2211 Physics A Summer 2007
Quiz #5
Solutions
gEarth = 9.8 m/s2 REarth = 6.37 106 m -11 2 2 G = 6.673 10 Nm /kg MEarth = 5.98 1024 kg Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) A solid sphere of mass M and radius R is pivoted on a horizontal axle that passes through the sphere a distance R/2 from the center. The sphere is held so the center is an angle below the axle, and released. What is the angular acceleration at the instant of release, in terms of any or all of R, M , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
The net torque on the sphere can be related to its angular acceleration by the rotational version of Newton's Second Law, = I . The only torque about the axle is due to the force of weight, which acts at the center of mass. So = I w = I rw M g sin = I
where is the angle between the r vector and the weight vector when placed tail to tail, or = 90 - .
2 The rotational inertia of a solid sphere about its center of mass is Icm = 5 M R2 . This sphere is not rotating about its center of mass, though, so the parallel axis theorem, I = Icm + M d2 can be used to find the rotational inertia about the axle. Since the axle has been displaced a distance R/2 from the center of mass,
I = Icm + M d2 = Putting these together rw M g sin = I
2 M R2 + M 5
R 2
2
=
2 1 + 5 4
M R2 =
13 M R2 20
R 2
M g sin (90 - ) =
13 M R2 20
and solving for the angular acceleration (remember that sin (90 - ) = cos ) = (R/2) M g cos 10 = 2 (13/20) M R 13 g cos R
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II. (16 points) A 45 kg scientific satellite is to be launched into a circular orbit an altitude two Earth-radii above the Earth's surface. The launch will take place from the South Pole, so the satellite will have no kinetic energy before it is launched. How much work must be done to place this satellite in orbit? . . . . . . . . . . . . . . . . . . . . . . . Use the Work-Energy Theorem, since it relates the work done by an external force (the answer to the question) to the change in kinetic and potential energy. In this case, there are no non-conservative forces in the Earth-satellite system, and the potential energy is that of Universal Gravitation. Wext + Wnc = K + U Wext =
2 1 2 mvf 2 - 1 mvi + 2
-GM m -GM m - rf ri
where M is the mass of the Earth and m is the mass of the satellite. Note that the zero of potential energy has been chosen at infinite separation. Since the satellite starts a distance one Earth radius from the center of the Earth, and ends up three Earth radii from the center (i.e., two Earth radii from the surface), ri = RE and rf = 3RE . The initial satellite kinetic energy is zero. So
2 Wext = 1 mvf - GM m 2
1 1 - 3RE RE
2 = 1 mvf + GM m 2
2 3RE
Turn for a moment to Newton's Second Law to find the final speed required for the circular orbit. The only force on the satellite in orbit is the force of gravity, which points toward the center of the orbit. Fr = FG = mar
2 vf GM m =m 2 rf rf
2 1 2 mvf
=
GM m GM m = 2rf 2 3RE
Substituting this into the expression derived from the Work-Energy Theorem
Wext =
GM m + GM m 2 3RE
2 3RE
= =
5 GM m 6 RE 5 6.673 10-11 Nm2 /kg2 5.98 1024 kg (45 kg) = 2.3 109 J 6 6.37 106 m
1. (6 points) In the problem above, how does the work required to place the satellite in the orbit compare to that needed to achieve an altitude of two Earth-radii without orbiting? . . . . . . . . . . . . . . . . . . . . . . . If the satellite just achieved an altitude of 2RE without orbiting, it would have a kinetic energy of zero at that altitude. More work must be done to give the satellite the kinetic energy required to maintain orbit. The work required to place the satellite in orbit is greater than that required to achieve the altitude.
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III. (16 the the the
points) A thin-walled hollow sphere of mass m and radius r is rolled from a height h so that it "loops loop," that is, it rolls around the track with a loop of radius R without slipping or losing contact with track, as illustrated. The radius r of the thin-walled hollow sphere is much smaller than the radius R of loop.
Find the minimum height h that will accomplish this, in terms of any or all of m, r, R, and physical or mathematical constants. (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . First, find the translational speed required for the sphere to maintain contact with the top of the loop. Use Newton's Second Law. At the minimum necessary speed, the normal force downward on the sphere at the top of the loop is zero. The only force on the sphere is the downward force of gravity.
