Concept explainers
(a)
The angular speed of the rod.
(a)
Answer to Problem 73AP
The angular speed of the rod is
Explanation of Solution
For calculation of gravitational energy, rigid body can be modeled as particle at center of mass.
Only conservative forces are acting on the bar, we have conservation of energy of the bar-Earth system. This can be solving by Conservation of energy.
Write the expression for conservation of energy as.
Here,
Substitute
Rearrange the above equation as.
Write the expression for final kinetic energy of the rod as.
Here,
The moment of inertia of rod is
Substitute
Here,
Rearrange the above equation as.
Write the expression for initial potential energy as.
Here,
Conclusion:
Substitute
Simplify the above equation as.
Thus, the angular speed of the rod is
(b)
The magnitude of
(b)
Answer to Problem 73AP
The magnitude of the angular acceleration is
Explanation of Solution
Redraw the figure P10.73 as.
Angular acceleration can be found using torque analysis.
Write the expression for torque of the rod as.
Here,
Write the expression for torque in terms of force and distance as.
Here,
The length is between a center of rotation and the pivot point where a force is applied that means distance for the pivot is
Substitute
Substitute
Rearrange the above expression for angular acceleration as.
Thus, the magnitude of the angular acceleration is
Conclusion:
(c)
The
(c)
Answer to Problem 73AP
The net acceleration for the system is
Explanation of Solution
Redraw the figure P10.73 for calculate the components of the acceleration as
Refer to figure, the horizontal acceleration is directed towards the centre. If an object is free to move due to fixed point. The object will follow the circular path. Since, the horizontal acceleration will act as centripetal acceleration.
Write the expression for horizontal acceleration as.
Here,
Negative sign shows the direction of acceleration towards the negative
Write the expression for centripetal acceleration for the rod as.
Here,
The radius of circular path for this system is defined as the half length of the rod that is
Substitute
Simplify the above equation as.
As the vertical acceleration will be same as tangential acceleration the rod follows the circular path.
Write the expression for vertical acceleration as.
Here,
Write the expression for tangential acceleration as.
Substitute
Simplify the above equation as.
Write the expression for net acceleration for the
Here,
Conclusion:
Substitute
This is centripetal acceleration; it is directed along the negative horizontal.
Substitute
This is centripetal acceleration; it is directed along the negative vertical.
Substitute
Simplify the above equation as.
Thus, the net acceleration for the system is
(d)
The components of the reaction force at the pivot.
(d)
Answer to Problem 73AP
The net force exerts on the rod at the pivot is
Explanation of Solution
The pivot exerts a force
Write the expression for force on the rod as.
Here,
Write the expression for horizontal force as.
Substitute
Rearrange the above equation as.
Write the expression for vertical force as.
Substitute
Simplify the above equation as.
Conclusion:
Substitute
Thus, the net force exerts on the rod at the pivot is
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Chapter 10 Solutions
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
- A 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_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_forwardThe angular momentum vector of a precessing gyroscope sweeps out a cone as shown in Figure P11.31. The angular speed of the tip of the angular momentum vector, called its precessional frequency, is given by p=/I, where is the magnitude of the torque on the gyroscope and L is the magnitude of its angular momentum. In the motion called precession of the equinoxes, the Earths axis of rotation processes about the perpendicular to its orbital plane with a period of 2.58 104 yr. Model the Earth as a uniform sphere and calculate the torque on the Earth that is causing this precession. Figure P11.31 A precessing angular momentum vector sweeps out a cone in space.arrow_forward
- A 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_forwardTwo astronauts (Fig. P10.67), each having a mass M, are connected by a rope of length d having negligible mass. They are isolated in space, orbiting their center of mass at speeds v. Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum of the two-astronaut system and (b) the rotational energy of the system. By pulling on the rope, one of the astronauts shortens the distance between them to d/2. (c) What is the new angular momentum of the system? (d) What are the astronauts new speeds? (e) What is the new rotational energy of the system? (f) How much chemical potential energy in the body of the astronaut was converted to mechanical energy in the system when he shortened the rope? Figure P10.67 Problems 67 and 68.arrow_forwardA uniform, hollow, cylindrical spool has inside radius R/2, outside radius R, and mass M (Fig. P10.47). It is mounted so that it rotates on a fixed, horizontal axle. A counterweight of mass m is connected to the end of a string wound around the spool. The counterweight falls from rest at t = 0 to a position y at time t. Show that the torque due to the friction forces between spool and axle is f=R[m(g2yt2)M5y4t2] Figure P10.47arrow_forward
- The reel shown in Figure P10.71 has radius R and moment of inertia I. One end of the block of mass m is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the reel. The reel axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its unstretched position and the reel is then released from rest. Find the angular speed of the reel when the spring is again unstretched. Figure P10.71arrow_forwardTwo astronauts (Fig. P10.67), each having a mass of 75.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.00 m/s. Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum of the two-astronaut system and (b) the rotational energy of the system. By pulling on the rope, one astronaut shortens the distance between them to 5.00 m. (c) What is the new angular momentum of the system? (d) What are the astronauts new speeds? (e) What is the new rotational energy of the system? (f) How much chemical potential energy in the body of the astronaut was converted to mechanical energy in the system when he shortened the rope? Figure P10.67 Problems 67 and 68.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_forward
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