(a)
The total torques due to the weight of the hand about the axis of rotation when the time reads
(a)
Answer to Problem 72AP
The total torques due to the weight of the hand about the axis of rotation when the time reads
Explanation of Solution
Given information: The mass of hour hand is
Formula to calculate the net torque produced by the clock’s hand is,
Here,
Formula to calculate the angular speed of the hour hands is,
Substitute
Thus, the angular speed of the hour hand is
Formula to calculate the angular speed of the minute hands is,
Substitute
Thus, the angular speed of the hour hand is
Let take
Write the expression for the angular position of the hour hands at time
Here,
Substitute
Write the expression for the angular position of the minute hands at time
Here,
Substitute
Substitute
Substitute
When clock shows time
Substitute
Thus, the net torque is
When clock shows time
Substitute
Thus, the net torque is
When clock shows time
Substitute
Thus, the net torque is
When clock shows time
Substitute
Thus, the net torque is
When clock shows time
Substitute
Thus, the net torque is
Conclusion:
Therefore, the total torques due to the weight of the hand about the axis of rotation when the time reads
(b)
The all the time nearest to second when total torque about the axis of rotation is zero by solving the transcendental equation.
(b)
Answer to Problem 72AP
The time corresponding to the zero torque is given as:
Time(hr) | Clock time |
0 | 12:00:00 |
0.515 | 12:30:55 |
0.971 | 12:58:19 |
1.54 | 1:32:31 |
1.95 | 1:57:01 |
2.56 | 2:33:25 |
2.94 | 2:56:29 |
Explanation of Solution
Given information: The mass of hour hand is
From equation (2), the expression for the total torque is given by,
Substitute
Since it is a transcendental equation, solving the equation numerically the values of time comes out to be 0, 0.515, 0.971, 1.54, 1.95……so on.
The time corresponding to the time is given as:
Time(hr) | Clock time |
0 | 12:00:00 |
0.515 | 12:30:55 |
0.971 | 12:58:19 |
1.54 | 1:32:31 |
1.95 | 1:57:01 |
2.56 | 2:33:25 |
2.94 | 2:56:29 |
Conclusion:
Therefore, time corresponding to the zero torque is given as:
Time(hr) | Clock time |
0 | 12:00:00 |
0.515 | 12:30:55 |
0.971 | 12:58:19 |
1.54 | 1:32:31 |
1.95 | 1:57:01 |
2.56 | 2:33:25 |
2.94 | 2:56:29 |
Want to see more full solutions like this?
Chapter 10 Solutions
PHYSICS:F/SCI...W/MOD..-UPD(LL)W/ACCES
- A constant net torque is applied to an object. Which one of the following will not be constant? (a) angular acceleration, (b) angular velocity, (c) moment of inertia, or (d) center of gravity.arrow_forwardA playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kg m2 and is rotating at 10.0 rev/min about a frictionless, vertical axle. Facing the axle, a 25.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?arrow_forwardFind the net torque on the wheel in Figure P10.23 about the axle through O, taking a = 10.0 cm and b = 25.0 cm. Figure P10.23arrow_forward
- A student sits on a freely rotating stool holding two dumbbells, each of mass 3.00 kg (Fig. P10.56). When his arms are extended horizontally (Fig. P10.56a), the dumbbells are 1.00 m from the axis of rotation and the student rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus stool is 3.00 kg m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.300 m from the rotation axis (Fig. P10.56b). (a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward. Figure P10.56arrow_forwardRigid rods of negligible mass lying along the y axis connect three particles (Fig. P10.18). The system rotates about the x axis with an angular speed of 2.00 rad/s. Find (a) the moment of inertia about the x axis, (b) the total rotational kinetic energy evaluated from 12I2, (c) the tangential speed of each particle, and (d) the total kinetic energy evaluated from 12mivi2. (e) Compare the answers for kinetic energy in parts (b) and (d). Figure P10.18arrow_forwardA disk with moment of inertia I1 rotates about a frictionless, vertical axle with angular speed i. A second disk, this one having moment of inertia I2 and initially not rotating, drops onto the first disk (Fig. P10.50). Because of friction between the surfaces, the two eventually reach the same angular speed f. (a) Calculate f. (b) Calculate the ratio of the final to the initial rotational energy. Figure P10.50arrow_forward
- Big Ben, the Parliament tower clock in London, has an hour hand 2.70 m long with a mass of 60.0 kg and a minute hand 4.50 m long with a mass of 100 kg (Fig. P10.17). Calculate the total rotational kinetic energy of the two hands about the axis of rotation. (You may model the hands as long, thin rods rotated about one end. Assume the hour and minute hands are rotating at a constant rate of one revolution per 12 hours and 60 minutes, respectively.) Figure P10.17 Problems 17, 49, and 66.arrow_forwardBig Ben (Fig. P10.17), the Parliament tower clock in London, has hour and minute hands with lengths of 2.70 m and 4.50 m and masses of 60.0 kg and 100 kg, respectively. Calculate the total angular momentum of these hands about the center point. (You may model the hands as long, thin rods rotating about one end. Assume the hour and minute hands are rotating at a constant rate of one revolution per 12 hours and 60 minutes, respectively.)arrow_forwardA long, uniform rod of length L and mass M is pivoted about a frictionless, horizontal pin through one end. The rod is released from rest in a vertical position as shown in Figure P10.65. At the instant the rod is horizontal, find (a) its angular speed, (b) the magnitude of its angular acceleration, (c) the x and y components of the acceleration of its center of mass, and (d) the components of the reaction force at the pivot. Figure P10.65arrow_forward
- If the system shown in Figure P8.37 is set in rotation about each of the axes mentioned in Problem 37, find the torque that will produce an angular acceleration of 1.50 rad/s2 in each case. Figure P8.37 Problems 37 and 38.arrow_forwardA 12.0-kg solid sphere of radius 1.50 m is being rotated by applying a constant tangential force of 10.0 N at a perpendicular distance of 1.50 m from the rotation axis through the center of the sphere. If the sphere is initially at rest, how many revolutions must the sphere go through while this force is applied before it reaches an angular speed of 30.0 rad/s?arrow_forwardA rigid, massless rod has three particles with equal masses attached to it as shown in Figure P11.37. The rod is free to rotate in a vertical plane about a frictionless axle perpendicular to the rod through the point P and is released from rest in the horizontal position at t = 0. Assuming m and d are known, find (a) the moment of inertia of the system of three particles about the pivot, (b) the torque acting on the system at t = 0, (c) the angular acceleration of the system at t = 0, (d) the linear acceleration of the particle labeled 3 at t = 0, (e) the maximum kinetic energy of the system, (f) the maximum angular speed reached by the rod, (g) the maximum angular momentum of the system, and (h) the maximum speed reached by the particle labeled 2. Figure P11.37arrow_forward
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill