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Native people throughout North and South America used a bola to hunt for birds and animals. A bola can consist of three stones, each with mass m, at the ends of three light cords, each with length ℓ. The other ends of the cords are tied together to form a Y. The hunter holds one stone and swings the other two above his head (Figure P11.41a, page 308). Both these stones move together in a horizontal circle of radius 2ℓ with speed v0. At a moment when the horizontal component of their velocity is directed toward the quarry, the hunter releases the stone in his hand. As the bola flies through the air, the cords quickly take a stable arrangement with constant 120-degree angles between them (Fig. P11.41b). In the vertical direction, the bola is in free fall. Gravitational forces exerted by the Earth make the junction of the cords move with the downward acceleration
Figure P11.41
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Chapter 11 Solutions
Physics for Scientists and Engineers with Modern Physics
- A child works on a project in art class and uses an outline of her hand on a sheet of construction paper to draw a turkey (Fig. P16.36). The teacher pins the turkey to the bulletin board in the front of the classroom by using a thumbtack. The student notices that if she flicks her finger on the end of the turkey, it oscillates back and forth with a frequency of about 1.65 Hz. If the rotational inertia of the paper turkey is 1.25 105 kgm2 and its mass is 0.005 kg, what is the distance between the thumbtack and the center of mass of the turkey? FIGURE P16.36arrow_forwardOne end of a cord is fixed and a small 0.400-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.50 m, as shown in the figure below. When θ = 23.0°, the speed of the object is 5.50 m/s. An object is swinging to the right and upward from the end of a cord attached to a horizontal surface. The cord makes an angle θ with the vertical. An arrow labeled vector v points in the direction of motion. (a) At this instant, find the magnitude of the tension in the string.N(b) At this instant, find the tangential and radial components of acceleration. at = m/s2 downward tangent to the circle ac = m/s2 inward (c) At this instant, find the total acceleration.inward and below the cord at °(d) Is your answer changed if the object is swinging down toward its lowest point instead of swinging up? YesNo (e) Explain your answer to part (d).arrow_forwardOne end of a cord is fixed and a small 0.400-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.50 m, as shown in the figure below. When θ = 23.0°, the speed of the object is 5.50 m/s. An object is swinging to the right and upward from the end of a cord attached to a horizontal surface. The cord makes an angle θ with the vertical. An arrow labeled vector v points in the direction of motion. (a) At this instant, find the magnitude of the tension in the string.Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. N(b) At this instant, find the tangential and radial components of acceleration. at = Your response differs from the correct answer by more than 100%. m/s2 downward tangent to the circle ac = Your response differs from the correct answer by more…arrow_forward
- One end of a cord is fixed and a small 0.700-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.00 m, as shown in the figure below. When θ = 21.0°, the speed of the object is 8.50 m/s. An object is swinging to the right and upward from the end of a cord attached to a horizontal surface. The cord makes an angle θ with the vertical. An arrow labeled vector v points in the direction of motion. (a) At this instant, find the magnitude of the tension in the string. N(b) At this instant, find the tangential and radial components of acceleration. at = m/s2 downward tangent to the circle ac = m/s2 inward (c) At this instant, find the total acceleration. inward and below the cord at °(d) Is your answer changed if the object is swinging down toward its lowest point instead of swinging up? YesNo (e) Explain your answer to part (d).arrow_forwardAn amusement park ride rotates around a fixed axis such that the angular position of a point on the ride follows the equation: θ(t) = a + bt2 – ct3 where a = 1.9 rad, b = 0.75 rad/s2 and c = 0.025 rad/s3.Randomized Variables a = 1.9 radb = 0.75 rad/s2c = 0.025 rad/s3 1) What is the magnitude of the angular displacement of the ride in radians between times t = 0 and t = t1? 