Lecture - 12 More examples of Newton’s laws EXAMPLE: Merry-go-round Little Jacob (15 kg) sits on the edge of a merrygo-round of radius 1.0 m while his big sister makes it turn… faster and faster. How fast can the system go before Jacob takes off if the coefficient of static friction between Jacob’s pants and the merry-go-round is 0.50? Little Jacob (15 kg) sits on the edge of a merry-go-round of radius 1.0 m while big sister makes it turn… faster and faster. How fast can the system go before Jacob takes off if the coefficient of static friction between Jacob’s pants and the merry-go-round is 0.50? Centrifugal force N (fictitious force), it is the direction Jacob tends to be pushed out fs mg To oppose the tendency, the force towards the center comes from the contact between Jacob and the merry go round: force of static friction 2 Static friction provides the needed radial acceleration: fS = mrω Maximum speed → Maximum static friction: ωMAX = µS g r = (0.50)(9.8 m/s2 ) 1.0 m 2 µS mg = mr ωMAX = 2.2 rad/s ≈ 1 3 turn/s Example: Flat Curve A car of mass m with constant speed v drives through a curve of radius R. What is the minimum value of the coefficient of static friction between the tires and the road for the car not to slip? Fnet = ma a fs v2 fs = m R fs ≤ µS N = µS mg v v2 m ≤ µS mg R Or, given ms, there is a maximum v: v2 µS ≥ gR v ≤ µ s gR Example: Bucket A stone of mass m sits at the bottom of a bucket. A string is attached to the bucket and the bucket is made to move in circles. What is the minimum speed that the bucket needs to have at the highest point of the trajectory in order to keep the stone inside the bucket? N R mg + N = ma mg v2 a needs to be a = R v2 mg + N = m R N R mg v2 mg + N = m R N= m( v2/R) - mg DEMO: Bucket with water • If v increases, N needs to be larger (if v becomes too large, since N is also the force on the bucket by the stone, the bottom of the bucket might end up broken…) • If v decreases, N needs to be smaller. But at some point, N will become zero! This is the condition for the minimum speed: 2 v min mg = m R ⇒ v min = gR The speed cannot get any smaller or the trajectory will not be a circle anymore (because the remaining forces – mg-- will produce an acceleration that is too strong for a circle of radius R ―at that speed) R If v < vmin, the stone will do something like this... ACT: Tension and mass Example Test Question Example Test Ques.on Example: Banked Curve A car of mass m with constant speed v drives through a banked curve of radius R. If there is no friction, what speed does the car need to maintain for it not to slide up or down the roadway? y x x: N sinθ = max = mv2/R Y: N cosθ – mg = may = 0 N θ a tanθ = v2 Rg ⇒ v = Rg tanθ If R = 400 m, v = 49 m/s = 109 mph mg v = velocity out of page θ= 31o R DEMO: Conical pendulum Example: Box on truck A box with mass m = 50 kg sits on a truck. The coefficients of friction between the box and the truck are μK = 0.2 and μS = 0.4. What is the maximum acceleration that the truck can have without the box slipping? a m a NB,T Direction of motion fS B,T Not slipping: static fS = mBa relative to the truck in the absence of friction WB,E fS MAX = mBa MAX N-W = 0 amax µSN µS mB g fS ,MAX = µS g = 0.4 g = 3.9 m/s2 = = = mB mB mB Drag forces For solid-fluid relative motion, friction force (called “drag force” or “resistance”) depends on the relative speed: fD = kv for low speeds fD = Dv 2 for high speeds k and D depend on the geometry and the materials. Terminal speed Acceleration of a suitcase that falls from a plane: fD mg − fD = ma Increases with v Eventually, fD = mg , so a = 0! When this happens, the system has reached its terminal speed: 2 mg − Dv terminal =0 v terminal mg = D mg This is how parachutes work! DEMO: Parachute EXAMPLE: Pulling yourself up A kid with mass m = 30 kg has designed a rough elevator to get to his tree-house. It’s made of a seat of mass M = 5 kg, a rope and a pulley. To use the elevator, you sit on the seat and pull on the rope as shown below. How strong is the kid pulling if the elevator is moving at constant speed? 2T − (m + M ) g = (m + M )a = 0 T = m +M 2 g T T 35 kg T = (9.8 m/s2 ) = 172 N 2 (M+m)g EXAMPLE: Box on another box A box of mass m1 = 1.5 kg is being pulled by a horizontal string with tension T = 45 N. It slides with friction (μK = 0.50, μS = 0.70) on top of a second box of mass m2 = 3.0 kg, which in turn sits on a frictionless floor. Find the acceleration of box 2. a1 T a2 ? μK = 0.5 m1 m2 frictionless T m1 N fK m2 fK m1g For box 2: fK = μKN = m2a2 (only force in x- direction) From box 1, we know that N - m1g = 0 a2 = µk N m2 = µk m1 m2 g = 0.50 1.5 kg 3.0 kg (9.8 m/s2 ) = 2.5 m/s2 The magnitude of the tension did not play any role! The tension just needs to be large enough so the boxes cannot move together. The magnitude of the tension did not play any role! The tension just needs to be large enough so the boxes cannot move together. If they moved together, the acceleration of both blocks would be: T 45 N m1 a= = = 10 m/s2 T m 1 + m2 4.5 kg fS The static friction would be the only horizontal force on m2: m2 m2a = fS But static friction has a maximum value: This imposes a lower limit to the coefficient of static friction: µS ≥ m2a m1 g = m2T = fS f S ≤ µSN = µSm 1 g m2a ≤ µSm1 g (3.0 kg)(45 N) = 2.0 > 0.7 kg)(9.8 m/s ) (m + m ) m gBlocks(4.5willkg)(1.5 not move together for this tension 1 2 1 2 Example Test Question

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