Justin72835 wrote:Next question: You have a solid cube of mass 'm' which is attached to a nearby wall using a massless, ideal spring of constant 'k'. If you launch an arrow of velocity 'v' and mass 'M' directly at the solid cube, what is the maximum compression of the spring if:
1) the arrow sticks into the solid after hitting it?
2) the arrow bounces off of the cube perfectly elastically?
1) kMv/(M+m) 2) 2kMv/(M+m)
You're definitely on the right track (I see that you already found the final velocities of the objects after the collision). Using the final velocity, you can find the kinetic energy of the object after the collision and set it equal to 1/2kx^2 and solve for x. This works because maximum compression occurs when the objects kinetic energy has been converted completely into potential energy.
You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.
a) What is the hollow sphere's total kinetic energy before rolling up the ramp?
b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."
University of Texas at Austin '23
Seven Lakes High School '19
You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.
a) What is the hollow sphere's total kinetic energy before rolling up the ramp?
b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.
There are 2 different kinetic energies to consider when solving for the total kinetic energy: Translational KE and Rotational KE.
To find the initial total KE, simply add the 2 energies.
[math]KE_t_o_t_a_l = 1/2mv^2 + 1/2Iw^2[/math]
The Moment of Inertia ([math]I[/math]) for a [b]hollow[/b] sphere is equal to [math]2/3MR^2[/math]
Given a mass of 4 kg and a radius of 0.25 m, [math]I[/math] = 0.167
Now to solve for [math]w[/math]:
[math]v = rw[/math]
[math]w = 23/0.25 = 92[/math]
Plugging back into the equation results in:
[math]KE_t_o_t_a_l = 1/2(4)(23)^2 + 1/2(0.167)(92)^2[/math]
The answer to part A is 1763 J.
The law of conversation of energy states that the total energy of a system remains constant. There are 3 different energies being used in this scenario: Translational Kinetic Energy, Rotational Kinetic Energy, and Gravitational Potential Energy. This gives us the equation:
[math]1/2mv_1^2 + 1/2Iw_1^2+mgh_1 = 1/2mv_2^2 + 1/2Iw_2^2+mgh_2[/math]
The initial potential energy is 0, and the final kinetic energies are also 0. This simplifies the equation down to:
[math]1/2mv_1^2 + 1/2Iw_1^2 = mgh_2[/math]
From the previous part, we already have the total initial KE. Substituting into our equation gives us:
[math]1763 = (4)(9.8)h_2[/math]
With basic algebra, we find that the answer to part B is 44.98 m.
Last edited by Riptide on Wed Jan 10, 2018 7:18 pm, edited 1 time in total.
You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.
a) What is the hollow sphere's total kinetic energy before rolling up the ramp?
b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.
There are 2 different kinetic energies to consider when solving for the total kinetic energy: Translational KE and Rotational KE.
To find the initial total KE, simply add the 2 energies.
[math]KE_t_o_t_a_l = 1/2mv^2 + 1/2Iw^2[/math]
The Moment of Inertia ([math]I[/math]) for a [b]hollow[/b] sphere is equal to [math]2/3MR^2[/math]
Given a mass of 4 kg and a radius of 0.25 m, [math]I[/math] = 0.167
Now to solve for [math]w[/math]:
[math]v = rw[/math]
[math]w = 23/0.25 = 92[/math]
Plugging back into the equation results in:
[math]KE_t_o_t_a_l = 1/2(4)(23)^2 + 1/2(0.167)(92)^2[/math]
The answer to part A is 1763 J.
The law of conversation of energy states that the total energy of a system remains constant. There are 3 different energies being used in this scenario: Translational Kinetic Energy, Rotational Kinetic Energy, and Gravitational Potential Energy. This gives us the equation:
[math]1/2mv_1^2 + 1/2Iw_1^2+mgh_1 = 1/2mv_2^2 + 1/2Iw_2^2+mgh_2[/math]
The initial potential energy is 0, and the final kinetic energies are also 0. This simplifies the equation down to:
[math]1/2mv_1^2 + 1/2Iw_1^2 = mgh_2[/math]
From the previous part, we already have the total initial KE. Substituting into our equation gives us:
[math]1763 = (4)(9.8)h_2[/math]
With basic algebra, we find that the answer to part B is 44.98 m.
Hey, great job! Now its your turn.
Last edited by Justin72835 on Wed Jan 10, 2018 7:37 pm, edited 1 time in total.
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."
University of Texas at Austin '23
Seven Lakes High School '19
You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.
a) What is the hollow sphere's total kinetic energy before rolling up the ramp?
b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.
There are 2 different kinetic energies to consider when solving for the total kinetic energy: Translational KE and Rotational KE.
To find the initial total KE, simply add the 2 energies.
[math]KE_t_o_t_a_l = 1/2mv^2 + 1/2Iw^2[/math]
The Moment of Inertia ([math]I[/math]) for a [b]hollow[/b] sphere is equal to [math]2/3MR^2[/math]
Given a mass of 4 kg and a radius of 0.25 m, [math]I[/math] = 0.167
Now to solve for [math]w[/math]:
[math]v = rw[/math]
[math]w = 23/0.25 = 92[/math]
Plugging back into the equation results in:
[math]KE_t_o_t_a_l = 1/2(4)(23)^2 + 1/2(0.167)(92)^2[/math]
The answer to part A is 1763 J.
