Ratio of angular speed with conservation of energy

In summary, the ball has potential energy when it is at h/3 from the bottom of the incline. The ball converts this potential energy into kinetic energy when it is at height 2h/3. The ball's ratio of kinetic energy at 2h/3 to kinetic energy at h/3 is 4.
  • #1
RoboNerd
410
11

Homework Statement


  1. A ball rolls down an incline plane without slipping. What is the ratio of its angular velocity at h/3 to its angular velocity at 2h/3?
    1) 1:2
  2. 2) 1:sqrt(2)
  3. 3) 1:1
  4. 4) sqrt(2):1
  5. 5) 2:1

Homework Equations


Conservation of energy with provisions for rotational and translational motion.

The Attempt at a Solution


Hi everyone!

I got this funky problem today. Here's what I tried to do:

at h/3 I have potential energy resulting from lowering the ball a height of 2*h/3 that is converted to kinetic energy.

Thus, I did the following
m * g * (2h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * I * omega^2]
m * g * (2h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * (2 / 5) * m * r^2 * (v^2 / r^2)]

Rearrange to get an expression with v, and then substitute v = r * omega to get
(20 * h * g)/ (21 * r^2) = omega^2 for the h/3 height

OK. then I said that at height 2h/3 I have potential energy from descending h/3 converted to kinetic energy
m * g * (h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * I * omega^2]
m * g * (h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * (2 / 5) * m * r^2 * (v^2 / r^2)]

Rearrange to get an expression with v and then substitute v = r * omega to get
(10 * g * h) / (21 * r^2 ) = omega squared.

Thus, my ratio would have been for the two omega squared terms to be:
(20 * h * g)/ (21 * r^2) = omega^2 for the h/3 height : (10 * g * h) / (21 * r^2 ) = omega squared. for the 2h/3 height

Simplifying the ratio for the omega squared gives 20 : 10 -----> 2:1.
Taking the square root to compare just for the omegas gives sqrt(2) : 1.

Thus, I got 4. The answer according to the writers is 2.

Why is this the case?

Thanks a lot for the help in advance!
 
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  • #2
It's not entirely clear what "at h/3" means. Is that h/3 from the bottom or from the top? Is there any further information that clarifies this?
 
  • #3
They did not provide any information. I believe that 'h' corresponds to the height at which the ball is initially launched on the ramp.
 
  • #4
RoboNerd said:
They did not provide any information. I believe that 'h' corresponds to the height at which the ball is initially launched on the ramp.
Sure, but that still doesn't nail down the meaning of "at h/3". It could still be measured from top or from bottom. Measured from bottom, your answer is correct.
 
  • #5
I have no idea... these authors write really crummy questions, which does not help as I have a final on monday. :-(
 
  • #6
RoboNerd said:
I have no idea... these authors write really crummy questions, which does not help as I have a final on monday. :-(
So you quoted the question exactly as given to you, word for word, and there is no diagram?
 
  • #7
Yes, sir!
 
  • #8
Yes, sir! I just assumed they had a ramp at an arbitrary angle, with it vertically ascending "h".
 
  • #9
RoboNerd said:
Yes, sir! I just assumed they had a ramp at an arbitrary angle, with it vertically ascending "h".
Well, either they were thinking of h/3 etc. as descent, i.e. measured from the top, or they messed up and gave the wrong answer.
 
  • Like
Likes RoboNerd
  • #10
OK, thanks so much for helping and confirming what I did. Really appreciated!
 

1. What is the relationship between angular speed and conservation of energy?

The ratio of angular speed to conservation of energy is known as the moment of inertia. It represents how difficult it is to change the rotational motion of an object. As the moment of inertia increases, the angular speed decreases, and vice versa. This is due to the conservation of energy principle, which states that energy cannot be created or destroyed, only transferred or transformed.

2. How does changing the moment of inertia affect the angular speed?

If the moment of inertia is increased, the angular speed will decrease, and if the moment of inertia is decreased, the angular speed will increase. This is because the conservation of energy requires that any change in energy must be accompanied by an equal and opposite change in another form of energy. In this case, the change in kinetic energy (angular speed) is counteracted by a change in potential energy (moment of inertia).

3. Can the angular speed and moment of inertia be calculated?

Yes, the angular speed can be calculated using the formula: angular speed = linear speed / radius. The moment of inertia can be calculated using the formula: moment of inertia = mass x radius^2. These calculations are based on the properties of the object, such as its mass, shape, and distribution of mass.

4. How is the conservation of energy principle applied in rotational motion?

In rotational motion, the conservation of energy principle is applied by considering the total energy of the system, which includes both kinetic energy (from the angular speed) and potential energy (from the moment of inertia). As the object rotates, the potential energy is converted into kinetic energy, and vice versa, while the total energy remains constant.

5. Are there any real-world applications of the ratio of angular speed with conservation of energy?

Yes, the concept of moment of inertia and its relationship with angular speed is crucial in many real-world applications, such as designing spinning objects like tops and gyroscopes, analyzing the performance of vehicles with rotating parts (e.g., wheels, propellers), and understanding the behavior of celestial bodies like planets and stars.

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