AP50 Fall 2014
Problem Set 4 Solutions 1) Estimate the size of an asteroid that you could escape from by jumping.
AP50 Fall 2014
AP50 Fall 2014
2) All 2) All the marbles While working on their Rube-Goldberg Machine, Alice and Bob play a Rube-Goldberg-ish game, in which they try to hit a small box using a spring-loaded marble gun, which is fixed rigidly to a table and projects a marble of inertia m horizontally from the edge of the table. The edge of the table is a height h above the top of the box. The spring has a spring constant k and the edge of the box is some unknown horizontal distance L away from the table. Bob compresses the spring a distance x and finds that the marble falls short of its target by a horizontal distance y. How far should Alice compress the s pring in order for the marble to land in the box? Let g denote the gravitational acceleration. Express your answer in terms of k, m, x, g, h , and y as needed but do not use the unknown distance
AP50 Fall 2014
AP50 Fall 2014
3) Loop-the-Loop A roller-coaster car initially at a position on the track a height h above the ground begins a downward run on a long, steeply sloping track and then goes into a circular loop-the-loop of radius R whose bottom is a distance d above the ground (see picture). Ignore friction. (a) What is the car’s speed when it reaches the bottom of the loop? (b) What is the magnitude of the normal force exerted on the car at that instant? (c) What is the speed when its position is one-quarter of the way around the loop? (d) What is the magnitude of the normal force exerted on it at the one-quarter position? (e) What is the car’s acceleration at the one-quarter position?
AP50 Fall 2014
AP50 Fall 2014
AP50 Fall 2014
4) I saw the sign You want to hang a 10-kg sign that advertises your new business. To do this, you use a pivot to attach the base of a 5.0-kg beam to a wall. You then attach a cable to the beam and to the wall in such a way that the cable and the beam are perpendicular to each other. The beam is 2.0 m long lon g and makes an angle of 37 degrees with the vertical. You hang the sign from the end of the beam to which the cable is attached. attached. See figure. (a) What must be the minimum tensile strength of the cable (the minimum amount of tension it can sustain) if it is not to snap? (b) Determine the horizontal and vertical components of the force exerted by the pivot on the beam.
I: Getting Started: We know that the system must be in mechanical equilibrium equilibrium because the sign is not moving. This means that both the sum of the forces acting on the beam and the sum torque caused by the forces on the beam are each equal to zero.
II: Devise Plan: Draw a free body diagram of the beam showing all of the forces acting on it. Also draw an extended free body diagram of the beam showing all of the forces acting on it to produce torques. Follow either one of the procedures in Figure 12.5 to either split the forces into components parallel and perpendicular to the lever arm, OR calculate the lever arms perpendicular to each force. We arbitrarily choose the first version, to split the forces into components, but either is fine. Use the fact that the beam is in mechanical equilibrium to find the unknown tension in the cable and the forces exerted by the pivot o n the beam (by summing forces and torque and setting equal to zero).
III: Execute Plan: We draw our free body diagram (see Ch 8.5) and choose the x-axis along the beam and the y-axis perpendicular (but other axes are ok as long as you are consistent). We draw an extended free body diagram (see Ch. 12.3) and break the forces on the beam into parallel and perpendicular components to calculate the torque but this is not the only way (see figure 12.5 for two methods to determine torques)
AP50 Fall 2014
The sign is not rotating when it is hanging, so we know that the sum of the torques caused by the forces about the pivot must be zero (We could choose a different reference point, however this is a natural choice since the force by the pivot on the beam is unknown. See procedure for drawing extended free body diagram in Ch. 12.3) We have broken all forces into components along or perpendicular to the beam, so the only forces that cause torques are those perpendicular to the beam, The force of the cable and the sign at the end of the beam (a length l away away from the pivot) and the force of Earth’s gravity on the beam (at the center of mass, a length l/2 away from the pivot)
In this equation we know everything except for the con tact force of the cable on the beam, which is what we are trying to solve for.
We can find the force of the sign on the beam in the y direction by using interaction pairs to find the portion of the weight of the sign that causes a torque around the pivot.
If we plug in this number for the mass of the sign and also plug in the weight of the beam, we can find the tension in the cable such that no overall torque results.
AP50 Fall 2014
To find the force of the pivot on the beam, set the sum of the forces on the beam = to zero.
Note that the problem asks for horizontal and vertical components. Given our choice of x and y directions, these are the two components of the force of the pivot on the beam, in the x direction (along the beam) and the y direction (perpendicular to the beam). If we instead want the forces along the wall (vertical) and perpendicular to the wall (horizontal), we can break our forces into components as shown below.
IV: Evaluate Result: We draw another diagram including the numbers that we found for all of the forces and choose a different reference point (in this case the other end o f the beam) to double check our calculations and make sure the sum of the torques is still zero. We also check that the forces sum to zero in both the x and y directions. This should catch any sign errors!
AP50 Fall 2014
5) Rolling away A 3.0-kg solid ball rolls without slipping down a ramp inclined to the horizontal at an angle of 30 degrees. What are (a) the acceleration of the ball’s center of mass and (b) the magnitude of the frictional force exerted on the ball?
I: Getting Started: We draw a sketch of the situation, a ball rolling down a ramp with 30 degree incline.
We draw a free body diagram and an extended free diagram of the ball and choose the axes and positive rotational direction shown in the diagrams. (r=ramp, o = object (ball), n = normal force, s = static friction force, x,y = components of force in that direction) We split the gravitational force of the Earth on the object (ball) into x and y components.
II: Devise Plan: We will find the acceleration of the ball’s center of mass and the magnitude of the frictional force on the ball by creating a system of two equations and two unknowns by using = for the translational motion and = , the condition for rotational motion without slipping.
