Physics 4A Chapte 8: Dynamics II Motion in a Plane
Conceptual Questions and Example Poblems fom Chapte 8 Conceptual Question 8.5 The figue below shows two balls of equal mass moving in vetical cicles. Is the tension in sting A geate than, less than, o equal to the tension in sting B if the balls tavel ove the top of the cicle with equal speed? 8.5. The diffeence in the tension between A and B is due to the centipetal foce ( F ) net =. Since the velocity v is the same fo both, the geate adius fo B means that the tension in case A is geate than fo case B. Conceptual Question 8.6 Ramon and Sally ae obseving a toy ca speed up as it goes aound a cicula tack. Ramon says, The ca s speeding up, so thee must be a net foce paallel to the tack. I don t think so, eplies Sally. It s moving in a cicle, and that equies centipetal acceleation. The net foce has to point to the cente of the cicle. Do you agee with Ramon, Sally, o neithe? Explain. 8.6. Neithe Sally no Raymond is completely coect. Both have patially coect desciptions, but ae missing key points. In ode to speed up, thee must be a nonzeo acceleation paallel to the tack. In ode to move in a cicle, thee must be a nonzeo centipetal acceleation. Since both of these ae equied, the net foce points somewhee between the fowad diection (paallel to the tack) and the cente of the cicle. Poblem 8.7 A 00 g block on a 50-cm-long sting swings in a cicle on a hoizontal, fictionless table at 75 pm. (a) What is the speed of the block? (b) What is the tension in the sting? 8.7. Model: Teat the block as a paticle attached to a massless sting that is swinging in a cicle on a fictionless table. Solve: (a) The angula velocity and speed ae ev π ad ω = 75 = 471. ad/min min 1 ev The tangential velocity is 3.9 m/s. (b) The adial component of Newton s second law is vt 1min = ω = (0. 50 m)(471. ad/min) = 3. 93 m/s 60 s
Thus F = T = (3. 93 m/s) T = (0. 0 kg) = 6. N 0. 50 m Poblem 8.13 Mass m1 on the fictionless table of the figue to the ight is connected by a sting though a hole in the table to a hanging mass m. With what speed must m1 otate in a cicle of adius if m is to emain hanging at est? 8.13. Model: Masses m 1 and m ae consideed paticles. The sting is assumed to be massless. Solve: The tension in the sting causes the centipetal acceleation of the cicula motion. If the hole is smooth, it acts like a pulley. Thus tension foces T 1 and T act as if they wee an action/eaction pai. Mass m 1 is in cicula motion of adius, so Newton s second law fo m 1 is 1 F = T1 =
Mass m is at est, so the y-equation of Newton s second law is F = T mg= 0N T = mg y Newton s thid law tells us that T 1= T. Equating the two expessions fo these quantities: Poblem 8.0 A olle coaste ca cosses the top of a cicula loop-the-loop at twice the citical speed. What is the atio of the nomal foce to the gavitational foce? 8.0. Model: Model the olle coaste ca as a paticle at the top of a cicula loop-the-loop undegoing unifom cicula motion. Notice that the -axis points downwad, towad the cente of the cicle. Solve: The citical speed occus when n goes to zeo and F G povides all the centipetal foce pulling the ca in the vetical cicle. At the citical speed vt = v and the centipetal foce is c mg = c /, theefoe vc = g. Since the ca s speed is twice the citical speed, Thus the nomal foce is n = 3 mg. Consequently, nf / G = 3. c m(4 v ) m(4 g) F = n + FG = = = = 4mg Poblem 8.4 A 500 g ball swings in a vetical cicle at the end of a 1.5-m-long sting. When the ball is at the bottom of the cicle, the tension in the sting is 15 N. What is the speed of the ball at that point? 8.4. Model: Model the ball as a paticle which is in a vetical cicula motion. Solve: At the bottom of the cicle, (0. 500 kg) v F = T FG = (15 N) (0. 500 kg)(9. 8 m/s ) = v= 5. 5 m/s (1. 5 m)
Poblem 8.9 An 85,000 kg stunt plane pefoms a loop-the-loop, flying in a 60-m-diamte vetical cicle. At the point whee the plane is flying staight down, its speed is 55 m/s and it is speeding up at a ate of 1 m/s pe second. (a) What is the magnitude of the net foce on the plane? You can neglect ai esistance. (b) What angle does the net foce make with the hoizontal? Let an angle above the hoizontal be positive and an angle below the hoizontal be negative. 8.9. Model: The plane is a paticle undegoing nonunifom cicula motion. Make the -t plane vetical; then thee will be on motion in the z-diection. Neglect ai esistance. Solve: We ae given the tangential acceleation and we find the centipetal acceleation fom a = v /. Use the second law to find the components of the foce. v (55 m/s) 6 F = m = (85000 kg) = 1.978 130 m 6 Ft = mat = (85000 kg)(1 m/s ) = 1.0 (a) Now find the magnitude of the net foce. 6 6 6 Fnet = F + Ft = (1.978 ) + (1.0 ) =. (b) Find the angle the net foce makes with the hoizontal. 6 1 Ft 1 1.0 θ = tan = tan = 7 F 6 1.978 The angle is negative because it is below the hoizontal. Assess: The plane s velocity is down and inceasing, so the net foce must be below the hoizontal. Poblem 8.40 A concete highway cuve of adius 70 m is banked at a 15 angle. What is the maximum speed with which a 1500 kg ubbe-tied ca can take this cuve without sliding? (µ s = 1.0) 8.40. Model: We will use the paticle model fo the ca, which is undegoing unifom cicula motion on a banked highway, and the model of static fiction.
