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PHYS_2014_Homework_6_soln

PHYS_2014_Homework_6_soln - PHYS2014 Benton Fall 2010 OSU...

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PHYS2014 Fall 2010 Benton OSU Physics Dept. 6-1 Physics 2014: General Physics I Solutions to Homework Assignment 6 Due: Friday, 15 October 2010 at the beginning of your Recitation session. 1. A 0.2 g bug is sitting at the edge of a 12” (30.48 cm) diameter phonograph turntable that is rotating at 33.33 rpm (revolutions per minute). What is the minimum coefficient of static friction, μ s , needed to keep the bug from sliding off? Solution: First convert all units to SI units. M bug = 0.0002 kg, r =0.1524 cm, 33.33 rpm = 0.55 rev/s. ω = 2 π f = 3.49 rad/s. The bug is being held onto the turntable only by the force of friction, so the force of friction must be equal to the centripetal force. 2 f s N s f s F F mg F m r m μ μ ω μ = = = g m = ( ) 2 2 2 2 3.49rad s 0.1524m 0.19 9.8m s s r r g ω ω μ = = = 2. A horse pulls a 700 kg cart at a constant speed of 2 m/s up a hill having a constant slope of 10 ° with respect to the horizontal. a) If the coefficient of rolling friction of the cart is 0.05, how much work does the horse do in 5 minutes? b) How much of that work is done to overcome friction? Solution: We’re asked to find the work done by the horse. W F s = ⋅Δ arrowrightnosp arrowrightnosp We’re not given the distance, but since the horse is moving at a constant speed we can find it easily: s v t W F v t Δ = Δ = ⋅ Δ arrowrightnosp arrowrightnosp arrowrightnosp arrowrightnosp Since the horse is moving at a constant speed, there is zero acceleration and the sum of the forces must equal zero. Let’s start by rotating the coordinate system by 10 ° counterclockwise.
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PHYS2014 Fall 2010 Benton OSU Physics Dept. 6-2 ( ) cos 0 cos sin cos sin cos sin cos sin y N g N x H f g f r H f g H r H r F F F F mg F F F F F mg F F F F mg mg F mg θ θ θ μ θ θ μ θ θ μ θ θ = = = = = = + = + = + The last equation is the amount of force the horse must exert to pull the cart. ( ) ( ) 2 cos sin 700 9.8m s 0.05cos10 sin10 2m s 300s 917,409J r W mg v t W kg μ θ θ = + Δ = + = ringoperator ringoperator The amount of work done to overcome friction will just be the force of friction multiplied by the distance Δ s. 2 cos 0.05 700kg 9.8m s cos10 2m s 300s 202,673J f f r W F s F v t W mgv t W μ θ = ⋅Δ = ⋅ Δ = Δ = = ringoperator arrowrightnosp arrowrightnosp arrowrightnosp arrowrightnosp The percentage of the work needed to pull the cart up the slope that goes into overcoming friction is then: 202,673J 100 22.1% 917,409J × = 3. a) Estimate the mass of the Sun based on the average distance between the Sun and the Earth and given the fact that the orbital period of the Earth is 365.25 days, b) Estimate the mass of the Earth based on the average distance between the Earth and the Moon and given the fact that the orbital period of the Moon is 27.3 days. Compare your estimates to the values printed on the back cover of your text book. Solution: a) From the back cover of our textbook, we find that the mean radial distance between the Earth and the Sun is 1.50 × 10 11 m. The angular velocity of the Earth around the sun is: θ = 10 ° N F arrowrightnosp f F arrowrightnosp g F arrowrightnosp H F arrowrightnosp
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PHYS2014 Fall 2010 Benton OSU Physics Dept.
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