Physics 2211 Quiz #3 Solutions Spring 2008 Unless otherwise ...
Physics 2211 Spring 2008
Quiz #3
Solutions
Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) The Earth is a planet with mass M and radius R. What would the period of the Earth’s rotation be if objects at the equator had an apparent weight that was one-third of the gravitational force on them? Express your answer in terms of any or all of M , R, and physical or mathematical constants. Note: Do NOT use the physical constant “g”, as it includes the effect of the Earth’s actual period. . . . . . . . . . . . . . . . . . . . . . . . This is a Newton’s Second Law problem with Uniform Circular Motion. Sketch a Free Body Diagram of the object at the equator. Choose the positive direction to be in the direction of the centripetal acceleration, toward the center of the Earth. The only forces acting on the object are the force of gravity and a normal force. The normal force is the object’s apparent weight. The force of gravity on the object will be calculated using the Law of Universal Gravitation. The acceleration is a centripetal acceleration. Note that the speed of the object, v, is the circumference of the Earth, 2πR, over the period, T . Letting M be the mass of the Earth, and m be the mass of the object, X
Fr = FG − n = mar
⇒
FG − 13 FG = m
v2 R
⇒
G
Mm 1 Mm m − 3G 2 = 2 R R R
µ
2πR T
¶2
Solve for T . Note that the mass of the object cancels. 2 Mm 3 G R2
4mπ 2 R = T2
⇒
2
T =
3 2
4π 2 R3 GM
r ⇒
T =
6π 2 R3 GM
Quiz #3 Solutions Page 1 of 5
1. (6 points) An object of mass M is attached to the end of a string and swings in a vertical circle having a radius r. At an instant when the string is at an angle θ above the horizontal, the tension in the string is T . What is the radial component of the weight? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . Sketch a Free Body Diagram of the object. Resolve the weight into radial and tangential components. From the FBD, wr = M g sin θ
II. (16 points) In the question above, what is the speed of the ball at this instant, in terms of any or all of M , r, θ, T , and physical or mathematical constants? . . . . . . . . . . . . . . . . . . . . . . . This is a Newton’s Second Law problem with Uniform Circular Motion. A Free Body Diagram was sketched in the previous part of the problem. Note that the r axis was selected with positive toward the center, in the direction of the radial or centripetal acceleration. X
Fr = T + wr = M ar
⇒
v2 T + M g sin θ = M r
r ⇒
v=
r (T + M g sin θ) M
Quiz #3 Solutions Page 2 of 5
III. (16 points) Bob stands atop a building and throws a rock with speed v0 at an angle θ below the horizontal. He releases the rock a height h above the level ground. In terms of any or all of v0 , θ, h, and physical or mathematical constants, how far from the base of the building does the rock land? Assume θ is between 0◦ and 90◦ . (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . This is a projectile motion problem. That is, it is a constant-acceleration kinematics problem in which the vertical acceleration is that of gravity, and the horizontal acceleration is zero. Choose a coordinate system. One choice is for the origin to be at the release point, with positive x away from the building and positive y downward. Let time be zero at the instant of release. Looking at the x direction, 2
xf = xi + v0x ∆t + 12 ax (∆t)
Remember that ax = 0, and with the chosen coordinate system, xi = 0. The answer to the question is xf , and v0x = v0 cos θ. So xf = v0 cos θ ∆t The vertical information can be used to find ∆t. yf = yi + v0y ∆t + 12 ay (∆t)
2
With the chosen coordinate system, yi = 0, yf = h, v0y = v0 sin θ, and ay = +g. Remember that g is a magnitude, so the “+” in “+g” isn’t really necessary. So, h = v0 sin θ ∆t + 12 g (∆t)
2
2 1 2 g (∆t)
⇒
+ v0 sin θ ∆t − h = 0
That’s a quadratic in ∆t, so ∆t =
−B ±
√
B 2 − 4AC 2A
where A = 12 g, B = v0 sin θ, and C = −h. The time must be positive (the ball hits the ground after release), so q ∆t =
−v0 sin θ ±
q
2
(v0 sin θ) − 4 (g/2) (−h) 2 (g/2)
=
−v0 sin θ +
2
(v0 sin θ) + 2gh g
Substituting into the expression for xf above q
xf = v0 cos θ
−v0 sin θ +
2
(v0 sin θ) + 2gh g
2. (6 points) In the problem above, Biff stands next to Bob atop the building and throws an identical rock at the same time. Biff throws his rock horizontally with a speed 2v0 . Whose rock hits the ground first? . . . . . . . . . . . . . . . . . . . . . . . The horizontal motion is irrelevant. Bob’s rock has a non-zero initial velocity downward, while that for Biff’s rock is zero. Bob’s rock hits first.
