Hovercraft B/C

[Adi1008]

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.

Part A (Click to Show Hidden Text)

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

Part B (Click to Show Hidden Text)

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]

Part C (Click to Show Hidden Text)

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.

That's correct! Your turn.

Some stuff I'd add for part B (Click to Show Hidden Text)

You determined that [math]cos\theta =2/3[/math] in your answer for part A. You can use this to simplify your answer to part B a bit.

[math]v=\sqrt{2gR(1-cos\theta)} = \sqrt{2gR\bigg(1-\frac{2}{3}\bigg)} = \sqrt{\frac{2}{3}gR}[/math]

[shrewdPanther46]

Sorry for interrupting the chain of Q and A, but I just wanted to thank you guys for the explanation on the last problem...

I've seen it before and I was always confused about how to solve it, now I understand lol

So thanks...

[Justin72835]

Sorry for interrupting the chain of Q and A, but I just wanted to thank you guys for the explanation on the last problem...

I've seen it before and I was always confused about how to solve it, now I understand lol

So thanks...

No problem!

That's correct! Your turn.

Alright, get ready for this one .

Normally when dealing with elastic potential energy and Hooke's Law, we assume that the springs we are dealing with are ideal and are of negligible mass. However, in the real world, springs have mass too. Consider the following scenario:

You have a spring with a spring constant of k and a mass M resting on a frictionless horizontal table and you attach one end of the spring to a nearby wall. At the other end of the spring, you attach a block of mass m. If you compress the spring by a distance of x, what is the maximum velocity v attained by the block (in terms of all the other variables)?

Things to note:
- the block and the end of the spring attached to it travel at the same velocity
- assume that the velocity at each point on the spring varies linearly with its length
- basically, try to sum up a bunch of tiny dKE's

[Adi1008]

Sorry for interrupting the chain of Q and A, but I just wanted to thank you guys for the explanation on the last problem...

I've seen it before and I was always confused about how to solve it, now I understand lol

So thanks...

No problem!

That's correct! Your turn.

Alright, get ready for this one .

Normally when dealing with elastic potential energy and Hooke's Law, we assume that the springs we are dealing with are ideal and are of negligible mass. However, in the real world, springs have mass too. Consider the following scenario:

You have a spring with a spring constant of k and a mass M resting on a frictionless horizontal table and you attach one end of the spring to a nearby wall. At the other end of the spring, you attach a block of mass m. If you compress the spring by a distance of x, what is the maximum velocity v attained by the block (in terms of all the other variables)?

Things to note:
- the block and the end of the spring attached to it travel at the same velocity
- assume that the velocity at each point on the spring varies linearly with its length
- basically, try to sum up a bunch of tiny dKE's

Answer (Click to Show Hidden Text)

The spring has a mass of M but not all of it is moving at the same speed as the end, which is moving with a velocity v. As a result, the spring's kinetic energy must be less than (1/2)mv^2. Suppose that the velocity of the spring at a certain point is u. Then,
[math]E = \int_{0}^{L} \frac{M u^{2}}{2L} dy=\frac{M}{2L}\int_{0}^{L}u^{2}dy[/math]. 

The speed at each point varies linearly with length, so [math]u^{2} = \bigg(\frac{vy}{L}\bigg)^{2}[/math]. Using that,

[math]\frac{Mv^{2}}{2L^{3}}\bigg(\frac{1}{3} L^{3}\bigg) = \frac{1}{2} \frac{M}{3} v^2[/math]. That will be how much kinetic energy the spring uses up when the spring-block system is released from rest. Equating the elastic potential energy with the kinetic energy of the spring and the kinetic energy of the block and solving for v gives the answer. v is the same because the end of the spring and the block travel at the same speed.

