: ml θ 2 = N. Here, N is positive, since the left-hand-side is clearly positive. In the tangential direction, there is friction, with F = µn. Note however we have to give the friction a minus sign because it opposes motion: ( m r θ + 2 ṙ θ = µn Put it all together: ml θ 2 = N, ml θ = µn. For the second part, divide the second equation by the first to get θ = µ θ 2. 3

4 Now let y = θ. The ODE becomes with solution or But y = θ, hence dy dt = µy2, ( y y t=0 = µt, y = y 0 + µt, y 0 = y(t = 0. θ = y 0 + µt. Use v = l θ and v 0 = y 0 l. This then gives as required. v(t = v 0 + µtv 0 l, 4

5 4. Consider a particle that is constrained on top of a semicircle (See Figure. Gravity points downwards. Suppose that the particle starts from rest. At what angle does the particle fall off the semicircle? Answer clue: φ = cos (2/3. Work in the θ coordinates, where θ is the standard angle pointing from the standard x-axis towards the North Pole. Start with the unconstrained equations of motion: m ( r r θ 2 ( m r θ + 2ṙ θ = U r, = r U θ, where U = mgy = mgr sin θ. Now put in the constraints into the equation of motion, such that ṙ = 0: m (0 r θ 2 = N r mg sin θ, mr θ = mg cos θ. There is no constraining force (i.e. non-conservative force in the tangential equation, so this can be rephrased as an energy-conservation law: Hence, E = 2 m θ 2 + mgr sin θ = E = E (t = 0 = mgr sin (π/2 = mgr. r θ 2 = 2g ( sin θ. Insert this result into the radial EOM, obtain N r = +mg sin θ mr θ 2, = mg sin θ 2mg( sin θ, = 2mg + 3mg sin θ. The particle falls off the semicircle when the force constraining it to the surface vanishes, i.e. N r = 0, or = sin θ. 2 3 It is customary to measure the angle in this problem form the vertical, φ = 2 π θ, hence cos φ = sin θ, and φ = cos 2 3. Please do not put this into your calculator and write θ = 0.73 Rad or θ 0.73 Rad, as both these answers are only finite-precision approximations to the correct answer and are therefore technically, wrong. 5

6 Figure : 6