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Dynamics II Motion in a Plane Review Problems Problem 1 A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired , exerts an 8.0 N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal loop that is 20 m above the launch point. At what horizontal point to the left of the hoop should you launch? AP Physics C 2 Problem 1 con’t AP Physics C 3 Problem 1 con’t Analysis: Essentially a projectile motion problem with positive vertical acceleration and constant horizontal velocity. Assume a particle model. Compute the time required to rise 20 m and use that to determine the horizontal distance. 8N- .5kg(9.8m / s2 ) = 6.2m / ss FT - mg = ma a = .5kg 1 20 = 6.2 t 2 2 t = 2.54 sec d = 2.54s(3m / s) = 7.62m AP Physics C 4 Problem 2 A highway curve of radius 500 m is designed for traffic moving at speeds of 90 kph. What is the correct banking angle of the road? Analysis: Nr provides the centripetal acceleration and Nz counters the gravitational force. mv2 Nsinθ = Ncosθ = mg r θ θ v2 tanθ = rg 252 = 0.128 7.3 500(9.8) AP Physics C 5 Problem 3 A 30 g ball rolls around a 40 cm diameter horizontal track at 60 rpm. What is the magnitude of the net force that the track exerts on the ball? Neglect rolling friction. F z Analysis: Force against the vertical element of the track provides centripetal acceleration and the horizontal supports the weight. Fr 2 60 rev 2π rad 1min Fr = mr 2 = 0.030 kg 0.20 m × × = 0.24 N 1 rev 60 s min AP Physics C 6 Problem 4 A student has 65 cm arm length. What is the minimum angular velocity, in rpm, for swinging bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm. Analysis: The minimum angular velocity for swinging a bucket of water in a vertical circle without spilling any water corresponds to the case when the speed of the bucket is critical. In this case, n = 0 N when the bucket is in the top position of the circular motion. mvc2 FrG = 0 N+mg = r = mrωc2 9.8 m / s2 ωc = g / r = = 3.13 rad / s = 30 rpm 1.00 m AP Physics C 7 Problem 5 A car is tested on a 200 m diameter track. If the car speeds up at a steady 1.5 m/s2, how long after starting is the magnitude of its centripetal accelerations equal to the tangential acceleration? Analysis: NUCM; we know at = 1.5 m/s2 from that we should be able to determine the angular velocity and acceleration, ω and α, and then the time. ar = ω r = 1.5 m / s 2 2 1.5 m / s2 1.5 m / s2 ω= = = 0.122 rad / s r 100 m at 1.5 m / s2 α= = = 1.5 ×10-2 s-2 . r 100 m ω = ωi + αΔt ω - ωi 0.122 s-1 - 0 s-1 Δt = = = 8.2 s α 0.015 s-2 AP Physics C 8 Problem 6 A popular pastime is to see who ca push an object closest to the edge of a table without its going off. You push the 100 g object and release it 2.0 m from the table edge. Unfortunately you push a little too hard. The object slides across, sails off the edge, falls 1.0 m to the floor, and lands 30 cm from the edge of the table. If the coefficient of kinetic friction is 0.50, what was the object’s speed as you released it? Analysis: After release velocity is affected by friction only. When it reaches the edge of the table it becomes a projectile motion problem with horizontal velocity constant. Best approaches appear to be to work backward from the projectile motion to the release velocity. AP Physics C 9 Problem 6 con’t Δy = 1 m and v1y = 0 y2 = y1 + v1y t2 - t1 + 21 -g t2 - t1 2 0 m = 1.0 m + 0 m- g 2 t2 - t1 t2 - t1 = 0.4518 s 2 Similarly in the x direction x2 = x1 + v1x t 2 - t1 2.30 m = 2.0 m + v1x 0.4518 s v1x = 0.664 m/s Given v1x we can find v0x given the acceleration. F x = -fk = max = -μkmg ax = -μkg = - 0.50 