Some Mechanics Problems (Could someone verify my work?)

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Could someone look over the problems I have before and tell me if my reasoning / answers look alright? I'd really appreciate it.

Which of the following statements are true? (Mark ALL that apply.)

i. If an object receives an impulse its kinetic energy must change.
ii. An object’s kinetic energy can change without receiving any impulse.
iii. An object can receive a net impulse without any work being done on it.
iv. A force may do work on an object without delivering any impulse.
v. None of the statements above is true.
For this problem I am going to say that that only iii is true. To prove that the others are wrong (or to give an example), I think of a block on a table. If a force pushes down perpendicular to the table, the block won't move (no work done), but there is an impulse because there was a force exerted for a certain amount of time.

A uniform bar of unknown mass has length 40 cm. When a mass of 1 kg is hung from one end of the bar, the system will balance at a point 15 cm from the end. The mass of the bar is:

i. 4 kg ii. 1/3 kg iii. 2 kg iv. 1/2 kg v. 3 kg
I set this up as follows:

[tex]\frac{5m}{8}=\frac{3m}{8}+1\implies m=4\text{kg}[/tex]

So the answer would be i.

Consider the following situation. A father and a daughter are both on roller blades in the middle of a large parking lot. They are initially skating together with a speed of 0.5 m/s. They push off of one another in such a way that the father remains stationary afterwards. The daughter weighs exactly half as much as her father. (You may assume that the daughter proceeds in the same direction after being pushed.) Which of the following statements about energy in this situation are true? (Mark ALL that apply.)

i. The final kinetic energy is twice the initial kinetic energy.
ii. Because there is no friction, kinetic energy is conserved.
iii. The final kinetic energy is less that the initial kinetic energy.
iv. The daughter’s final kinetic energy is 9 times her initial kinetic energy.
v. The father transfers his kinetic energy to the daughter.
Using conservation of momentum, I come up with the daughter's final KE being 9 times the original. So answer iv is correct. The first choice is false (it is three times the initial), and the second and third choices are obviously false as well. The last seems false because only the momentum is transferred. I am going to say only choice iv is correct.

A bicycle wheel rolls with speed v without slipping on a horizontal surface. Which of the following statements are true? (Mark ALL that apply.)

i. No two points on the rim of the wheel have the same speed.
ii. The angular velocity about the center is equal to the angular velocity about the contact point.
iii. No two points on the rim of the wheel have the same velocity.
iv. The velocity of the contact point is zero.
v. None of the statements above is true.
Choice 2 seems true, because the angular velocity does not change with the radius. Three is also true by looking at the parametric equations of a cycloid (also, the direction). 4 is true because it is momentarily at rest at the bottom (as seen in a cycloid). I am going to say answer choices ii and iv are true.

Thank you very much for your help :smile:.
 

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  • #2
Galileo
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apmcavoy said:
I'd really appreciate it.
For this problem I am going to say that that only iii is true. To prove that the others are wrong (or to give an example), I think of a block on a table. If a force pushes down perpendicular to the table, the block won't move (no work done), but there is an impulse because there was a force exerted for a certain amount of time.
Are you sure there is a force acting on the block? If it did it should accelerate according to Newton's 2nd law, but it doesn't. Ofcourse, what you should consider is the NET force. The net force is zero because of the normal force. Recheck the definition of impulse.

I set this up as follows:
[tex]\frac{5m}{8}=\frac{3m}{8}+1\implies m=4\text{kg}[/tex]
So the answer would be i.
I can't see where you got that equation from.
If the rod is balanced, the net torque about the pivot point is zero. Equate the torque from the center of mass and the 1kg mass. I got 3 kg.

Using conservation of momentum, I come up with the daughter's final KE being 9 times the original. So answer iv is correct. The first choice is false (it is three times the initial), and the second and third choices are obviously false as well. The last seems false because only the momentum is transferred. I am going to say only choice iv is correct.
Totally correct. :smile:
The reason kinetic energy is not conserved is because this is not a closed system in a sense. When they push against each other, work is done! This work is precisely equal to the gain in kinetic energy of the system.
 
  • #3
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I can't see where you got that equation from.
If the rod is balanced, the net torque about the pivot point is zero. Equate the torque from the center of mass and the 1kg mass. I got 3 kg.
What do you mean by that last sentence? I know that torque is [itex]\tau=\|\vec{r}\times\vec{F}\|[/itex], so in this case it is [itex]\tau=rF[/itex]. What do you mean by equate the torque from the center of mass and the 1kg mass?

Are you sure there is a force acting on the block? If it did it should accelerate according to Newton's 2nd law, but it doesn't. Ofcourse, what you should consider is the NET force. The net force is zero because of the normal force. Recheck the definition of impulse.
Right, the net force exerted over a period of time. In this case then, the answer would be i only, right?

