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Centripetal Acceleration 13 Examples with full solutions Example 1 A 1500 kg car is moving on a flat road and negotiates a curve whose radius is 35m. If the coefficient of static friction between the tires and the road is 0.5, determine the maximum speed the car can have in order to successfully make the turn. 35m Example 1 – Step 1 (Free Body Diagram) F N Acceleration direction F ac F +y +x This static friction is the only horizontal force keeping the car moving toward the centre of the arc (else the car will drive off the road). G f Example 1 - Step 2 (Sum of Vector Components) F N +y ac F F Vertical Components F F N y 0 FG 0 FN m g Static Friction G Horizontal Components F x mac F f mac We have an acceleration in x-direction s FN m a c FN m g 0 f v2 s FN m R v 2 s FN R m s mg R m v s g R From Vertical Component +x Example 1 - Step 3 (Insert values) F N +y ac F F v s g R 0.50 9.8 13.0958 m 35 m 2 s m s 13.0958 m s 47 km h 1km 1000 m 3600 s 1h G f +x Example 2 A car is travelling at 25m/s around a level curve of radius 120m. What is the minimum value of the coefficient of static friction between the tires and the road to prevent the car from skidding? 120 m Example 2 – Step 1 (Free Body Diagram) F N Acceleration direction F ac F +y +x This static friction is the only horizontal force keeping the car moving toward the centre of the arc (else the car will drive off the road). G f Example 2 - Step 2 (Sum of Vector Components) F N +y ac F F Vertical Components F N F y 0 FG 0 FN m g Static Friction From Vertical Component G Horizontal Components F x mac F f mac s FN m a c FN m g 0 f v2 s FN m R 2 m v s FN R 2 m v mg R v 2 gR We have an acceleration in x-direction +x Example 2 - Step 3 (Insert values) F N +y ac F F s s v 2 We require the minimum value gR v 2 gR 2 m 2 5 s m 9.8 120 m 2 s 0.53 G f +x Example 3 An engineer has design a banked corner with a radius of 200m and an angle of 180. What should the maximum speed be so that any vehicle can manage the corner even if there is no friction? 180 200 m Example 3 – Step 1 (Free Body Diagram) The normal to the road F Acceleration First the car direction 18 F N co s 1 8 F N sin 1 8 ac Now for gravity +y F +x N Components of Normal force along that axis we (we Notice ensured haveone no axis staticwas along acceleration friction force in direction this example G (question did not require one) Example 3 - Step 2 (Sum of Vector Components) F N F N c o s 1 8 18 +y F N s in 1 8 ac +x F Horizontal Components Vertical Components F F Ny y F 0 F FG 0 x mac Nx mac F N sin 18 m a c F N cos 18 m g 0 FN G mg cos 18 v2 mg sin 18 m cos 18 R g tan 18 From Vertical Component v v We have an acceleration in the x-direction 2 R R g tan 18 Example 3 - Step 3 (Insert values) F 18 N F N c o s 1 8 +y F N s in 1 8 ac +x F v R g tan 18 v m 200 m 9.8 tan 18 2 s 25.2357 m s 25.2357 m s 91 km h 1km 1000 m 3600 s 1h G Example 4 An engineer has design a banked corner with a radius of 230m and the bank must handle speeds of 88 km/h. What bank angle should the engineer design to handle the road if it completely ices up? ? 