9.8 m/s2 = -4.9 m/s2 2 2 v1x = v0x + 2ax x1 - x0 : 0.664 m/s 2 AP Physics C = v20x +2-4.9 m/s2 2.0 m v0x = 4.5 m/ s 10 Problem 7 A motorcycle daredevil plans to ride up a 2.0 m high, 20° ramp, sail across a 10 m wide pool filled with crocodiles, and land at ground level on the other side. Unfortunately, the motorcycle engine dies just as he starts up the ramp. He is going 11 m/s at that instant, and the coefficient of rolling friction is 0.02. Does he make it? Analysis: Two part problem: First is a ramp problem to determine velocity at end of ramp. Second part is a projectile motion problem determining how far right he will travel while dropping 2.0 m. Different sets of axes will be required for each part. AP Physics C 11 Problem 7 con’t First find the acceleration parallel to the ramp: (Fnet )|| = -fr - mgsin20° = -μrn - mgsin20° = -μrmgcos20° - mgsin20° = ma0 Dividing by M and solving for a0 gives: a0 = -g(μr cos20° + sin20°) = -(9.8 m / s2 )((0.02)cos20° + sin20°) = -3.536 m / s2 The length of the ramp is 2 m/ sin 20° = 5.85 m Using v12 = v02 + 2a0 (s1 - s0 ) v1 = (11.0 m / s)2 + 2(-3.536 m / s2 )(5.85 m) = 8.92 m / s AP Physics C 12 Problem 7 con’t Now we have the velocity at the beginning of projectile motion. Axes are now horizontal and vertical. Velocity components are: v1x = v1cos20° = 8.38 m s v1y = v1sin20° = 3.05 m S Time in the air is given by: y2 = 0 m = y1 + v1y t + 21 a1y t 2 For Δt = t 2 - t1 = 2.0 m + (3.05 m / s) t - (4.90 m / s2 ) t 2 t = 1.021s X2 is given by: 8.38 m/s (Δt) = 8.56 m. Looks like crocodile food!! AP Physics C 13 Problem 8 A concrete highway curve of radius 70 m is banked at a 15° angle. What is the maximum speed with which a 1500 kg rubber tired car can take this curve without sliding? The summation of r and z forces gives: mv 2 Fr = fkcosθ + nsinθ = r Fz = ncosθ - fk sinθ - FG = 0 Maximum speed is when the static friction force reaches its maximum value fk max = μkn. Substituting this into the above equations gives: mv 2 n μscos15° + sin15° = r n cos15° - μssin15° = mg Dividing the bottom into the top gives: AP Physics C μs + tan15° v2 μ + tan15° = v = gr s = 34 m s 1- μs tan15° gr 1- μs tan15° 14 Problem 9 A conical pendulum is formed by attaching a 500g ball to a 1.0 m long string, then allowing the mass to move in a horizontal circle of radius 20 cm. a. Find the tension is the string and b. the angular speed of the ball in rpm. Analysis: The mass moves in a horizontal circle of radius The acceleration and the net force vector point to the center of the circle, not along the string. The only two forces are the string tension, which does point along the string, and the gravitational force. These are shown in the free-body diagram. Newton’s second law for circular motion is F = Tcosθ - F = Tcosθ - mg = 0 N mv F = Tsinθ = ma = r z G 2 r From FZ: AP Physics C r 0.500 kg 9.8 m / s mg T= = cosθ cos11.54° 2 = 5.00 N 15 Problem 10 A 500 g ball moves in a vertical circle on a 102 cm long string. If the speed at the top of the 4.0m/s, then the speed at the bottom will be 7.5 m/s. Find (a) the gravitational force acting on the ball, (b) The tension in the string at the top of the circle and (c) the tension in the string at the bottom of the circle. (a) The gravitational force is: FG = mg = 0.500 kg 9.8 m / s2 = 4.9 N. (b) The tension in the string at the top of the circle is: 4.0 m / s 2 v2 2 T1 = m - g = 0.500 kg - 9.8 m / s = 2.9N 1.02 m r AP Physics C 16 Problem 10 con’t (c) Similarily at the bottom of the circle the tension is: mv 2 Fr = T2 - FG = r 2 7.5 m / s v2 2 T2 = m g + = 0.500 kg 9.8 m / s + = 32 N r 1.02 m AP Physics C 17