2) Determine an equation for the angular acceleration of the ride as a function of time, α(t). Write your answer using the symbols a, b, and c, instead of their numerical values. 3) What is the angular acceleration in rad/s2 when the ride is at rest at t = t1?arrow_forwardA physics lab instructor is working on a new demonstration. She attaches two identical copper balls with mass m = 0.180 g to threads of length L as shown in the figure. There are two strings in the figure. The top of each string is connected to the ceiling, and both strings are connected at the same point. The bottom of each string is connected to a spherical mass labeled m. Both strings have length L and hang at an angle of ? to the vertical, with the two strings on opposite sides of the vertical. Both balls have the same charge of 7.60 nC, and are in static equilibrium when ? = 4.75°. What is L (in m)? Assume the threads are massless. m (b) What If? The charge on both balls is increased until each thread makes an angle of ? = 9.50° with the vertical. If both balls have the same electric charge, what is the charge (in nC) on each ball in this case?arrow_forward
- An amusement park ride rotates around a fixed axis such that the angular position of a point on the ride follows the equation: θ(t) = a + bt2 – ct3 where a = 1.9 rad, b = 0.75 rad/s2 and c = 0.025 rad/s3.Randomized Variables a = 1.9 radb = 0.75 rad/s2c = 0.025 rad/s3 1) Determine an equation for the angular speed of the ride as a function of time, ω(t). Write your answer using the symbols a, b, and c, instead of their numerical values. 2)Besides at t = 0, at what time t1 is the ride stopped? Give your answer in seconds. 3) What is the magnitude of the angular displacement of the ride in radians between times t = 0 and t = t1?arrow_forwardAnswer either this question or the next. L M Ax 3. A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table of height h = 0.85 m. You may assume the string is massless, and drag is negligible. a. Calculate the speed of the bob just before it strikes the box (i.e., when the pendulum string hangs straight down). b. Calculate the speeds of the bob and the box just after they collide elastically. c. Determine the impulse that acts on the box during the collision. d. Determine how far from the bottom edge of the table, Ax, the box lands.arrow_forwardPenny is adjusting the position of a stand up piano of mass mp = 150 kg in her living room. The piano is lp = 1.6 m in length. The piano is currently at an angle of θp = 45 degrees to the wall. Penny wants to rotate the piano across the carpeted floor so that it is flat up against the wall. To move the piano, Penny pushes on it at the point furthest from the wall. This piano does not have wheels, so you can assume that the friction between the piano and the rug acts at the center of mass of the piano.Randomized Variables mp = 150 kglp = 1.6 mθp = 45 degrees a) Write an expression for the minimum magnitude of the force Fs in N Penny needs to exert on the piano to get it moving. Assume the corner of the piano on the wall doesn't slide and the static friction between the rug and the piano is μs. Fs,min = b) The coefficient of kinetic friction between the carpet and the piano is μk = 0.27. Once the piano starts moving, calculate the torque τp in N⋅m that Penny needs to apply to keep…arrow_forward
- You have a cylinder. You don't know what its internal structure looks like, but you plan to roll it down a ramp, as in this week's procedure. The ramp is 1 m long, and is elevated at an angle of 15°. The mass of the cylinder is 450 g and its diameter is 2.1 cm.After you release the cylinder, it rolls down the ramp without slipping, gaining speed. How much total energy (in J)does the block have at the bottom of the ramp?arrow_forwardI'm not sure how to work this problem. The problem gives an answer of Itot=Icm+Md^2=18.1 kg*m^2 for part A. The problem also gives an answer of a=6.41 rad/s^2 for part B but does not show how to work the problems to get to these answers.arrow_forwardA small ball of mass 740.0 g is placed in a tube that is bent into a circular arc of radius R= 67.5 cm.The friction between the ball and the walls of the tube is negligible. The ball has an iron core. A magnet is used to raise the ball until it makes an angle of 5.00 degrees with the vertical, and then released from rest. Where will the ball be 1.20 seconds after being released? Express your answer in terms of the angle,θ, that the ball makes with the vertical.arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning