# Calculate the angular velocity for each case

• leafy
In summary: A more useful conservation law would be momentum conservation. Then you would not need to consider the impulse. Conservation of momentum is more useful for this problem.
leafy
Homework Statement
Consider a wheel of large moment of inertia and radius. A mass is first allowed to descend with the wheel for a short distance as shown. It is then compared to the second case where it is free fall dropped for the same distance and make inelastic impact. The energy is the same for both cases.

Moment of inertia I = 10,000
Radius of the wheel r = 1000
Mass m = 2
Drop height h = 3
Relevant Equations
Mgh = .5mv^2
Mgh = .5(I)(w^2)
Ft = mv
Ftr = Iw
First case, descends with the wheel:
mgh = .5(I)(w^2) ———- GPE converted to wheel energy
w = .1095. ———- rotation result is .1095

Second case, allow to free fall and impulse:
mgh = .5(m)(v^2). ———- GPE converted to kinetic energy
v = 7.746 ———- result in speed of 7.746
Ft = mv ———— impulse when hit the wheel
Ftr = I(w) ———— impulse converted to angular momentum
w = 1.5491. ———— rotation result is 1.5491

is this correct?

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leafy said:
Homework Statement:: Consider a wheel of large moment of inertia and radius. A mass is first allowed to descend with the wheel for a short distance as shown. It is then compared to the second case where it is free fall dropped for the same distance and make inelastic impact. The energy is the same for both cases.
Have you quoted the problem statement exactly as given to you? As stated, it is not clear. For example, it says the energy is the same in both cases. But it's not clear what particular energy is the same.

leafy said:
Moment of inertia I = 10,000
Radius of the wheel r = 1000
Mass m = 2
Drop height h = 3
Always include the units when giving numerical values of physical quantities.

leafy said:
First case, descends with the wheel:
mgh = .5(I)(w^2) ———- GPE converted to wheel energy
w = .1095. ———- rotation result is .1095
The final KE is not just the KE of the wheel. There is also the final KE of the mass m. The GPE is converted into the final total KE of the wheel and mass m.

Include units.

leafy said:
Second case, allow to free fall and impulse:
mgh = .5(m)(v^2). ———- GPE converted to kinetic energy
v = 7.746 ———- result in speed of 7.746
Ft = mv ———— impulse when hit the wheel
The impulse experienced by the mass m equals the change in momentum of m. The change in momentum is not equal to mv because m does not come to rest when it sticks to the wheel. Just after the collision, m will be moving with the wheel. If you take this into account, then your approach should yield the correct answer. However, a nice way to approach this problem is to use a conservation law. Then you do not need to consider the impulse.

Last edited:
The gravitational potential energy is mgh is the same for both cases

The units are standard metric

Since the wheel is made with so much intertia, final KE of the mass can be ignored or 0.

Again, the intertia of the wheel so big, the change of the mass momentum could consider be mv.

Of course if you approach this problem with energy conservation, I believe the energy of the wheel is halves for the impusle case. But the impulse approach is many times greater energy that overwhelm the error margin.

Actually I think you're right, because the radius is also large that we can't ignored the final KE of the mass. Thanks

leafy said:
The gravitational potential energy is mgh is the same for both cases
OK. But did you quote the problem statement exactly as given to you?

leafy said:
The units are standard metric
So, the radius of the wheel is 1000 meters. That's big

It is a good idea to include units in your calculations. Checking the consistency of units during a calculation is a good way to catch careless errors.

leafy said:
Actually I think you're right, because the radius is also large that we can't ignored the final KE of the mass. Thanks
Right. The rotational inertia of the particle, mr2, is much larger than the moment of inertia of the wheel. For the second part, I find that the particle m loses very little speed when it collides with the wheel.

Last edited:
leafy said:
Of course if you approach this problem with energy conservation, ...
For part 2, I was thinking of a different conservation law than energy. Conservation of energy is not very useful for this part, since there is some internal energy ("heat") created in the collision.

## 1. What is angular velocity?

Angular velocity is a measure of the rate at which an object rotates or revolves around a fixed point. It is usually represented by the symbol ω (omega) and is measured in radians per second (rad/s).

## 2. How is angular velocity calculated?

Angular velocity is calculated by dividing the change in angular displacement by the change in time. It can also be calculated by dividing the linear velocity by the radius of the circular path.

## 3. What are the units of angular velocity?

The units of angular velocity are radians per second (rad/s). However, it can also be expressed in revolutions per minute (rpm) or degrees per second (deg/s).

## 4. How does angular velocity relate to linear velocity?

Angular velocity and linear velocity are related through the radius of the circular path. As the radius increases, the linear velocity also increases, while the angular velocity remains constant.

## 5. Why is angular velocity important?

Angular velocity is important in understanding the motion of objects that rotate or revolve, such as planets, satellites, and wheels. It is also used in many engineering applications, such as designing gears and calculating the speed of rotating machinery.

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