Physics Past Paper Solutions (Chapter 6) PDF

# Torque ## 6.17 Torque - ←こ - 2=-Fll₂-F'l+mg sind/2 - since displacement and force are in opposite directions. - FI/2=h0 - F=kx = k\theta - F'=kx' = k\theta - ⇒ τ = +k\theta.ΔL - k\theta.l + mg l\theta/2 - τ = [-k1²/4 - k1²/4 + mgl/2] - τ = -k1²/4 - k1²/4 + mgl/2 - θ = 0 - θ²+ω²θ=0 ⇒ ω² =...

# Torque ## 6.17 Torque - ←こ - 2=-Fll₂-F'l+mg sind/2 - since displacement and force are in opposite directions. - FI/2=h0 - F=kx = k\theta - F'=kx' = k\theta - ⇒ τ = +k\theta.ΔL - k\theta.l + mg l\theta/2 - τ = [-k1²/4 - k1²/4 + mgl/2] - τ = -k1²/4 - k1²/4 + mgl/2 - θ = 0 - θ²+ω²θ=0 ⇒ ω² = k1²/4 + mgl/2 / ml²/3 ## 6.18 - τ=γF = Iα = Iθ" - τ = -mgsin\theta l/2 - Mgsin\theta l - τ = (mg/2 - Mgl ) θ = I θ" - I[ (mg/2 +Mgl) θ = 0 - θ = 0 - I[θ" + ω² θ = 0 - ω = √(mg/2 + Mgl)/I - T = 2π/ω ### Cases #### Case 1 - I = Irod + Idist - MOI about the pivot when no rotations of the disc - 1/2 m1² + 1/2 Ml² - Put it in equation 1 to get the answer. #### Case 2 - I = ml²/3 + Ml² + MR² - When disc starts to rotate - Put it in equation 1 to get the answer. ## 6.23 - The length of the tape = x + y + ΔY/2 * circumference - When unwrapping, the length of the rope increases by length = ΔRO - (α) At any time - x + y = constant + ΔRO - x' + y' = ΔR'O - ΔR'O = a + Δ - a = ang. accel - (β) a, A, and α = ? - Equation of motions - 1. Translational: mg + T = ma and Mg +T = MA - 2. Rotational: TΔ = Iα = MR²α - Torque: - T = 1/2 *MRα = 1/2 * MCA + A using equation 2 - Use this expression of T in equation 2 ## 6.24 - Initial accel" of drum A? - Torque on both the drum (which are connected) are the same = TR (using RxF) - any displacement x because of rotation of the drums x = 10 + (RO+RO) * a = 2ΔRo, where ΔR = length of the tape increase in length of the tape when both drums are at rest when drums start rotating - equation of motion of drum A: Mg - T = Ma and Torque: TR = 1/2 * MR^2 * α - equation of motion of drum B: TR = 1/2 * MR^2 * α and T = 1/2 * MR α = 1/2 * MA (using 1/2 * MRα = MCA + A) - From 2, 1/2*MA = Mg - MA - ⇒ a = 4/3 * g. This is the accel of drum a. ## 6.25 - What is l after which it roll back? - Initial energy → Translational motion Ei = 1/2 mu² 1/2 * Iω² → Rotational Motion and I = 2/5 * MR^2 - I = the moment of inertia of a sphere about an axis through its center - for Rotation without slipping, ΔRO = V0 - Using it in equation, Ei = 1/2 *mu² + 1/2 * 1/5 * Mu^2 = 7/10 * Mu^2. - The marble is momentarily stationary at final height. - Conservation of: - Ef = 0 + MgΔh = MgLsinθ = Ei = 7/10 * Mu^2 - l = 7/10 * ( Mu^2/ Mgsinθ) = 7/10 * (V0²/gsinθ) ## 6.26 - Ef = 1/2 * mu^2 + 1/2 * Iω^2 = Ei = MgΔh - As for rolling without slipping, RO = v - Ef = 1/2 * Mu^2 + 1/2 * I (V^2/R^2) = ( 1/2 * M+ 1/2 * I/R^2) * v^2 - v = √(Mgh/ (1/2 * M + 1/2 * I/R^2)) - Howe, MOI of sphere, Isphere = 2/5* MR^2 and MOI of cylinder Icylinder = 1/2 * MR^2 - Since Isphere < Icylinder - ⇒ vsphere > Vcylinder ⇒ sphere is faster ## 6.27 - Coefficient of friction b/w yo-yo & table is μ - Max'm value of F for which yo-yo will roll without slipping? - Horizontal equation of motion: translation F - f = Ma' = MA (1) - Vertical equation of motion: translation N - Mg = 0 ⇒ N = Mg (2) - Rotational equation of motion: Rf - bF = Iω' = 1/2 * MR^2 * α = 1/2* MRA, where R = aΔ - Using the in equation 1 - f = F - MA = Rf/2 - (MA/2) + bF = Rf - f = (MA/2) + bF = MA + b(MA + f) * f(1 - b) = MA /2 + bMA