[math]\frac{1}{2} kx^2 = \frac{1}{2} \frac{M}{3} v^2 + \frac{1}{2} mv^2 \Rightarrow v = \sqrt{\frac{kx^2}{\frac{M}{3}+m}}[/math]

[Justin72835]

Answer (Click to Show Hidden Text)

The spring has a mass of M but not all of it is moving at the same speed as the end, which is moving with a velocity v. As a result, the spring's kinetic energy must be less than (1/2)mv^2. Suppose that the velocity of the spring at a certain point is u. Then,
[math]E = \int_{0}^{L} \frac{M u^{2}}{2L} dy=\frac{M}{2L}\int_{0}^{L}u^{2}dy[/math]. 

The speed at each point varies linearly with length, so [math]u^{2} = \bigg(\frac{vy}{L}\bigg)^{2}[/math]. Using that,

[math]\frac{Mv^{2}}{2L^{3}}\bigg(\frac{1}{3} L^{3}\bigg) = \frac{1}{2} \frac{M}{3} v^2[/math]. That will be how much kinetic energy the spring uses up when the spring-block system is released from rest. Equating the elastic potential energy with the kinetic energy of the spring and the kinetic energy of the block and solving for v gives the answer. v is the same because the end of the spring and the block travel at the same speed.

[math]\frac{1}{2} kx^2 = \frac{1}{2} \frac{M}{3} v^2 + \frac{1}{2} mv^2 \Rightarrow v = \sqrt{\frac{kx^2}{\frac{M}{3}+m}}[/math]

Sorry for taking so long to get back to this... and that's correct! Your turn!

[Riptide]

Answer (Click to Show Hidden Text)

The spring has a mass of M but not all of it is moving at the same speed as the end, which is moving with a velocity v. As a result, the spring's kinetic energy must be less than (1/2)mv^2. Suppose that the velocity of the spring at a certain point is u. Then,
[math]E = \int_{0}^{L} \frac{M u^{2}}{2L} dy=\frac{M}{2L}\int_{0}^{L}u^{2}dy[/math]. 

The speed at each point varies linearly with length, so [math]u^{2} = \bigg(\frac{vy}{L}\bigg)^{2}[/math]. Using that,

[math]\frac{Mv^{2}}{2L^{3}}\bigg(\frac{1}{3} L^{3}\bigg) = \frac{1}{2} \frac{M}{3} v^2[/math]. That will be how much kinetic energy the spring uses up when the spring-block system is released from rest. Equating the elastic potential energy with the kinetic energy of the spring and the kinetic energy of the block and solving for v gives the answer. v is the same because the end of the spring and the block travel at the same speed.

[math]\frac{1}{2} kx^2 = \frac{1}{2} \frac{M}{3} v^2 + \frac{1}{2} mv^2 \Rightarrow v = \sqrt{\frac{kx^2}{\frac{M}{3}+m}}[/math]

Sorry for taking so long to get back to this... and that's correct! Your turn!

Aright it’s gotten a little stale so ima start it back up. Water leaves a hose lying on the ground at a speed of 8.4 m/s. If the nozzle of the hose is lifted 2 meters above the ground, what speed does the water come out of the hose?

[Justin72835]

Aright it’s gotten a little stale so ima start it back up. Water leaves a hose lying on the ground at a speed of 8.4 m/s. If the nozzle of the hose is lifted 2 meters above the ground, what speed does the water come out of the hose?

Answer (Click to Show Hidden Text)

Since you didn't give an initial pressure I'm just going to assume that the water in the hose is at atmospheric pressure. Alright so to do this problem you can apply Bernoulli's Principle:

[math]P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1=P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2[/math]

P1 and P2 are the same (atmospheric pressure) so you can cancel those out. You know that h1 is 0 meters so you can cancel out that term. Lastly, since the water's density remains constant throughout the process, you can cancel out rho. Now you can plug the rest of the values in.

[math]\frac{1}{2}(8.4)^2=\frac{1}{2}v_2^2 + g(2)[/math]

Solving for v2 give 5.6 m/s.