Note that we need to use the coefficient of static fiction µ s, which is 1.0 fo ubbe on concete. Solve: Newton s second law fo the ca is s cos sin F = f θ + n θ = Fz = ncosθ fssinθ FG = 0 N Maximum speed is when the static fiction foce eaches its maximum value ( fs) max = µ sn. Then µ s + = n µ s n( cos15 sin15 ) Dividing these two equations and simplifying, we get s µ s + tan15 v s tan15 v g µ + = = 1 µ tan15 g 1 µ tan15 Assess: The above value of 34 m/s 70 mph is easonable. (cos15 sin15 ) = mg s (1. 0 + 0. 68) = (9. 80 m/s )(70 m) = 34 m/s (1 0. 68) Poblem 8.47 A conical pendulum is fomed by attaching a ball of mass m to a sting of length L, then allowing the ball to move in a hoizontal cicle of adius. The figue below shows that the sting taces out the suface of a cone, hence the name. Find an expession fo the tension T in the sting. 8.47. Use the paticle model fo the ball, which is undegoing unifom cicula motion. We ae given L,, and m, so ou answes must be in tems of those vaiables. L is the hypotenuse of the ight tiangle. The ball moves in a hoizontal cicle of adius = Lcos θ. The acceleation and net foce point towad the cente of the cicle, not along the sting.
Solve: (a) Apply Newton s second law in the z -diection. Fom the ight tiangle cos θ = L / L. (b) Apply Newton s second law in the -diection. mg Fz = T cosθ mg = 0 T = cosθ mg T = = cosθ mgl L F = Tsin θ = mω = mω ( Lsin θ) T = mω L. Set the two expessions fo T equal to each othe, cancel m and one L, and solve fo ω. (c) Inset L = 1. 0 m, = 0. 0 m and m = 0. 50 kg. mgl g = mω L ω = L L mgl (0. 50 kg)(9. 8 m/s )(1. 0 m) T = = = 50N. L (1. 0 m) (0. 0 m) g 9. 8 m/s 1ev 60 s ω = = = 3. 163 ad/s = 30 pm L (1. 0 m) (0. 0 m) π ad 1min Assess: Notice that the mass canceled out of the equation fo ω, but not fo T, so the 500 g was necessay infomation. Poblem 8.56 A 100 g ball on a 60-cm-long sting is swung in a vetical cicle about a point 00 cm above the floo. The tension in the sting when the ball is at the vey bottom of the cicle is 5.0 N. A vey shap knife is suddenly inseted, as shown in the figue below, to cut the sting diectly below the point of suppot. How fa to the ight of whee the sting was cut does the ball hit the floo? 8.56. Model: Model the ball as a paticle swinging in a vetical cicle, then as a pojectile.
Solve: Initially, the ball is moving in a cicle. Once the sting is cut, it becomes a pojectile. The final cicula-motion velocity is the initial velocity fo the pojectile. The fee-body diagam fo cicula motion is shown at the bottom of the cicle. Since T > F G, thee is a net foce towad the cente of the cicle that causes the centipetal acceleation. The - equation of Newton s second law is ( Fnet) = T FG = T mg = 0. 60 m vbottom = ( T mg) = [5. 0 N (0. 10 kg)(9. 8 m/s )] = 4. 91 m/s m 0. 100 kg As a pojectile the ball stats at y 0 = 1. 4 m with v ˆ 0 = 4. 91 i m/s. The equation fo the y-motion is 1 1 1= 0 m = 0 + 0y ( ) = 0 1 y y v t g t y gt This is easily solved to find that the ball hits the gound at time t1 = y0 g = 0. 535 s Duing this time inteval it tavels a hoizontal distance x1= x0 + v0xt1 = (4. 91 m/s)(0. 535 s) =. 63 m So the ball hits the floo.6 m to the ight of the point whee the sting was cut. Poblem 8.59 A 100 g ball on a 60-cm-long sting is swung in a vetical cicle about a point 00 cm above the floo. The sting suddenly beaks when it is paallel to the gound and the ball is moving upwad. The ball eaches a height of 600 cm above the floo. What was the tension in the sting an instant befoe it boke? 8.59. Model: Model the ball as a paticle undegoing cicula motion in a vetical cicle.
Solve: Initially, the ball is moving in cicula motion. Once the sting beaks, it becomes a pojectile. The final ciculamotion velocity is the initial velocity fo the pojectile, which we can find by using the kinematic equation 1 0 y 1 0 0 0 v = v + a ( y y ) 0 m /s = ( v ) + ( 9. 8 m/s )(4. 0 m 0 m) v = 8. 85 m/s This is the speed of the ball as the sting boke. The tension in the sting at that instant can be found by using the -component of the net foce on the ball: F = v 0 y (8. 85 m/s) T = m T = (0. 100 kg) = 13 N 0. 60 m