The law of conversation of energy states that the total energy of a system remains constant. There are 3 different energies being used in this scenario: Translational Kinetic Energy, Rotational Kinetic Energy, and Gravitational Potential Energy. This gives us the equation:
[math]1/2mv_1^2 + 1/2Iw_1^2+mgh_1 = 1/2mv_2^2 + 1/2Iw_2^2+mgh_2[/math]
The initial potential energy is 0, and the final kinetic energies are also 0. This simplifies the equation down to:
[math]1/2mv_1^2 + 1/2Iw_1^2 = mgh_2[/math]
From the previous part, we already have the total initial KE. Substituting into our equation gives us:
[math]1763 = (4)(9.8)h_2[/math]
With basic algebra, we find that the answer to part B is 44.98 m.
Also, I want to add a little bit to your explanation.
The cool thing about these types of problems is that you can always cancel stuff out and/or combine like terms.
[math]KE_{total}=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/math]
Looking at the second term:
[math]\frac{1}{2}I\omega^2=\frac{1}{2}(kmR^2)v^2/R^2=\frac{1}{2}kmv^2[/math]
...where k is the coefficient in front of the respective moment of inertia equation.
This gives:
[math]KE_{total} = \frac{1}{2}mv^2(1+k)[/math]
From this simple expression, you can do a multitude of different operations, including the problem from part B. :D
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."
University of Texas at Austin '23
Seven Lakes High School '19
You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.
a) What is the hollow sphere's total kinetic energy before rolling up the ramp?
b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.
There are 2 different kinetic energies to consider when solving for the total kinetic energy: Translational KE and Rotational KE.
To find the initial total KE, simply add the 2 energies.
[math]KE_t_o_t_a_l = 1/2mv^2 + 1/2Iw^2[/math]
The Moment of Inertia ([math]I[/math]) for a [b]hollow[/b] sphere is equal to [math]2/3MR^2[/math]
Given a mass of 4 kg and a radius of 0.25 m, [math]I[/math] = 0.167
Now to solve for [math]w[/math]:
[math]v = rw[/math]
[math]w = 23/0.25 = 92[/math]
Plugging back into the equation results in:
[math]KE_t_o_t_a_l = 1/2(4)(23)^2 + 1/2(0.167)(92)^2[/math]
The answer to part A is 1763 J.
The law of conversation of energy states that the total energy of a system remains constant. There are 3 different energies being used in this scenario: Translational Kinetic Energy, Rotational Kinetic Energy, and Gravitational Potential Energy. This gives us the equation:
[math]1/2mv_1^2 + 1/2Iw_1^2+mgh_1 = 1/2mv_2^2 + 1/2Iw_2^2+mgh_2[/math]
The initial potential energy is 0, and the final kinetic energies are also 0. This simplifies the equation down to:
[math]1/2mv_1^2 + 1/2Iw_1^2 = mgh_2[/math]
From the previous part, we already have the total initial KE. Substituting into our equation gives us:
[math]1763 = (4)(9.8)h_2[/math]
With basic algebra, we find that the answer to part B is 44.98 m.
Hey, great job! Now its your turn.
Thanks! Next question.
Given that the Earth's distance from the Sun is on average 1.5 x 10^11 m, determine the mass of the Sun using Kepler's laws.
There are 2 different kinetic energies to consider when solving for the total kinetic energy: Translational KE and Rotational KE.
To find the initial total KE, simply add the 2 energies.
[math]KE_t_o_t_a_l = 1/2mv^2 + 1/2Iw^2[/math]
The Moment of Inertia ([math]I[/math]) for a [b]hollow[/b] sphere is equal to [math]2/3MR^2[/math]
Given a mass of 4 kg and a radius of 0.25 m, [math]I[/math] = 0.167
Now to solve for [math]w[/math]:
[math]v = rw[/math]
[math]w = 23/0.25 = 92[/math]
Plugging back into the equation results in:
[math]KE_t_o_t_a_l = 1/2(4)(23)^2 + 1/2(0.167)(92)^2[/math]
The answer to part A is 1763 J.
The law of conversation of energy states that the total energy of a system remains constant. There are 3 different energies being used in this scenario: Translational Kinetic Energy, Rotational Kinetic Energy, and Gravitational Potential Energy. This gives us the equation:
[math]1/2mv_1^2 + 1/2Iw_1^2+mgh_1 = 1/2mv_2^2 + 1/2Iw_2^2+mgh_2[/math]
The initial potential energy is 0, and the final kinetic energies are also 0. This simplifies the equation down to:
[math]1/2mv_1^2 + 1/2Iw_1^2 = mgh_2[/math]
From the previous part, we already have the total initial KE. Substituting into our equation gives us:
[math]1763 = (4)(9.8)h_2[/math]
With basic algebra, we find that the answer to part B is 44.98 m.
