Problem # 10:
We begin with the given assumption that the total gravitational flux due to a mass m is 4 `pi G m, as indicated at the top of the figure below.
We will find the force on the mass m2 by first using flux to find the gravitational field at a distance r from m1. We will then multiply m2 by this acceleration in order to find the force on m2.
The flux from the mass m1 is 4 `pi G m1.
A sphere of radius r centered at the location of m1 will thus have flux 4 `pi G m1 spread over area 4 `pi r^2. Since flux is the product of the field strength and the area, we find that the gravitational field is g = 4 `pi G m1 / (4 `pi r^2) = G m1 / r^2.
This gravitational field strength is the acceleration experienced by an object at this position. Since the mass m2 will therefore be accelerated at rate g = G m1 / r^2, the force will be F = m2 g = m2 ( G m1 / r^2). Rearranging this last expression we obtain F = G m1 m2 / r^2, which is Newton's expression of his Law of Universal Gravitational.
Problem #8
If a beam accelerates at 1.2 rad/s^2 from angular position `theta = .4 rad to `theta = 4.3 rad, starting from rest, then we know that initial angular velocity is zero, change angular position is 4.3 rad - .4 rad = 3.9 rad, and angular acceleration is 1.2 rad/s^2. Expressing these quantities and symbols we have `omega0 = 0, `d`theta = 3.9 rad and `alpha =- 1.2 rad/s^2.
If we had a uniform acceleration situation with linear quantities we would have v0, `ds and a, and would use the equation vf^2 = v0^2 + 2 a `ds to determine the final velocity. Here we use the rotational analog `omegaf^2 = `omega0^2 + 2 `alpha `d`theta of this equation. We easily obtain the result `omegaf = 3.1 rad/s.
At distance r from the center of the Earth, your centripetal acceleration would be v^2 / r, as indicated by the figure below. Your gravitational attraction to the Earth would be 9.8 m/s^2 ( rE / r) ^ 2. We will set the two expressions equal and solve for r.
First we note that all quantities other than r are known, with the exception of v. We can easily express v as a function of r by noting that the time required to complete one revolution around the circle will be the 24-hour rotational period of the Earth. We see thus that v = 2 `pi r / 24 hours = 2 `pi r / 87,000 sec (approximately; you should find the exact number of seconds in a day and correct results obtained here).
We see therefore that the centripetal acceleration will be approximately (2 `pi r / 87,000 sec) ^ 2 / r.
Now setting this expression for the centripetal acceleration equal to the gravitational field strength, we obtain the equation in the second line below. In the third line we squarely quantity in parentheses and multiply both sides by r; in the fourth line we multiply both sides by r and divide by the coefficient of r^2 on the left to obtain an expression for r^3.
We complete dissolution by taking the cube root of both sides. In the last line we determine the cube root of 7.4 * 10 ^ 10 m / 4 `pi, obtaining approximately 1.8 * 10^3 (meters)^(1/3) as the coefficient of rE ^ (2/3).
Problem #5: A projectile with initial velocity 50 meters/second in a direction 60 degrees above horizontal will have initial vertical velocity component 43 m/s and initial horizontal velocity component 25 m/s.
The y motion will thus be characterized by initial velocity 43 m/s, displacement -17 m and acceleration - 9.8 m/s^2. We will be able to solve the equation in the fifth line of the figure below for `dt. We will obtain a positive than in negative value for `dt; only the positive value has meaning in this context (the negative value of `dt indicates that a 'phantom projectile' passed through ground level at the indicated time before the projectile was launched, and subsequently followed the same path as the projectile, presumably continuing on forever after it and the projectile reach ground level at the positive value of `dt).
We can then easily find the x range of the projectile by multiplying the initial x velocity, which by assumption does not change, by `dt.
Problem #6:
Escape velocity was found to be vEsc = `sqrt(2 G M / r), while orbital velocity was found to be vOrb = `sqrt(G M / r). We note again that these two velocities differ by factor equal to the square root of 2. We also note weight is the product of mass and gravitational acceleration, where gravitational acceleration can be found from g = flux / area. We will obtain g = G M / r^2.
Problem #2:
If the object is to be stationary, they will be in equilibrium and the total torque about the axis of rotation will be zero.
The torque of the weight and of the tension T will be calculated by considering the moment arm to be the distance from the axis of rotation to the line of action of the force. We therefore obtain respective moment arms .87 m and 1.73 m.
The torques will therefore be -W ( .87 m) and T ( 1.73 m), using the counterclockwise direction as the positive direction for rotation. The condition of rotational equilibrium will thus be as expressed in the second line below. We easily obtain tension T = 58 N (approximately; note that if W = 118 N, T must be 59 N, not 58 N).