230 m Example 4 – Step 1 (Free Body Diagram) The normal to the road F F N sin ac Now for gravity +y F +x F N co s Acceleration First the car direction N Components of Normal force along that axis we (we Notice ensured haveone no axis staticwas along acceleration friction force in direction this example G (question did not require one) Example 4 - Step 2 (Sum of Vector Components) F N FN c o s +y F N sin ac +x F Horizontal Components Vertical Components F F Ny y F 0 F FG 0 x mac Nx mac F N sin F N cos m g 0 FN G mg cos mg cos s in g tan From Vertical Component m ac v2 m R v 2 R ta n 1 v2 Rg We have an acceleration in the x-direction Example 4 - Step 3 (Insert values) F N FN c o s +y F N sin ac +x F tan 1 v2 R g km 1000 m 1h 88 h km 3 6 0 0 s 1 tan m 230 m 9.8 2 s 2 15 G Don’t forget to place in metres per second Example 5 A 2kg ball is rotated in a vertical direction. The ball is attached to a light string of length 3m and the ball is kept moving at a constant speed of 12 m/s. Determine the tension is the string at the highest and lowest points. Example 5 – Step 1 (Free Body Diagram) Note: any vertical motion problems that do not include a solid attachment to the centre, do not maintain a constant speed, v and thus (except at top and bottom) have an acceleration that does not point toward the centre. (it is better to use energy conservation techniques) Bottom Top FT ac FT F G ac When the ball is at the top of the curve, the string is “pulling” down. F When the ball is at the bottom of the curve, the string is “pulling” up. G In both cases, gravity is pulling down Example 5 - Step 2 (Sum of Vector Components) FT ac +y ac FT F +x G F Top F y Bottom mac F F T F G mac FT m g m a c FT m g m a c m ac g v2 m g r G y mac F T F G mac Note: acceleration is down (-) FT m g m a c FT m a c m g m g ac 2 v mg r Example 5 - Step 3 (Insert values) FT ac +y ac FT F +x G F Top v2 FT m g r 2 kg 7 6 .4 N 2 m 12 m s 9 .8 2 3m s G Bottom 2 v FT m g r m 2 kg 9 .8 2 s 1 1 5 .6 N 2 m 12 s 3m Example 6 A conical pendulum consists of a mass (the pendulum bob) that travels in a circle on the end of a string tracing out a cone. If the mass of the bob is 1.7 kg, and the length of the string is 1.25 m, and the angle the string makes with the vertical is 25o. Determine: a) the speed of the bob b) the frequency of the bob Example 6 – Step 1 (Free Body Diagram) Let’s decompose our Tension Force into vertical and horizontal components FT FT co s 2 5 25 ac FT sin 2 5 +y +x It’s easier to make the x axis positive to the left F G Example 6 - Step 2 (Sum of Vector Components) FT FT co s 2 5 25 ac FT sin 2 5 +y +x F Horizontal F x Vertical mac F y 0 FT x m a c F Ty F G 0 FT sin 25 m a c FT cos 25 m g 0 FT sin 25 m v 2 r FT mg cos 25 G Example 6 - Step 3 (Insert values for velocity) FT FT co s 2 5 25 ac +y FT sin 2 5 +x F FT sin 2 5 m mg co s 2 5 2 v 2 r g r ta n 2 5 v r sin 2 5 m g tan 2 5 v v v m 9.8 1.25 m sin 25 tan 25 2 s 2 r G 1.554 m s 1.55 m s The speed of the bob is about 1.55 m/s Example 6 - Step 3 (Insert values for frequency) FT FT co s 2 5 25 ac FT sin 2 5 +y +x FT sin 25 m a c f FT sin 25 m 4 rf 2 mg sin 25 4 rf 2 cos 25 g tan 25 4 rf 2 f 2 g tan 25 4 r 2 2 2 g tan 25 4 r 2 m 9.8 tan 25 2 s 2 4 1.25 m sin 25 0.46811 H z 0.468 H z The frequency of the bob is about 0.468Hz F G Example 7 A swing at an amusement park consists of a vertical central shaft with a number of horizontal arms. Each arm supports a seat suspended from a cable 5.00m long. The upper end of the cable is attached to the arm 3.00 m from the central shaft. Determine the time for one revolution of the swing if the cable makes an angle of 300 with the vertical Example 7 – Step 1 (Free Body Diagram) FT FT co s 3 0 ac 30 30 FT sin 3 0 +y +x F G Example 7 - Step 2 (Sum of Vector Components) FT FT co s 3 0 ac 30 