III: Execute Plan: First, we focus on the translational motion of the ball down the ramp in the x direction.
This equation has two unknowns, so we can’t solve it directly and now look at the rotational motion (rolling without slipping).
We can solve for the rotational acceleration, by summing the forces acting on the ball to produce torque and thus, rotational acceleration as it rolls down the ramp (eq 12.22). We use rotational inertia, I = 2/5mR 2 since we have a ball (Table 11.3)
AP50 Fall 2014
Substituting this result for the angular acceleration into eq. 12.23 (the eqn for the acceleration of the center of mass for rolling without slipping) and setting it equal to the equation for the acceleration of the center of mass using translational forces, we can solve for the static friction force of the ramp on the object (ball):
Plugging in numbers, we find the static friction force and center of mass acceleration.
IV: Evaluate Result: We had another equation for the acceleration, so let’s try to solve for it using that equation to double check our math and make sure we get the same thing.
To check the value of the friction force we found, we can also look at how the equation we found for friction behaves as we change the value of different variables. As m increases, the static friction force increases. As we expect, a heavier ball has more friction. We also see that if g were to decrease, the friction force would also decrease. We expect that in a low-gravity environment (like on the moon) that we would experience less friction so this makes sense. Finally, if we increase the angle of the ramp, the friction force would increase. We realize though that there is some limit to this as we have done calculations for static friction only. At some point, the ball would start to slip and we would instead have kinetic friction.
AP50 Fall 2014
6) Mission to Mars To travel between Earth and any other planet requires consideration of such things as expenditure of fuel energy and travel time. To simplify the calculations, one chooses a path such that the position in Earth's orbit where the launch occurs and the position in the other planet's orbit when the spacecraft arrives define a line that passes through the Sun, as shown for the Earth-Mars transfer in the figure. The path the spacecraft travels is an ellipse that has the Sun at one focus. Such a path is called a Hohmann transfer orbit, and the major axis of the ellipse, 2a, is the sum of the radii of the Earth's and the other planet's orbits around the Sun. Take ms = 1.99 x 10 30 kg, mm = 6.42 x 10 23 kg, and me = 5.97 x 10 24 kg, ae = 1.50 x 1011 m, and am = 2.28x1011 m and assume the planets have circular orbits. (a) What is the energy of the system comprising the Sun and a 1000kg space probe in a Hohmann transfer orbit? (b) What is the speed of the probe in this orbit, as a function of r, the probe's radial distance from the Sun? (c) What is the probe's speed relative to the Sun as the probe enters the transfer orbit?
I: Getting Started: We already have a sketch of what a Hohmann transfer would look like. We draw another diagram depicting the elliptical orbit o f that transfer and labeling important parameters.
II: Devise Plan: The sun and probe define the system in this problem. This system is closed and isolated so neither energy nor momentum change. The mass of the sun is much larger than the mass of the probe so we assume the center of the sun is the center of mass of o f the system. We can therefore find the total energy of the system by adding the gravitational potential energy and the kinetic energy of the spacecraft. For part a, using conservation of angular momentum and energy we can find the energy of the system. For part b, if we set the energy we found in part a equal to the sum of the kinetic and gravitational energy we can solve for the speed of the probe as a function of r. In part c, we just need to plug p lug in the radius of the earth’s orbit around the sun into the equation we calculated in part b since we know that is when it will launch (by definition of the Hohmann orbit).
AP50 Fall 2014
III: Execute Plan: momentum at (a) As shown in example 13.5, we know that the probe’s probe’s angular momentum launch (when it is injected from low Earth orbit into the transfer orbit) will be the same as at its arrival (into Mars orbit from the transfer orbit). As also a lso shown, we can use that relation to write an expression for the kinetic energy of the probe and therefore the energy of the system both at launch and at arrival. The mechanical energy must be the same so we can set them equal and find: ℎ =
2
Using this result from example 13.5, we substitute in the values given in the problem to find the energy of the probe-sun system as:
(b) We know that the energy of the system is the gravitational potential energy plus the kinetic energy of the spacecraft.
We want v(r) and above we have an expression for the total mechanical energy of the system so we can set them equal and solve for v.
AP50 Fall 2014
(c) We want to know the probe’s speed relative to the sun as the probe enters the transfer orbit. Looking back at our diagrams, we see that the Hohmann transfer orbit we are using is defined both by the distance of Earth from the sun and by the distance of Mars from the sun. We enter the transfer orbit from low earth orbit and therefore, we must set se t the radial distance from the sun, r, in the equation we found in part b to the distance of Earth from the sun. This will give us the velocity we need to enter the Hohmann orbit.
IV: Evaluate Result: In part (a), we found that the total mechanical energy is negative. This makes sense because we know the total mechanical energy becomes positive only when the probe would have enough energy to “escape” to infinity. If it is negative, it is bound to the body it is orbiting around (here, the sun). It does not have enough kinetic energy to escape. In this problem we’re talking about a bound elliptic al orbit so we know this must be true. (Ch 13.7) In parts (b) & (c), we use an equation which we calculated. We can easily check the units (though this is not the best check, it’s something easy to try first if we think we’re wrong).
If we increase M in this equation (orbit around a more massive planet), the probe would have a larger velocity. This is because it would have more gravitational gravitational energy. We see from this equation that v is the fastest when r is the smallest on this orbit. We see that this is true in Figure 13.12 where in regions where the probe is closest to the body it is orbiting around, it must go faster to effectively sweep out more area of an arc (since angular momentum must be constant everywhere on the orbit). Also, the earth travels around the sun at 108,000 km/hr = 30,000m/s and we know our probe needs to go slightly faster to get into the Hohmann transfer orbit so our answer seems reasonable.