Fr = w = mar
mg = m
v2 R
v=
gR
Next, look at the Work-Energy There Theorem. are no nonconservative forces, and external forces do no work, in the Earth-sphere system. The potential energy is that of local gravity. There are both translational and rotational kinetic energies. Wext + Wnc = K + U Wext =
2 1 2 mvf 2 - 1 mvi + 2 2 1 2 If 2 - 1 Ii + (mgyf - mgyi ) 2
The sphere has no translational or rotational kinetic energy initially, as it is released from rest. Let the zero of potential energy be at the bottom of the loop. Then yi = h and yf = 2R.
2 2 1 0 = 2 mvf + 1 If + (mg2R - mgh) 2
The rotational inertia of a thin-walled hollow sphere about its center is circumference without slipping, f - vf /r. Remember that vf = gR.
2 mg (h - 2R) = 1 mvf + 2 1 2
2 3
mr2 . Since it is rolling about its
2 mr2 3
vf r
2
=
5 5 2 mvf = mgR 6 6
Solving for h, g (h - 2R) = 5 gR 6 h - 2R = 5 R 6 h= 17R 6
2. (6 points) How does the minimum height h for the thin-walled hollow sphere compare to that of an object that does not roll, but instead slides without friction? . . . . . . . . . . . . . . . . . . . . . . . Because not all of the initial gravitational potential energy is transformed into translational kinetic energy for a rolling object (some is transformed into rotational kinetic energy), a rolling object must start out with more gravitational potential energy than a sliding object to achieve the same translational speed in the end. h is greater for the thin-walled hollow sphere than for the sliding object.
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3. (10 points) Taking the zero point of gravitational potential energy to be at infinite separation, what must be true of a space probe if it is eventually to escape the Earth's gravity? . . . . . . . . . . . . . . . . . . . . . . . To escape from the Earth's gravity, the probe must have at least enough kinetic energy achieve infinite separation. At that point, its gravitational potential energy is zero, and its kinetic energy must be at least zero. Its mechanical energy must be greater than or equal to zero. Unfortunately (or perhaps fortunately), the question mistakenly does not ask what is necessary to achieve infinite separation, only what is true if it can achieve it. Therefore, full credit is issued for . . . Its kinetic energy must be greater than or equal to zero. which is always true, and Its kinetic energy must be greater than or equal to its gravitational potential energy. which is true with that choice of of zero for gravitational potential energy.
4. (10 points) Two uniform solid spheres have equal mass and each is spinning on an axis through its center. Sphere B only has half the radius of sphere A, but sphere B is spinning twice as fast as sphere A. How does the angular momentum magnitude of sphere B compare to that of sphere A? . . . . . . . . . . . . . . . . . . . . . . . Recall that L = I for an object spinning about a symmetry axis. Recall, too, that for objects of circular cross-section, rotational inertias depend on the square of the radius. So for sphere B is twice a much as for sphere A, but I for sphere B is only one-fourth as much as for sphere A. Putting these factors together . . . The angular momentum magnitude of sphere B is half as much as sphere A.
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5. (10 points) A solid sphere and a thin spherical shell have the same mass M and radius R, and turn on frictionless horizontal axles. A rope is wrapped around the diameter of each sphere and tied to a block. The blocks have the same mass m, and are held the same height h above the ground. Both blocks are released simultaneously. The ropes do not slip. Compare the times required for the blocks to hit the ground. . . . . . . . . . . . . . . . . . . . . . . . The solid sphere has more of its mass near the axis of rotation, so it has a lower moment of rotational inertia, and will have a greater angular acceleration. The block tied to the solid sphere hits the ground first.
6. (10 points) Five forces act on a rod, as illustrated. Forces i through iv have magnitude F . Force v has magnitude 2F . Forces i, ii, and v act at the midpoint of the rod, while forces iii and iv act at the end. Rank the five torque magnitudes these forces produce about the pivot point on the left, in order from greatest to least. . . . . . . . . . . . . . . . . . . . . . . . The line of action for force iii passes through the pivot, so it produces no torque. Forces iv and v produce the same torque as each other, as force v has twice the magnitude of force iv, but force iv is twice as far from the pivot. Force ii produces less torque than forces iv and v (same distance as force v from the pivot, but less force), and force i produces even less torque than force ii, since it isn't perpendicular to the rod. iv = v > ii > i > iii
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