Kleppner an introduction to mechanics solutions manual

After the elastic bounce, M moves upward with p speed v0. Here are two methods for finding the upward speed of the marble after it collides with the superball. To demonstrate the effect, it may be easier to use a coin instead of a marble, but a coin may experience greater air resistance. Method 1: This method is algebraic, using the conservation laws for momentum and for mechanical energy.

The top sketch shows the superball immediately after it has bounced off the floor. A gap greatly exaggerated in the sketch is shown between the marble, which is still moving downward with speed v0 , and the superball, which is moving upward with speed v0. The lower sketch shows the system immediately after the marble has collided with the superball. The initial momentum just before the collision upper sketch is Pi , and the final momentum just after lower sketch is P f.

Ki is the initial kinetic energy upper sketch , and K f is the kinetic energy just after the collision lower sketch. To an observer on M moving upward , the marble just before the collision ap- pears to be approaching with speed 2v0. Each car has mass M. The upper sketch is before the collision, and the lower sketch is after the collision.

Both momentum P and mechanical energy kinetic energy K are conserved in the elastic collision. P has both x and y components. The collision is inelastic, so momentum P is conserved, but but mechanical energy E is not. Squaring Eqs. After fission, the product 97 Sr and Xe nuclei move apart back-to- back, with equal and opposite momentum P. After fusion, the products have equal and opposite momentum P. At threshold, the energy in the C system is just enough to form the products.

If vn,C is moving parallel to V, the neutron will always be moving in the forward direction in L, and its energy is not restricted.

The speed of the neutron in C is Vin this case, and it is moving antiparallel to V. Pushing the piston down does work on the gas, raising its temperature by increasing the average speed of the gas molecules.

Consider the situation when the walls are stationary. Then convert back to the lab frame by adding V. To convert from C to L, add Vc to every velocity vector, as shown for mass m bottom sketch. To an observer on the drum, the sand appears to fly out radially.

Angular momentum L about the pivot is conserved, because no external torques are acting on the system. The ring is then momentarily at rest, but it is not necessarily back to its initial position. At tangential grazing, m is traveling at speed v and its trajectory is perpendicular to R. In this case, the package does not graze, but sails over the planet.

Problem 7. Using Eqs. Note the exact analogy with linear acceleration under a constant force. Momentum p and mechanical energy E are not conserved, because external force is acting on m.

Momentum is not conserved, because T dt , 0. Mechanical energy is conserved, because the force T on m is perpendicular to v, and hence does no work. Both springs act to restore the rod toward equilibrium. At this point, the system is no longer stable, and the motion ceases to be harmonic. The second term is the angular momentum of a mass M concentrated at the end of the rod. The third term is the angular momentum of the disk. The period is T , Eq. When the disk is mounted by a frictionless bearing, it cannot rotate and contributes no rotational angular momentum.

The term 21 MR2 should then be omitted from Eq. Hence the angular momentum of the system is conserved. The new angular momentum L0 equals the initial L.

Note that FV and F H do no work on the system, because the displacement is 0. Only the friction force f contributes. Let l0 be the initial length of the tape. At the start, the object has only gravitational potential energy, and it gains kinetic energy as it rolls down the plane.

Writing Eq. However, the solution requires that the Yo-yo be on the verge of slipping. To keep the device from rotating, the hand must apply an opposite torque. Thus the angular momentum of the device is not conserved, so analyze the problem. The mass has gravitational potential energy and kinetic energy, and the cone has rota- tional kinetic energy. The marble has gravitational potential energy E pot , translational kinetic energy Etrans , and rotational kinetic energy Erot.

One way is to note the analogy between Eq. At the instant shown in the sketch, the pivot point is n. The horizontal displacement of the center of mass must be less than the displacement of n. The motion is described most naturally as a combination of a uniform translation of the center of mass and a uniform rotation about the center of mass. Linear momentum, angular momentum, and mechanical energy are conserved.

Linear momentum is also not conserved in the collision, because of the force exerted at the point of impact. The force at the point of impact exerts no torque about that point, but the wheel has angular momentum of translation.