Quiz #3 Solutions Page 3 of 5
3. (10 points) An airplane pilot wishes to fly directly north. Because of a wind from the west, however, if he points the nose of his aircraft directly north, he finds himself flying over the ground at an angle θ east of north (Fig. 1). To compensate, the pilot points the nose of his aircraft at an angle φ west of north, so he flies over the ground directly to the north (Fig. 2). How does the angle φ compare to the angle θ, and how does the time required to reach his destination compare to the time required in still air? . . . . . . . . . . . . . . . . . . . . . . . Note that the missing side of the velocity triangle is the velocity of the air over the ground. The velocity of the plane through the air is a leg of the triangle in Fig. 1, while it is the hypotenuses in Fig. 2. Therefore, the velocity of the plane over the ground in Fig. 2 must be less than the velocity of the plane through the air in Fig. 2 and in Fig. 1. The angle φ, then, must be greater than the angle θ. Since the velocity of the plane over the ground in Fig. 2 is only a component of the velocity of the plane though the air, then the plane travels more slowly toward its destination then it would in still air. φ > θ and the time is greater than that required in still air.
4. (10 points) A fuzzy die is attached to the review-view mirror of a car by a string. The car is driving around in a circle to the left in the illustration, at a constant speed. What are all the forces acting on the die? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The die is not supported by a surface, so there is no normal force. The centrifugal force is “fictitious”, (i.e., a consequence of using a non-inertial frame), and the centripetal force is just the sum of actual forces toward the center. The forces acting on the die, then, are Tension and weight.
Quiz #3 Solutions Page 4 of 5
5. (10 points) A merry-go-round is spinning with a constat angular velocity. A person walks from the center towards the edge, and does not slide. How does the magnitude of the force of static friction, f , on the person vary as their distance from the center, r, increases? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . From the Free Body Diagram, X
Fr = fs = mar = m
v2 = mrω 2 r
The mass, m, and the angular velocity, ω, are constant. As the person approaches the edge, increasing r, the force of static friction must increase linearly.
6. (10 points) Two spheres of mass m and 9m have a center-to-center separation distance `. If there is a point at which there is no net gravitational force, how far is it from the center of the sphere with mass m? . . . . . . . . . . . . . . . . . . . . . . . Since gravity affects all mass, the only way there can be no net gravitational force on an object is if the forces on it add to zero (i.e., the individual forces cannot be zero). The forces from these two spheres would have to equal and opposite to add to zero. Since gravity is a inverse square law, there will be no net force on an object between the spheres and three times farther from the 9m sphere than from the m sphere. That can be accomplished at a point `/4 from the sphere with mass m, and 3`/4 from the sphere with mass 9m. More formally, let the distance from the sphere with mass m be x. The distance from the sphere with mass 9m will then be ` − x. Setting the force magnitudes equal
G
M (m) M (9m) =G 2 x2 (` − x)
⇒
2
(` − x) = 9x2
⇒
` − x = 3x
⇒
` = 4x
⇒
x = `/4
Quiz #3 Solutions Page 5 of 5
Quiz #3
Solutions
Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) The Earth is a planet with mass M and radius R. What would the period of the Earth’s rotation be if objects at the equator had an apparent weight that was one-third of the gravitational force on them? Express your answer in terms of any or all of M , R, and physical or mathematical constants. Note: Do NOT use the physical constant “g”, as it includes the effect of the Earth’s actual period. . . . . . . . . . . . . . . . . . . . . . . . This is a Newton’s Second Law problem with Uniform Circular Motion. Sketch a Free Body Diagram of the object at the equator. Choose the positive direction to be in the direction of the centripetal acceleration, toward the center of the Earth. The only forces acting on the object are the force of gravity and a normal force. The normal force is the object’s apparent weight. The force of gravity on the object will be calculated using the Law of Universal Gravitation. The acceleration is a centripetal acceleration. Note that the speed of the object, v, is the circumference of the Earth, 2πR, over the period, T . Letting M be the mass of the Earth, and m be the mass of the object, X
Fr = FG − n = mar
⇒
FG − 13 FG = m
v2 R
⇒
G
Mm 1 Mm m − 3G 2 = 2 R R R
µ
2πR T
¶2
Solve for T . Note that the mass of the object cancels. 2 Mm 3 G R2
4mπ 2 R = T2
⇒
2
T =
3 2
4π 2 R3 GM
r ⇒
T =
6π 2 R3 GM
Quiz #3 Solutions Page 1 of 5
1. (6 points) An object of mass M is attached to the end of a string and swings in a vertical circle having a radius r. At an instant when the string is at an angle θ above the horizontal, the tension in the string is T . What is the radial component of the weight? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . Sketch a Free Body Diagram of the object. Resolve the weight into radial and tangential components. From the FBD, wr = M g sin θ
II. (16 points) In the question above, what is the speed of the ball at this instant, in terms of any or all of M , r, θ, T , and physical or mathematical constants? . . . . . . . . . . . . . . . . . . . . . . . This is a Newton’s Second Law problem with Uniform Circular Motion. A Free Body Diagram was sketched in the previous part of the problem. Note that the r axis was selected with positive toward the center, in the direction of the radial or centripetal acceleration. X