We could now easily determine the force required from the constraining bolt at the axis of rotation. Since neither the weight nor the tension has a horizontal component we already have horizontal equilibrium and do not need any horizontal force from the bolt. However our vertical forces are not in equilibrium, since the weight has greater magnitude than the tension. If we let Fvert stand for the vertical force exerted by the bolt we have Fvert - 118 N + 59 N = 0, so Fvert = 59 N. Put simply, to ensure vertical equilibrium the bolt asked a supply 59 Newtons of force to help the 59 N tension counter the 118 N downward force of gravity.
As an exercise in using vector components, we replace the weight and tension vectors with their components parallel to and perpendicular to the beam. These components are easily calculated, and their approximate values are indicated in the figure below. Our x direction will be taken to be the direction of the beam, and if Fbx is taken to be the force of the bolt in this x direction we obtain Fbx - 59 N + 29 N = 0. Similarly in the wide direction, perpendicular to the beam, we see that Fby - 100 N + 50 N = 0.
We easily see that Fbx = 29 N and Fby = 50 N (approximately). The result of these two forces would be a force of magnitude 59 N at an angle of 60 degrees to the x direction. This force would therefore be 59 N, directed 60 degrees above the direction of the beam; since the beam is that 30 degrees with respect to horizontal, we see that this force is 59 N in the vertical direction. This agrees with our previous analysis.
Problem #3:
We begin by replacing the weight and applied force vectors by their components parallel and perpendicular to the incline, with the direction parallel to the incline being taken as the x direction.
We easily obtain the components indicated below.
The condition of y equilibrium requires a normal force N exerted by the plane on the object. We obtain the equation | N | - 270 N - 46 N = 0. We easily see that | N | = 316 N; this force arises as an elastic response of the incline to the perpendicular components of the gravitational force and the applied force F.
The magnitude of the frictional force can be at most equal to `mu | N | = .15 | N |, which for the magnitude of the normal force in this situation will be 46 Newtons. Since the sum of the other forces in the x direction is -117 N, corresponding to a net force in the direction down the incline, we see that the frictional force resisting this motion will be + 46 N and that the resulting sum of the forces in the x direction will be approximately - 70 N. Since we have y equilibrium this will be the net force.
Problem #1:
Though it is not necessary for this problem, we note that to fall 80 cm with an initial y velocity of 0 requires time `dt = .4 s, approximately, calculated from the initial y velocity, the y acceleration and the y displacement as in the figure below. Had we been given the displacements of the two balls in the horizontal plane, we could have then determine their after-collision velocities from these displacements and the calculated time interval.
It might have been better here to have used z for the vertical direction instead of y, since we will soon be talking about the x and y velocities in the horizontal plane. The use of y here might cause some confusion. Just note that what we have denoted as the y direction in this figure, and are used to denoting as y, should not be confused with the y direction of the rest of the problem, which is in the horizontal plane.
In this problem we are given the velocities of the two balls after collision. We take the x direction to be the original direction of motion of the ball and the y direction to be perpendicular to this direction. Using this convention we see that after collision the velocity components of the first ball are v1y = -13 cm/s and v1x = 47 cm/s, and those of the second ball are v2y = 4 cm/s and v2x = 70 cm / s (approximately).
The after-collision momentum of the first ball will therefore have components p1y = -2.6 kg cm/s and p1x = 9.4 kg cm/s. Note that we are using mixed units here, with kg for mass and cm/s for velocity; this isn't a particularly good idea, and we should be prepared at any time to change our units as appropriate to the situation. However, the numbers given here are intuitively comfortable and as it will turn out, we will end up with perfectly good results using these somewhat strange units. If this was an impulse-momentum situation, however, our careless use of units would be fatal.
Using m2 for the unknown mass of the second object, we obtain the after-collision x and y total momenta indicated in the second and third lines below.
The before-collision total x and y momenta are, respectively, .2 kg ( v1B) and 0, where v1B represents the unknown velocity of the first ball before collision.
If we set the before-collision values of px and py equal to their respective after-collision values, we will obtain two equations in the unknowns m2 and v1B.
We begin by solving the equation for the y momenta, as in the last two lines below, for m2.
From the equation we easily obtain m2 = 65 g.
We can understand this result intuitively by noting that after collision, the y velocity of the second ball is 40/13 times as great as that of the first. Since there was originally no y momentum, it follows that the second ball must be 13/40 as massive as that of the first; 13/40 ( 200 g ) = 65 g.
It is now easy to find v1B. Setting the x momentum after collision equal to the x momentum before, and using .065 kg for the mass of the second, we obtain and equation that is easily solved for this unknown.