Hence, angular momentum is conserved in the collision. However, it has angular momentum MvR from translation of the axis refer to Note 7. Sketch a shows the system at the start, and at the instant just before the collision with the track. Mechanical energy E f is conserved following the collision, when the wheel is on the track. The wheel comes to rest when the spring is compressed to b0. Mechanical energy E f is con- served as the wheel moves off the track onto the smooth surface.

The velocity v changes with time because of the spring force, but the angular velocity remains constant, because there is no torque on the wheel about its center of mass. Thus the mechanical energy is equally divided between translation and rotation. As shown is sketch c , Lrot has re- versed its direction, so the total angular momentum is 0, and it remains 0 after the second collision. The second collision has dissipated all the remaining mechanical energy!

Because the wall and floor are frictionless, the force Fw exerted by the wall on the plank and the force F f exerted by the floor are normal to the surfaces, as shown in the lower sketch. Let y0 be the initial height of the center of mass above the floor. We treated L as the angular momentum of a body with moment of inertia from the parallel axis theorem.

From Secs. Without the flywheel, the torque due to friction f is balanced by the torque due to the unequal loading N1 and N2. The flywheel thus needs to produce a counterclockwise torque on the car to balance the clockwise torque from f 0. If the car turns in the opposite direction, both the torque and the direction are reversed, so equal loading remains satisfied. This rotation is caused by precession of of its spin angular momentum due to the torque induced by the tilt.

The coin is accelerating, so take torques about the center of mass. The criterion is equivalent to being able to twirl a lariat vertically as well as horizontally. The hoop is vertical, so gravity exerts no torque. The blow by the stick is short, so the peak of force F is large; f can be neglected during the time of impact.

The spin- ning wheels increase the tilt angle by only about a degree, not a substantial effect. There is no applied torque, so the net rate of change is 0. When making such approximations, be sure to include all terms up to the highest order retained. In part a , MA is the fictitious force, and in part b , it is MA0. The directions of the fictitious forces are shown in the sketches.

English units are used. In both parts, the approach is to take torques about the point of contact of the rear tire with the road. One advantage is that the friction forces do not appear in the torque equations. The same result is obtained using the general result Eq. The Coriolis force on the train is directed toward the west, so the force on the tracks is toward the east. For points at rest, a is radial, directed toward the axis, as shown in the sketch.

Note that x is dimensionless. In the rotating system, a fictitious centrifugal force Fcent acts on M. Fcent is directed radially outward from the axis of rotation. Take torques about the pivot point a. Consider, however, the torque equation without using the small angle approximation. L2 L2 1! For a stable circular orbit, Ue f f must have a minimum at some radius r0. See Problem The satellite does not move far during the brief firing time, so the potential energy is essentially unchanged.

These conditions are satisfied best at the closest point perigee. Here are two ways of finding vmax. In equilibrium, the total force on m is 0. Neglecting perturbations, the configuration is therefore unchanging during the rotation. Let A be the major axis. The equation of motion for a critically damped oscillator is given in Eq. As indicated in the sketch, its speed speed decreases slightly from v0 to v1 during a half cycle.

The escapement provides an impulse every period to make up for the loss. Brackets h i denote time averages. Let time averages be denoted by brackets h i. The mass of the pendulum is m, and the mass of the falling weight is M. The power to the clock by the descending weight compensates the power dissipated by friction. The equation of motion for y1 Eq. At long times, x1 and x2 move together so that their separation is constant and there is therefore no damping. To evaluate the initial conditions, express y1 and y2 in terms of x1 and x2.

This is a consequence of the linearity of Eqs. In Eq. The transient eventually becomes negligibly small, leaving only the steady- state behavior. The times to traverse the arms are v2!

Then 2 4. To find tg , consider two events in the S 0 rest frame of the slab. In the S lab frame, light enters the slab at x1 and time t1 , and leaves at x2 at time t2. At the same time t, the farmer in frame S observes end A at xA. From the Lorentz transformation Eq. How does this event, call it C, look to the farmer in frame S? Note that v, the relative velocity of the S and S 0 frames, is a constant. We use cookies on our website to give you the most relevant experience by remembering your preferences and repeat visits.

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