30 FT sin 3 0 +y +x F Horizontal F x Vertical F mac y 0 FT x m a c F Ty F G 0 FT sin 30 m a c FT cos 30 m g 0 FT sin 30 m 4 r 2 T 2 FT mg cos 30 G Example 7 - Step 2 (Sum of Vector Components) FT FT co s 3 0 ac 30 30 FT sin 3 0 +y 3.00 m 5.00 m sin 30.0 +x FT sin 30 m mg cos 3 0 sin 30 m g tan 30 4 r 2 F 2 T T 4 r 2 2 3 .00 m 5.00 m sin 30 2 T 4 r 2 T 4 m 9.8 tan 30 2 s 6.1948 s 2 6.19 s 4 r 2 T g t an 3 0 G The period is 6.19s Example 8 A toy car with a mass of 1.60 kg moves at a constant speed of 12.0 m/s in a vertical circle inside a metal cylinder that has a radius of 5.00 m. What is the magnitude of the normal force exerted by the walls of the cylinder at A the bottom of the circle and at B the top of the circle Example 8 – Step 1 (Free Body Diagram) Note: any vertical motion problems that do not include a solid attachment to the centre, do not maintain a constant speed, v and thus (except at top and bottom) have an acceleration that does not point toward the centre. (it is better to use energy conservation techniques) Bottom Top F ac F N F G When the car is at the top of the curve, the normal force is “pushing” down. N ac F When the ball is at the bottom of the curve, the normal force is “pushing” up. G In both cases, gravity is pulling down Example 8 - Step 2 (Sum of Vector Components) F N ac +y ac F N F +x G F Top F F N y Bottom mac F F G mac FN m g m a c FN m g m a c m ac g v2 m g r G F Note: acceleration is down (-) N y mac F G mac FN m g m a c FN m a c m g m g ac 2 v mg r Example 8 - Step 3 (Insert values) F N ac +y ac F N F +x G F Top v2 FN m g r 2 m 12.0 m s 1.60 kg 9.8 2 5.00 m s 30.4 N G Bottom 2 v FT m g r m 1.60 kg 9.8 2 s 61.8 N m 12.0 s 5.00 m 2 Example 9 A 0.20g fly sits 12cm from the centre of a phonograph record revolving at 33.33 rpm. a) What is the magnitude of the centripetal force on the fly? b) What is the minimum static friction between the fly and the record to prevent the fly from sliding off? Example 8 – Step 1 (Free Body Diagram) F Fc N Fs F G Example 9 - Step 2 (Sum of Vector Components) F N Fc Fs F G a. F m ac Fs m 4 r f 2 2 4 2.0 10 k g 4 2. 9 1 0 4 N Convert to correct units 2 1 rev 1 m in 2 0.12 m 33 3 m in 6 0 s Example 9 - Step 2 (Sum of Vector Components) F N Fc Fs F b. F m ac Fs m 4 r f 2 2 s m g 2.9 10 4 2.9 10 4 s G N N mg 2.9 10 2.0 10 0. 1 5 4 4 N m kg 9.8 2 s Example 10 A 4.00 kg mass is attached to a vertical rod by the means of two 1.25 m strings which are 2.00 m apart. The mass rotates about the vertical shaft producing a tension of 80.0 N in the top string. a) What is the tension on the lower string? b) How many revolutions per minute does the system make? Example 10 – Step 1 (Free Body Diagram) F T1 sin ac 5 3 .1 +x FT2 1.25 m 53.1 5 3 .1 +y 1.00 m F G Example 10 - Step 2 (Sum of Vector Components) F T1 ac 5 3 .1 +y 5 3 .1 + FT2 Horizontal F x mac F T 1 x F T 2 x m a cx FT 1 cos 53.1 FT 2 cos 53.1 m a c F G Vertical F y 0 F T1 F T1 F G 0 FT 1 sin 53.1 FT 2 sin 53.1 m g 0 Example 10 - Step 2 (Sum of Vector Components) F T1 ac 5 3 .1 +y 5 3 .1 + FT2 a) 80.0 N sin 53.1 FT 2 sin 53.1 4.00 kg 9.8 F G N 0 kg N 4.00 kg 9.8 80.0 N sin 53.1 kg FT 2 sin 53.1 FT 2 30.98 N 31 N Example 10 - Step 2 (Sum of Vector Components) F T1 ac 5 3 .1 +y 5 3 .1 + F FT2 b) FT 1 cos 53.1 FT 2 cos 53.1 m a c 2 2 FT 1 cos 53.1 FT 2 cos 53.1 m 4 rf f 2 f F T1 cos 53.1 FT 2 cos 53.1 4 rm 2 0 .7 4 9 8 0 .7 4 9 8 80.0 N cos 53.1 30.98 N c os 53.1 4 1.25 m cos 53. 