Physics

Physics · mechanics

Problem

ii. What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision? Given constants: g=9.8 N/kgG=6.67×10−11 N⋅m2/kg2k=1/4πϵ0=8.99×109 N⋅m2/C2km=μ0/4π=10−7 T⋅m/Ac=3.00×108 m/skB=1.38×10−23 J/KNA=6.02×1023( mol)−1R=NAkB=8.31 J/(mol⋅K)σ=5.67×10−8 J/(s⋅m2⋅K4)e=1.602×10−19C1eV=1.602×10−19 Jh=6.63×10−34 J⋅s=4.14×me=9.109×10−31 kg=0.511MeV/c2(1+x)n≈1+nx for ∣x∣≪1sin⁡θ≈θ−16θ3 for ∣θ∣≪1cos⁡θ≈1−12θ2 for ∣θ∣≪1\begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} M⊙=1.989×1030 kgL⊙=3.828×1026 WRearth =1.5×1011 mλmax⁡=500 nm\begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} Context question: a. Suppose you drop a block of mass mm vertically onto a fixed ramp with angle θ\theta with coefficient of static and kinetic friction μ\mu. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is vv and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision? Context answer: u=v(sin⁡θ−μcos⁡θ)u=v(\sin \theta-\mu \cos \theta) Context question: ii. What is the minimum μ\mu such that the speed of the block right after the collision is 0 ? Context answer: μ=tan⁡θ\mu=\tan \theta Context question: b. Now suppose you drop a sphere with mass mm, radius RR and moment of inertia βmR2\beta m R^{2} vertically onto the same fixed ramp such that it reaches the ramp with speed vv. i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case? Context answer: u=vsin⁡θ1+βu=\frac{v \sin \theta}{1+\beta}

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