Thanks for your help :smile:
 
  • #4
Galileo
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apmcavoy said:
What do you mean by that last sentence? I know that torque is [itex]\tau=\|\vec{r}\times\vec{F}\|[/itex], so in this case it is [itex]\tau=rF[/itex]. What do you mean by equate the torque from the center of mass and the 1kg mass?
The total torque about the pivot point is a sum of two contributions. One is ofcourse from the gravitational force acting on the mass at a distance of 15 cm. The other is the torque from gravity acting in the center of mass of the bar (5 cm from the pivot point), this torque has the opposite direction of the other one. The system balances, so the total torque should be zero.
[tex]\N = \vec r_1 \times \vec F_1+\vec r_2 \times \vec F_2 =0[/tex]
which, in this case, becomes simply (watch the signs):
[tex]gm_1r_1-gm_2r_2=0 \iff m_1r_1=m_2r_2[/tex]

Right, the net force exerted over a period of time. In this case then, the answer would be i only, right?
Thanks for your help :smile:
What about iii and iv? Any thoughts about those?
The impulse equals the change in momentum and the work done equals the change in kinetic energy. Remember that momentum is a vector, while energy is a number. Can you think of a way to change one without the other? (i.e. impuls without work or vice versa?) this may also alter your view on answer i
 
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  • #5
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OK that makes sense about the torque; I see how you came up with 3:

[tex]5m=15\implies m=3\text{kg}[/tex]

Now, about the impulse / work question:

I know that momentum is a vector and KE is a scalar, but it still seems that iii and iv are false. The impulse is the net force exerted over a period of time and the work is the net force exerted over a distance, right? If a force is applied for a certain time, the object must move right? Because of this, if an impulse is delivered, work is being done. Now if work is done, the force must be exerted for a period of time, thus an impulse is being delivered as well. If I changed the time a force was exerted, it would also change the distance. If I changed the distance, it would change the time (assuming the force doesn't change). Where am I going wrong here?

Thanks a lot for the help.
 
  • #6
Galileo
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apmcavoy said:
I know that momentum is a vector and KE is a scalar, but it still seems that iii and iv are false. The impulse is the net force exerted over a period of time and the work is the net force exerted over a distance, right? If a force is applied for a certain time, the object must move right? Because of this, if an impulse is delivered, work is being done.
Let's look at this a little more closely. If a force is delivered and the object is moving, is work always being done? Let's consider a displacement over a small distance [itex]\vec d[/itex]. The work done when the object moves this small distance is:
[tex]W=\vec F \cdot \vec d[/tex]
Is this always nonzero when F and d are nonzero?
Now can you think of a motion that allows momentum to change while no work is being done?
 
  • #7
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Galileo said:
Let's look at this a little more closely. If a force is delivered and the object is moving, is work always being done? Let's consider a displacement over a small distance [itex]\vec d[/itex]. The work done when the object moves this small distance is:
[tex]W=\vec F \cdot \vec d[/tex]
Is this always nonzero when F and d are nonzero?
Now can you think of a motion that allows momentum to change while no work is being done?
[itex]W=\vec{F}\cdot\vec{d}[/itex] is zero when F and d are perpendicular. So are you saying that if a force is applied perpendicular to the direction of motion that there is an impulse? I can't seem to see how that works because 1. the object would either move it the direction of the force or 2. a normal force would cancel out the force. Sorry for being difficult here. I appreciate your help.

Thanks again.
 
  • #8
Galileo
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apmcavoy said:
[itex]W=\vec{F}\cdot\vec{d}[/itex] is zero when F and d are perpendicular. So are you saying that if a force is applied perpendicular to the direction of motion that there is an impulse? I can't seem to see how that works because 1. the object would either move it the direction of the force or 2. a normal force would cancel out the force. Sorry for being difficult here. I appreciate your help.
Thanks again.
What if the force is always perpendicular to the direction of the motion?
 
  • #9
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Galileo said:
What if the force is always perpendicular to the direction of the motion?
Would the centripetal force of an object in circular motion fall into this category?
 
  • #10
Galileo
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Sure does.
 
  • #11
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OK that clears things up. Going back to the original problem, this means that answer i is false. This leaves me with iii and iv as true answers. Is this correct?

Thank you.
 
  • #12
Galileo
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I agree about answer iii. In circular motion, no work is done on the object (F is perpendicular to v everywhere), so no change in energy, however, since the direction of motion changes continuously its momentum does too.
However, this does not show that work can be done on an object without changing its momentum. So iv is still open.

Hint: Can [itex]T=\frac{1}{2}mv^2=\frac{p^2}{2m}[/itex] change while [itex]\vec p=m\vec v[/itex] stays constant?
 
  • #13
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Galileo said:
I agree about answer iii. In circular motion, no work is done on the object (F is perpendicular to v everywhere), so no change in energy, however, since the direction of motion changes continuously its momentum does too.
However, this does not show that work can be done on an object without changing its momentum. So iv is still open.
Hint: Can [itex]T=\frac{1}{2}mv^2=\frac{p^2}{2m}[/itex] change while [itex]\vec p=m\vec v[/itex] stays constant?
No, it cannot (because the mass is constant). So right now, after looking at 4 again, it seems that 1 must be true as well as 3. 2 and 4 I'm not totally sure about, but I'm leaning towards "false" looking at the equation you posted.
 
  • #14
Galileo
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Doh! Why do you change your mind on statement i?:grumpy:
You just said last post that you refuted it, because in circular motion an object it receives impulse but no kinetic energy. And that's correct.

I agree that kinetic energy cannot change without changing the momentum.
But you can change momentum while keeping the kinetic energy constant. This is because momentum has magnitude AND direction. Circulair motion is exactly when you keep the magintude of p constant whilst changing the direction. That is why I reminded you of the fact that p is a vector and KE is a scalar.
 
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