1 4.00 kg 2 rev 6 0 s s m rev 4 4 .9 9 m in cos 53.1 FT 2 cos 53.1 T1 4 r m rev s F 2 G 45 rev m in Example 11 The moon orbits the Earth in an approximately circular path of radius 3.8 x 108 m. It takes about 27 days to complete one orbit. What is the mass of the Earth as obtained by this data? Example 11 – Step 1 (Free Body Diagram) F G Example 11 - Step 2 (Sum of Vector Components) F G Horizontal F x mm a c F a mm a c GM E mm r 2 GM E mm r 2 GM r E 2 M E 2 ac ac r G GT 2 4 r 3 2 GT m m ac m m ac 4 r r 2 2 4 2 3.8 10 m 2 11 N m 6.67 10 2 kg 8 3 24 h 3600 s 27 d 1 d 1h 2 5.97 10 k g 24 2 The mass of the Earth is about 6.0x1024 kg Example 12 Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she only sides on the seat as the ride moves. a) Determine the Normal Force she would experience from the bottom of the seat when she is at the lowest point on the ride. b) Determine the Normal Force she would experience from the bottom of the seat when she is at the highest point on the ride. c) Determine the Net Force she would experience from the bottom of the seat when she is at the mid-point on the ride with her height equal to the axis. Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she only sides on the seat as the ride moves. a) Determine the Normal Force she would experience from the bottom of the seat when she is at the lowest point on the ride. With this ride, gravity always points down, the normal (seat) force always points up, and the centripetal acceleration is always toward the centre. F c m ac FN Fg m a c FN Fg m a c W m ac Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she only sides on the seat as the ride moves. Determine the Normal Force she would experience from the bottom of the seat when she is at the highest point on the ride. With this ride, gravity always points down, the normal (seat) force always points up, and the centripetal acceleration is always toward the centre. F c m ac FN Fg m a c FN Fg m a c W m ac Example 12 Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she only sides on the seat as the ride moves. Determine the Net Force (force of seat) she would experience from the bottom of the seat when she is at the mid-point on the ride with her height equal to the axis. Fc m a c FN m g F N et Fc F F N et Fc 2 N FN m ac 2 2 mg tan 2 1 mg g 1 tan m ac ac Example 13 (Hard Question) An engineer has design a banked corner with a radius of R and an angle of β. What is the equation that determines the velocity of the car given that the coefficient of friction is µ ? Example 13 – Step 1 (Free Body Diagram) The normal to the road F ac F N sin F Friction +y F +x F N co s Acceleration First the car direction G N f Components of Normal force along axis (we ensured one axis was along acceleration direction We have friction going down by assuming car Now forwants to slide up. This gravity will provide an equations for the maximum velocity Example 13 – Step 2 (Components)F N F N co s F N sin F F F F Ny FG F f G Horizontal Components Vertical Components y 0 f 0 F F Nx F x mac fx mac F N cos m g F f sin 0 F N sin F f cos m a c F N cos m g uF N sin 0 F N sin F N cos m a c From Vertical m F N co s F N sin g ac +y + x Example 13 – Step 2 (Components)F N F N co s F N sin F F F N co s F N sin v gR F N sin F N cos F N cos F N sin g R sin cos cos sin + x f G F N cos F N sin v 2 F N sin F N cos g R Solve for v v +y F N sin F N cos m a c g 2 Sub into Horizontal From Vertical m ac Minimum velocity (slides down) v gR sin cos cos sin π Flash