Hey, great job! Now its your turn.
Thanks! Next question.
Given that the Earth's distance from the Sun is on average 1.5 x 10^11 m, determine the mass of the Sun using Kepler's laws.
Convenient units give 1 solar mass, or[math]2 \times 10^{30} \mathrm{kg}[/math].
[math]\frac{365 \times 24 \times 60^{2}}{(1.5 \times 10^{11})^{3}} = \frac{4\pi^{2}}{6.67 \times 10^{-11} \times M_{Sun}} \Rightarrow M_{Sun} = 2.009 \times 10^{30} \mathrm{kg}[/math]
Note that it's a slight overestimate, since the actual mean separation between the Earth and the Sun is a bit less than 1.5e11 meters (it's about 1.496e11 meters)
Stanford University
University of Texas at Austin '22
Seven Lakes High School '18
Beckendorff Junior High '14
Justin72835 wrote:
Hey, great job! Now its your turn.
Thanks! Next question.
Given that the Earth's distance from the Sun is on average 1.5 x 10^11 m, determine the mass of the Sun using Kepler's laws.
Convenient units give 1 solar mass, or[math]2 \times 10^{30} \mathrm{kg}[/math].
[math]\frac{365 \times 24 \times 60^{2}}{(1.5 \times 10^{11})^{3}} = \frac{4\pi^{2}}{6.67 \times 10^{-11} \times M_{Sun}} \Rightarrow M_{Sun} = 2.009 \times 10^{30} \mathrm{kg}[/math]
Note that it's a slight overestimate, since the actual mean separation between the Earth and the Sun is a bit less than 1.5e11 meters (it's about 1.496e11 meters)
Nice! You're definitely right on the overestimation, since I only gave 2 digits for the mass. Hit us with the next one.
Given that the Earth's distance from the Sun is on average 1.5 x 10^11 m, determine the mass of the Sun using Kepler's laws.
Convenient units give 1 solar mass, or[math]2 \times 10^{30} \mathrm{kg}[/math].
[math]\frac{365 \times 24 \times 60^{2}}{(1.5 \times 10^{11})^{3}} = \frac{4\pi^{2}}{6.67 \times 10^{-11} \times M_{Sun}} \Rightarrow M_{Sun} = 2.009 \times 10^{30} \mathrm{kg}[/math]
Note that it's a slight overestimate, since the actual mean separation between the Earth and the Sun is a bit less than 1.5e11 meters (it's about 1.496e11 meters)
Nice! You're definitely right on the overestimation, since I only gave 2 digits for the mass. Hit us with the next one.
Consider a bead on a sphere (with radius r) at the very top. A small nudge is given to the bead and it slides down the surface of the sphere.
(a) At what angle from the center of the sphere does the bead fly off?
(b) At what speed will the bead fly off?
(c) Does it take longer for the bead to slide from the top to the point where it flies off on a sphere or a straight line?
Stanford University
University of Texas at Austin '22
Seven Lakes High School '18
Beckendorff Junior High '14
Convenient units give 1 solar mass, or[math]2 \times 10^{30} \mathrm{kg}[/math].
[math]\frac{365 \times 24 \times 60^{2}}{(1.5 \times 10^{11})^{3}} = \frac{4\pi^{2}}{6.67 \times 10^{-11} \times M_{Sun}} \Rightarrow M_{Sun} = 2.009 \times 10^{30} \mathrm{kg}[/math]
Note that it's a slight overestimate, since the actual mean separation between the Earth and the Sun is a bit less than 1.5e11 meters (it's about 1.496e11 meters)
Nice! You're definitely right on the overestimation, since I only gave 2 digits for the mass. Hit us with the next one.
Consider a bead on a sphere (with radius r) at the very top. A small nudge is given to the bead and it slides down the surface of the sphere.
(a) At what angle from the center of the sphere does the bead fly off?
(b) At what speed will the bead fly off?
(c) Does it take longer for the bead to slide from the top to the point where it flies off on a sphere or a straight line?
Really nice question.
The bead falls off the sphere when the normal force between the two objects is 0 N. Taking into account mechanical energy and centripetal force, we get:
[math]\frac{1}{2}mv^2=mgR-mgRcos\theta=mgR(1-cos\theta)[/math]
[math]\frac{v^2}{R}=gcos\theta[/math]
Note that gcos(theta) is just the component of gravity supplying the centripetal acceleration. Now we can do some substitution:
[math]\frac{1}{2}mgcos\theta=mgR(1-cos\theta)[/math]
After cancelling mg and rearranging, we get:
[math]cos\theta=2/3[/math] or [math]cos\theta=2/3[/math] =====> or 48.2 degrees
From part A:
[math]\frac{1}{2}mv^2 = mgR(1-cos\theta)[/math]
Rearranging and cancelling:
[math]v=\sqrt{2gR(1-cos\theta)}[/math]
It takes longer to go along the sphere. The distance traveled along the sphere is greater while the final velocity stays the same. Therefore, a greater amount of time is spent accelerating.
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."
University of Texas at Austin '23
Seven Lakes High School '19
