Solution Set 4

1. To tighten a spark plug, it's recommended that a torque of 15 N m be applied. If a mechanic tightens the spark plug with a wrench that is 25 cm long, what is the force needed?
Solution:
We know that torque, t, is equal to the force applied, F, multiplied by the length of the lever, or, length of the moment arm, or, length of the handle, or whatever you want to call it, r. the relationship looks like this:
t = r F

Here, we are asked to find the force, F. If we re-write the relationship above, we have:

F = t/r. Do we know the torque, t? Yes, it is given as 15 N m. Do we know the length of the moment arm, r? Yes, it is given as 25 cm = 0.25 m. So all we have to do is “plug and chug” to get the answer, and we don't have to search for clues to find anything else:

F = t/r = (15)/(0.25) = 60 N.
2. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is
a. horizontal
b. 22.5° below the horizontal?
Solution:
In reality, torque is the length of the moment arm, r, multiplied by the force, F, then multiplied by the sine of the angle between them, Ɵ. Or, in other ways of putting it,

t = r F sin Ɵ. If it is a right angle, Ɵ = 90°, then the sin 90° = 1.0, and we don't worry about it.
a. t = r F = (0.65 m)(1.61)(9.8) = 10.2557 N-m, or, 10.3 N-m.
b. Since the angle is 22.5°, or (90° – 22.5°) from the axis, then the torque is t = r F sin Ɵ = 10.3 sin 67.5° = 9.52 N.

3. Suppose a torque rotates your body about one of three different axes of rotation: case A, an axis through your spine; case B, an axis through your hips; and case C, an axis through your ankles. Rank these three axes of rotation in increasing order of the angular acceleration produced by the torque. Indicate ties where appropriate.
Solution:
The least torque will be through the spine, the greatest through the ankles.

4. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
Solution:
One of our relationships is t = I a, and to find torque, we need the moment of inertia, I, and the angular acceleration, a. Do we have them? Yes! The moment of inertia around a uniform rod of length 3.15 meter and mass of 8.42 kg, is I = 1/12 m L2, or, I = (0)(8.42)(3.15)2. = (0.083)(8.42)(9.9225) = 6.96 kg m2.

Now we can find the torque: t = I a = (6.96)(0.302) = 2.1 N m.

5. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg.
a. What is the angular acceleration of the fishing reel?
b. How much line does the fish pull from the reel in ¼ sec.?
Solution:
a. We are looking for acceleration and can use: t = I a = r F, so a = r F/I. Since I = ½ MR2 for a cylinder, then = 2F/Mr = 4.4/(0.55) = 8 rad/sec2.
b. First we use θ = θ0 + w0 t + ½ a t2 = (in radians), but initial angle and initial angular velocity are both 0.0. So, we rewrite it to be: θ = ½ a t2 = (0.50)(8)(0.25)2 = 0.25 radian. Since distance traveled is s = r , then s = (0.055m)(0.25) = 0.01375 or about 0.014m.

6. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length. What horizontal force applied to the top of the crate will start the crate to tip?
Solution: This is a torque problem.
7. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.
Solution: Angular momentum in this case is L = I w. where I = 2/5 M r2 = (0.4)(6 x 1024 kg)(6.4 x 106 m)2 = (2.4 x 1024)(4.1 x 1013) = 9.8 x 1037 kg m2 /s2 .
8. As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
Solution: 3.17/5.46 = 0.58.
9. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
Solution: W = ½ I w2, where I = ML2/12 = (0.44)(0.53)2/12 = 0.124/12 = 0.01 kg-m 2 , and w = 7.4 rad/s. So, W = ½ I w2 = (0.5)(0.01)(7.4)2 = 0.28 Joule.
10. A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
Solution:
Using F = G m M/r2, we plug and chug:
a. F = [(6.67 x 10-11)(6.1)(7.2) / (0.75)2] = [(292 x 10-11)/(0.5625)] = 519 x 10-11 = 5.2 x 10-9 N.
b. Re-write F = G m M/r2 to be r2 = G m M/F, we now plug and chug, so r2 = [(6.67 x 10-11)(6.1)(7.2)]/(2.0 x 10-9) = (292 x 10-11)/(2.0 x 10-9) = (2.92 x 10-9)/(2.0 x 10-9) = (2.92)/(2.0) = 1.46. So, if r2 = 1.46, then r = ?(1.46) = 1.2 m.
11. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
Find the distance.
Solution:
a. First, knowing Newton's second law is F = m a, and that in this case, F = m g, and knowing both F and g, we can re-write it as m = F/g to get the mass of the object: Or, m = (2.2 N)/(9.8 m/s2) = 0.224 kg. And now, Using F = G m M/r2, we re-write it as r2 = G m M/F and then plug and chug: r2 = [(6.67 x 10-11)(0.224)(6 x 1024)/(2.2)] = (8.96 x 1013) / (2.2) = 4.1 x 1013 . So, if = 4.1 x 1013, then
r = ?(4.1 x 1013) = 6.4 x 106 m.
b. Duh. a = g = - 9.8 m/s2.
12. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
Solution: Geosynchronous means it will revolve about Earth in 24 hours, or 86,400 seconds. Also, if it is 3.58 x 107 m above Earth's surface, then it is 3.58 x 107 m + 6.4 x 106 m 4.22 x 107 m = from the Earth's core. Speed is distance over time, or, v = c/P, where c = 2 p r = (2)(3.14)(4.22 x 107 m) = 26.5 x 107 m = 2.65 x 108 m. And P = 86,400 sec = 8.64 x 104 s. So, v = c/P, = (26.5 x 107)/(8.64 x 104) = 3.1 x 103 m/s.
13. Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
Solution: Here we use Kepler's 3rd Law, which is: P2 = k a3, where P is the period in seconds that we are looking for, a is the distance from the center, a = 9378 km = 9.378 x 106 m. And k = (4 p2 / G M) where G = 6.67 x 10-11. and for Mars, M = 6.4 x 1023 kg. Thus, k = (4)(p2) / (6.67 x 10-11)(6.4 x 1023) = (39.5)/(4.27 x 1013) = 9.25 x 10-13. So now we can plug and chug: P2 = k a3 = (9.25 x 10-13)(9.378 x 106 )3 = (9.25 x 10-13)(824.8 x 1018) = (9.25 x 10-13)(8.25 x 1016) = 76.3 x 103 = 7.6 x 104. Thus, if P2 = 7.6 x 104 , then P = ?(7.6 x 104 ) = 2.76 x 102 sec. (This is about 4 minutes, which sounds way too low. Maybe I need to do it again.
14. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
Solution:
a. Increase
b. Increase
15. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
Solution: Speed is distance over time, or, the circumference of the orbit divided by the about 80 years: v = 2 p r / P = (2)(3.14)(1.72 x 1012 m)/(2.52 x 109 seconds) = (10.8 x 1012)/(2.52 x 109 s) = 4.3 x 103 m/s.
16. Find the Escape Velocity, vesc, for the planet Mercury.
Solution:
Escape velocity is, ve = ?(2GM/r), with G = 6.67 x 10-11, M = 3.3 x 1023 kg, and r = 2.44 x 106 m. So, now we know that ve = ?(2GM/r) = ve = ?[(2)(6.67 x 10-11)(3.3 x 1023 kg)/(2.44 x 106 m)] = ?[(4.4 x 1013)/(2.44 x 106 m)] = ?(1.8 x 107) = 4.2 x 103 m/s.
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