Fr = T + wr = M ar
⇒
v2 T + M g sin θ = M r
r ⇒
v=
r (T + M g sin θ) M
Quiz #3 Solutions Page 2 of 5
III. (16 points) Bob stands atop a building and throws a rock with speed v0 at an angle θ below the horizontal. He releases the rock a height h above the level ground. In terms of any or all of v0 , θ, h, and physical or mathematical constants, how far from the base of the building does the rock land? Assume θ is between 0◦ and 90◦ . (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . This is a projectile motion problem. That is, it is a constant-acceleration kinematics problem in which the vertical acceleration is that of gravity, and the horizontal acceleration is zero. Choose a coordinate system. One choice is for the origin to be at the release point, with positive x away from the building and positive y downward. Let time be zero at the instant of release. Looking at the x direction, 2
xf = xi + v0x ∆t + 12 ax (∆t)
Remember that ax = 0, and with the chosen coordinate system, xi = 0. The answer to the question is xf , and v0x = v0 cos θ. So xf = v0 cos θ ∆t The vertical information can be used to find ∆t. yf = yi + v0y ∆t + 12 ay (∆t)
2
With the chosen coordinate system, yi = 0, yf = h, v0y = v0 sin θ, and ay = +g. Remember that g is a magnitude, so the “+” in “+g” isn’t really necessary. So, h = v0 sin θ ∆t + 12 g (∆t)
2
2 1 2 g (∆t)
⇒
+ v0 sin θ ∆t − h = 0
That’s a quadratic in ∆t, so ∆t =
−B ±
√
B 2 − 4AC 2A
where A = 12 g, B = v0 sin θ, and C = −h. The time must be positive (the ball hits the ground after release), so q ∆t =
−v0 sin θ ±
q
2
(v0 sin θ) − 4 (g/2) (−h) 2 (g/2)
=
−v0 sin θ +
2
(v0 sin θ) + 2gh g
Substituting into the expression for xf above q
xf = v0 cos θ
−v0 sin θ +
2
(v0 sin θ) + 2gh g
2. (6 points) In the problem above, Biff stands next to Bob atop the building and throws an identical rock at the same time. Biff throws his rock horizontally with a speed 2v0 . Whose rock hits the ground first? . . . . . . . . . . . . . . . . . . . . . . . The horizontal motion is irrelevant. Bob’s rock has a non-zero initial velocity downward, while that for Biff’s rock is zero. Bob’s rock hits first.
Quiz #3 Solutions Page 3 of 5
3. (10 points) An airplane pilot wishes to fly directly north. Because of a wind from the west, however, if he points the nose of his aircraft directly north, he finds himself flying over the ground at an angle θ east of north (Fig. 1). To compensate, the pilot points the nose of his aircraft at an angle φ west of north, so he flies over the ground directly to the north (Fig. 2). How does the angle φ compare to the angle θ, and how does the time required to reach his destination compare to the time required in still air? . . . . . . . . . . . . . . . . . . . . . . . Note that the missing side of the velocity triangle is the velocity of the air over the ground. The velocity of the plane through the air is a leg of the triangle in Fig. 1, while it is the hypotenuses in Fig. 2. Therefore, the velocity of the plane over the ground in Fig. 2 must be less than the velocity of the plane through the air in Fig. 2 and in Fig. 1. The angle φ, then, must be greater than the angle θ. Since the velocity of the plane over the ground in Fig. 2 is only a component of the velocity of the plane though the air, then the plane travels more slowly toward its destination then it would in still air. φ > θ and the time is greater than that required in still air.
4. (10 points) A fuzzy die is attached to the review-view mirror of a car by a string. The car is driving around in a circle to the left in the illustration, at a constant speed. What are all the forces acting on the die? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The die is not supported by a surface, so there is no normal force. The centrifugal force is “fictitious”, (i.e., a consequence of using a non-inertial frame), and the centripetal force is just the sum of actual forces toward the center. The forces acting on the die, then, are Tension and weight.
Quiz #3 Solutions Page 4 of 5
5. (10 points) A merry-go-round is spinning with a constat angular velocity. A person walks from the center towards the edge, and does not slide. How does the magnitude of the force of static friction, f , on the person vary as their distance from the center, r, increases? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . From the Free Body Diagram, X
Fr = fs = mar = m
v2 = mrω 2 r
The mass, m, and the angular velocity, ω, are constant. As the person approaches the edge, increasing r, the force of static friction must increase linearly.
6. (10 points) Two spheres of mass m and 9m have a center-to-center separation distance `. If there is a point at which there is no net gravitational force, how far is it from the center of the sphere with mass m? . . . . . . . . . . . . . . . . . . . . . . . Since gravity affects all mass, the only way there can be no net gravitational force on an object is if the forces on it add to zero (i.e., the individual forces cannot be zero). The forces from these two spheres would have to equal and opposite to add to zero. Since gravity is a inverse square law, there will be no net force on an object between the spheres and three times farther from the 9m sphere than from the m sphere. That can be accomplished at a point `/4 from the sphere with mass m, and 3`/4 from the sphere with mass 9m. More formally, let the distance from the sphere with mass m be x. The distance from the sphere with mass 9m will then be ` − x. Setting the force magnitudes equal
G
M (m) M (9m) =G 2 x2 (` − x)
⇒
2
(` − x) = 9x2
⇒
` − x = 3x
⇒
` = 4x
⇒
x = `/4
Quiz #3 Solutions Page 5 of 5
