Energy: ( E = \frac12 I \omega^2 + \textconst ), but rolling: ( I_\texthoop = mR^2 ), point mass at bottom adds ( m(0)^2 )? No, at bottom it’s distance ( R ) from center, but ( I_\texttotal ) about center: hoop ( mR^2 ) + point mass ( mR^2 = 2mR^2 ). Kinetic energy ( = \frac12 (2mR^2)\omega^2 + \frac12 (2m)v_\textcm^2 ). But ( v_\textcm = \omega R ) → total ( K = mR^2\omega^2 + m\omega^2R^2 = 2mR^2\omega^2 ). Potential energy: ( U = mg(2R) ) relative to bottom point mass. Energy conservation from initial (mass at same height as center, ( U = mgR )): ( mgR = mg(2R) + 2mR^2\omega^2 ) → ( -mgR = 2mR^2\omega^2 ) impossible — contradiction means constant ( \omega ) not possible without external torque. But if they ask for ω given rolling without slipping at bottom: Equilibrium: torque from gravity on point mass relative to contact point = 0 at bottom? Actually at bottom, torque due to gravity relative to contact point: ( mg \times 0)? Wait, at bottom, point mass is directly below contact point? No, contact point is at bottom of hoop. Point mass at bottom of hoop coincides with contact point? Then ( I= mR^2 + m(0)^2)? This degenerates. Problem likely means: constant ω means ( \alpha=0 ), so ( \tau_\textnet=0 ) → no gravity torque if mass at bottom? Yes. So ω arbitrary in principle. But energy conservation from start gives specific ω: initial U = mgR, final U = mg·0 (mass at bottom), ΔU = -mgR → ΔK = +mgR = 2mR^2ω^2 → ω = sqrt(g/(2R)). Problem 2 – Solution 1. Force balance on piston: ( pS = p_0 S + Mg + kx ) where ( x = (V_0-V)/S ) for volume V relative to V0? Better: equilibrium: ( pS = p_0 S + Mg + k \Delta x), with ( \Delta x = (V_0 - V)/S ) if we set V0 as reference? Actually, spring relaxed when V=0 → Δx = V/S when volume = V. So ( p = p_0 + \fracMgS + \frackVS^2 ).
Power = ( \mathcalE^2/R ) = ( (B^2\omega^2 L^4)/(4R) ). Mechanical power = ( \tau \omega ) → ( \tau = B^2\omega L^4/(4R) ). russian physics olympiad
At minimum deviation, rays emerge parallel (collimated). Lens focal length 20 cm. Object 10 cm before prism: The collimated beam from prism means rays incident on lens are parallel. A parallel beam is focused at the focal point: image at 20 cm after lens. Magnification: for an extended object, each point of object produces parallel beam at different angle → lens forms image at same plane (focal plane). Size: Angular size of object from prism’s position? But simpler: The prism at min deviation acts like parallel displacement, object distance to lens effectively: object to prism (10 cm) + prism to lens (30 cm) = 40 cm. Lens formula: 1/f = 1/u + 1/v → 1/20 = 1/40 + 1/v → 1/v = 1/20 – 1/40 = 1/40 → v=40 cm (image beyond focal point). Magnification = v/u = 40/40 = 1 → image height = 1 cm. Grading Rubric (per problem) | Part | Points | |------|--------| | Correct setup of equations | 3 | | Correct algebra/calculus | 3 | | Final numeric/analytic answer | 2 | | Proper physical reasoning | 2 | Energy: ( E = \frac12 I \omega^2 +
– initial: ( p_0 V_0 = RT_0 ) (1 mole), but p0 here is equilibrium pressure, not atm. Use p(V) above: At V0: ( p_0' = p_0 + Mg/S + kV_0/S^2 ). At V=2V0: ( p_2 = p_0 + Mg/S + 2kV_0/S^2 ). Ideal gas: ( pV = RT ) → ( T_f = p_2 (2V_0)/R ). From initial ( T_0 = p_0' V_0 / R ) → ( T_f/T_0 = 2p_2/p_0' ). Substitute p2 and p0'. But ( v_\textcm = \omega R ) →
( Q = \Delta U + W ), ( \Delta U = \frac32 R(T_f - T_0) ). ( W = \int_V_0^2V_0 p(V) dV = \int_V_0^2V_0 \left( p_0 + Mg/S + \frackVS^2 \right) dV ) = ( \left(p_0 + Mg/S\right) V_0 + \frack2S^2 (3V_0^2) ). So ( Q = \frac32 R(T_f - T_0) + \left(p_0 + Mg/S\right) V_0 + \frac3kV_0^22S^2 ). Problem 3 – Solution 1. Emf Element dr at distance r: ( d\mathcalE = B v dr = B \omega r dr ). Integrate: ( \mathcalE = \int_0^L B\omega r dr = \frac12 B\omega L^2 ).