[Riptide]

Aright it’s gotten a little stale so ima start it back up. Water leaves a hose lying on the ground at a speed of 8.4 m/s. If the nozzle of the hose is lifted 2 meters above the ground, what speed does the water come out of the hose?

Answer (Click to Show Hidden Text)

Since you didn't give an initial pressure I'm just going to assume that the water in the hose is at atmospheric pressure. Alright so to do this problem you can apply Bernoulli's Principle:

[math]P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1=P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2[/math]

P1 and P2 are the same (atmospheric pressure) so you can cancel those out. You know that h1 is 0 meters so you can cancel out that term. Lastly, since the water's density remains constant throughout the process, you can cancel out rho. Now you can plug the rest of the values in.

[math]\frac{1}{2}(8.4)^2=\frac{1}{2}v_2^2 + g(2)[/math]

Solving for v2 give 5.6 m/s.

Wow that was quick. Nice job! Sorry for not mentioning the pressure, you made the right assumption for keeping them the same. Your turn!

[Justin72835]

Aright it’s gotten a little stale so ima start it back up. Water leaves a hose lying on the ground at a speed of 8.4 m/s. If the nozzle of the hose is lifted 2 meters above the ground, what speed does the water come out of the hose?

Answer (Click to Show Hidden Text)

Since you didn't give an initial pressure I'm just going to assume that the water in the hose is at atmospheric pressure. Alright so to do this problem you can apply Bernoulli's Principle:

[math]P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1=P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2[/math]

P1 and P2 are the same (atmospheric pressure) so you can cancel those out. You know that h1 is 0 meters so you can cancel out that term. Lastly, since the water's density remains constant throughout the process, you can cancel out rho. Now you can plug the rest of the values in.

[math]\frac{1}{2}(8.4)^2=\frac{1}{2}v_2^2 + g(2)[/math]

Solving for v2 give 5.6 m/s.

Wow that was quick. Nice job! Sorry for not mentioning the pressure, you made the right assumption for keeping them the same. Your turn!

Alright. Nathan Chen is an 18-year-old American figure skater competing at the 2018 Winter Olympics. During one of his tricks, he begins spinning around at a steady 2π rad/s. Bringing his arms in toward his chest, Chen is able to decrease his moment of inertia by half. After doing this, how many revolutions does Chen complete after 3 seconds of spinning? Assume that because the ice is frictionless and that his angular velocity does not decrease.

[Riptide]

Answer (Click to Show Hidden Text)

Since you didn't give an initial pressure I'm just going to assume that the water in the hose is at atmospheric pressure. Alright so to do this problem you can apply Bernoulli's Principle:

[math]P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1=P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2[/math]

P1 and P2 are the same (atmospheric pressure) so you can cancel those out. You know that h1 is 0 meters so you can cancel out that term. Lastly, since the water's density remains constant throughout the process, you can cancel out rho. Now you can plug the rest of the values in.

[math]\frac{1}{2}(8.4)^2=\frac{1}{2}v_2^2 + g(2)[/math]

Solving for v2 give 5.6 m/s.

Wow that was quick. Nice job! Sorry for not mentioning the pressure, you made the right assumption for keeping them the same. Your turn!

Alright. Nathan Chen is an 18-year-old American figure skater competing at the 2018 Winter Olympics. During one of his tricks, he begins spinning around at a steady 2π rad/s. Bringing his arms in toward his chest, Chen is able to decrease his moment of inertia by half. After doing this, how many revolutions does Chen complete after 3 seconds of spinning? Assume that because the ice is frictionless and that his angular velocity does not decrease.

By the conservation of angular momentum, Iw = Iw (of course assuming frictionless surfaces). When the moment of inertia is cut in half, the angular velocity doubles, resulting in a angular velocity of 4π rad/s. After 3 seconds, he has traveled 12π radians, or 6 revolutions (1 revolution = 2π yay unit circle).