Concept explainers
You’re working for a playground equipment company, which wants to know the rotational inertia of its swing with a child on board: the combined mass is 20 kg. You observe the child twirling around in the swing, twisting the ropes as shown in Fig. 13.37. As a result, child and swing rise slightly, with the rise h in cm equal to the square of the number of full turns. When the child stops twisting, the swing begins torsional oscillations. You measure the period at 6.91 s. What do you report for the rotational inertia of the child-swing system?
FIGURE 13.37 Problem 84
Want to see the full answer?
Check out a sample textbook solutionChapter 13 Solutions
Essential University Physics: Volume 1 (3rd Edition)
Additional Science Textbook Solutions
University Physics Volume 2
College Physics (10th Edition)
College Physics: A Strategic Approach (4th Edition)
Applied Physics (11th Edition)
The Cosmic Perspective Fundamentals (2nd Edition)
Physics: Principles with Applications
- An engineer has an odd-shaped 10 kg object and needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant k = 0.50 N. m. If this torsion pendulum oscillates through 20 cycles in 50 s, what is the rotational inertia of the object?arrow_forwardThe wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the wheel is 0.6m. The radius of gyration KG=0.4 m. The spring's unstretched length is Lo=1.0 m. The stiffness coefficient of the spring is k=2.0 N/m. The wheel is released from rest at the state 1 when the angle between the spring and the vertical direction is 0-30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is 8=0°. The spring's length at the state 2 is L2=4 m. (7) The instantaneous center of zero velocity (IC) is L₂ State 2 O A Point A B. Point O O C. Point G H State 1arrow_forwardA 76 kg solid sphere with a 16 cm radius is suspended by a vertical wire. A torque of 0.56 N·m is required to rotate the sphere through an angle of 0.67 rad and then maintain that orientation. What is the period of the oscillations that result when the sphere is then released?arrow_forward
- The wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the wheel is 0.6m. The radius of gyration KG=0.4 m. The spring's unstretched length is Lo=1.0 m. The stiffness coefficient of the spring is k-2.0 N/m. The wheel is released from rest at the state 1 when the angle between the spring and the vertical direction is 8-30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is 0=0°. The spring's length at the state 2 is L2=4 m. _(kg-m²) (two decimal (9) The mass moment of inertial about the IC center is IIC=_ places) L₂ State 2 State 1arrow_forwardThe wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the wheel is 0.6m. The radius of gyration kG=0.4 m. The spring’s unstretched length is L0=1.0 m. The stiffness coefficient of the spring is k=2.0 N/m. The wheel is released from rest at the state 1 when the angle between the spring and the vertical direction is θ=30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is θ=0°. The spring’s length at the state 2 is L2=4 m. (10) The kinetic energy at the state1?________ (N·m) (two decimal places)arrow_forwardThe wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the wheel is 0.6m. The radius of gyration kG=0.4 m. The spring’s unstretched length is L0=1.0 m. The stiffness coefficient of the spring is k=2.0 N/m. The wheel is released from rest at the state 1 when the angle between the spring and the vertical direction is θ=30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is θ=0°. The spring’s length at the state 2 is L2=4 m. (11) The angular velocity at the state 2?_______(rad/s) (two decimal places)arrow_forward
- A simple pendulum consists of a 0.8-kg bob connected to a massless inextensible cord with a length L = 1.4 m. The bob is set into motion and its angular displacement is given by 0(t) = 0.11cos(wt), where e is in radians and t is in seconds. Take g = 9.8 m/s^2, determine the mechanical energy of this pendulum. O 0.085 J 0.090 J 0.046 J 0.066 J 0.052 J laresion al frictionless horizontal surface with aarrow_forwardConsider a string that rotates around the end of a motor. The string has length L = 0.20m and mass M = 0.015 kg. What (constant) torque must the motor exert on the string for it to bring it from rest to full speed in 0.50 s? Full speed is 1200rpm. Igore wind resistance. The moment of inertia for motion of a rod about the end is I=1/3ML^2.arrow_forwardThe system is released from rest with the spring initially stretched 4.5 in. Calculate the velocity v of the cylinder after it has dropped 1.5 in. The spring has a stiffness of 6.3 lb/in. Neglect the mass of the small pulley. k = 6.3 Ib/in 118 Ib Answer: v = ft/secarrow_forward
- A simple pendulum consists of a 0.8-kg bob connected to a massless inextensible cord with a length L = 1.1 m. The bob is set into motion and its angular displacement is given by 0(t) = 0.11cos(wt), where 0 is in radians and t is in seconds. Take g = 9.8 m/s^2, determine the mechanical energy of this %3D pendulum. O 0.090 J O 0.085 J O 0.066 J O 0.046 J O 0.052 J moncillates on a frictionless horizontal surface in simple 2Ywould thearrow_forwardAn engineer has an odd-shaped10 kg object and needs to find its rotationalinertia about an axis through itscenter of mass.The object is supportedon a wire stretched along the desiredaxis. The wire has a torsion constantk = 0.50 Nm. If this torsion pendulum oscillates through 20 cycles in50 s, what is the rotational inertia of the object?arrow_forwardThe piston inside a car cylinder osscilates up and down with Simple Harmonic Motion at 5400 cylces per minute. It travels upwards through a distance of 24cm and downwards through 24cm each cycle. What is the amplitude of the SHM? (m) Answer: 0.24 X What is the angular frequency of the motion? (rad/s) Answer: 565.5 The total distance moved is 24cm. The equillibrium position is in the middle. Hence the amplitude is equal to half of 24 cm, 0.12m. x Answer: 135.72 What is the maximum velocity of the piston? (m/s) x cylinder piston + The frequency is the number of cycles (or revolutions) per second. Hence the frequency is 5400/60= Hz. Then use w=2nf to caculate the angular frequeny. Use the formula V=Aw to calculate maximum velocity. SHMarrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON