Solved Problems

6. Work and energy

Problem 2

The law of universal gravitation states that any celestial body of mass produces an attractive force on any other body of mass , given by the expression:

where is the universal gravitational constant, is the distance between the two bodies and is the radial unit vector, which points from the body of mass to the body of mass . (a) Determine the expression for the gravitational potential energy due to the body of mass . (b) Taking into account the result of the previous part, how can the result in equation 6., , be justified for the gravitational potential energy of an object on Earth?

(a) The infinitesimal displacement vector, in spherical coordinates, is:

where and are two angles (measured from the positive axis and on the plane, from the positive axis) and the three unit vectors , and are perpendicular to each other. Thus, the dot product of the gravitational force and the infinitesimal displacement is equal to:

Since it only depends on one variable, we conclude that the line integral of does not depend on the integration curve and the gravitational force is conservative. The potential energy corresponding to the gravitational field is equal to minus any primitive of the radial component of the force

(b) In the neighborhood of any value , the Taylor series expansion of is:

The first term is a constant, which can be ignored because any arbitrary constant can be added to the potential energy. In the second term, replacing by the radius of the Earth, is then the height measured from the surface of the Earth and is the value of the acceleration of gravity near the Earth's surface. Ignoring the other terms in the series, which are very small for values of smaller than the radius of the Earth, we are left with only the second term of the series which leads to .

Problem 3

In a pole vault, a 70 kg athlete uses a uniform 4.5 kg stick, 4.9 m long. The athlete's jump has three phases: first the athlete runs with its center of gravity at 1 m above the ground and with the center of gravity of the pole at 1.5 m above the ground in height, until reaching a speed of 9 m/s at the instant when he sticks the tip of the pole to the ground. In the second phase, the energy of the run is transferred to the pole, which deforms and then recovers its shape in a vertical position, raising the athlete to a height close to the height of the bar. Finally, the athlete pushes the pole towards the ground, gaining some additional energy that makes his center of gravity go higher and up to 5.8 m above the ground, enough to pass the bar at 5.6 m. Assuming there are no energy losses, compute the mechanical energy transferred to the jumper when he pushes the pole in the direction of the ground.

Pole vault

While the athlete runs with the pole, the mechanical energy of the athlete-pole system (kinetic translation energy plus gravitational potential energy) is:

And the athlete reaches the highest point in his jump, when he and the pole are at rest, there is only gravitational potential energy

The energy transferred to the athlete when he pushes the pole in the direction of the ground is then the difference between the final and the initial mechanical energy

Problem 4

Solve problem 7 of chapter 4 by applying the work and mechanical energy theorem. Is the force exerted by the block on the cone, while the cone penetrates the block, conservative?

The cone is at rest in the initial and final positions, when it was 30 cm above the block and after it penetrates 5 cm inside the block. Hence, there is no variation of kinetic energy and the decrease of the mechanical energy equals the decrease of the gravitational potential energy between those two positions:

That decrease of mechanical energy equals the work done by the nonconservative forces which, neglecting the air resistance, is equal to the work done by the force excerted by the block on the cone, from the position when the cone touches the block ( ), to the position where it stops (  cm):

And we thus obtain the value of the constant . Since the units of are newton, the units of are then N/m2. The force exerted by the block on the cone is nonconservative, because it only acts while the cone is going into the block; once block has gone inside the block, if it went back to the force would not have the same value. The force of the block depends on the velocity, because when the cone moves down (negative velocity) it is different than when it goes up (positive velocity) and the force is zero when the cone stops (zero velocity).

Problem 7

A sphere of radius rolls without sliding, into a semicircular gutter of radius , which is on a vertical plane (see figure).

  1. Show that, in terms of the derivative of the angle , the kinetic energy of the sphere is
  2. Neglecting the resistance of the air, the mechanical energy is constant and its derivative with respect to time is zero. Differentiate the expression of the mechanical energy with respect to time and equate it to zero to find the expression of the angular acceleration in terms of angle.
  3. Among what values ​​must the mechanical energy be so that the sphere remains oscillating within the gutter?
  4. From the result of part b, determine the expression for , at the limit when the radius of the sphere is much smaller than the radius of the gutter ( ) and explain why the result is different from the result obtained for the simple pendulum in problem 6.
Sphere in a circular gutter

(a) The center of mass of the sphere moves in an arc of circle with angle and radius ; therefore, the velocity of the center of mass of the sphere is:

And since the sphere rolls without slipping, the point of the sphere in contact with the gutter has zero velocity. The difference of velocities of the center of mass and the contact point, divided by the radius of the sphere, gives the angular velocity of the sphere:

The total kinetic energy of the sphere is then

Substituting the expression for the moment of inertia of a sphere about its center (see table 5.1) and the expressions for the velocity of the center of mass and the angular velocity, we get

(b) The mechanical energy is:

and its derivative with respect to time is:

Equating it to zero we get,

(c) The energy is minimum when the sphere remains in the lowest point of the gutter ( ) at rest ( ):

and the energy is maximum when the sphere reaches point A ( ) with zero velocity ( ):

(d) The absolute value of is smaller by a factor of 5/7, due to the part of the gravitational potential energy that is transformed into kinetic energy of rotation of the sphere. The kinetic energy of rotation is always 2/5 of the kinetic energy of translation, independently from the value of . Thus, in the limit 2/7 of the gravitational potential energy will also be converted into kinetic energy of rotation and only the remaining 5/7 make increase.

From the point of view of forces, in the case of the pendulum there are no forces opposing the motion of the center of mass, while in this case the static frictional force is opposing the motion of the center of mass. However, that force does not do any work because the point of the sphere where it is applied does not move and the frictional force does not reduce the mechanical energy of the sphere; it just makes the energy provided by gravity to be distributed into translation and rotation kinetic energies.

Problem 9

Solve problem 8 of chapter 5 with the principle of conservation of mechanical energy.

The following figure shows the initial and final positions of the box. C is the center of mass and E the axis of rotation along the hinges.

Initial and final positions in problem 6.9

Since the velocity of E is zero, the velocity of the center of mass is then equal to , where is the angular velocity of the box, and the kinetic energy of the box is:

Substituting the expression for the moment of inertia of a parallelepiped (table 5.1) and the value of , we get

By the principle of conservation of mechanical energy, the kinetic energy in the final position must be equal to the decrease of the gravitational potential energy, which is equal to the weight times the height that the center of mass descended:

Equating the last two equations, we get the final value of the angular velocity