Rotational Motion Question -

In summary, the conversation discussed estimating the energy stored in the rotational motion of a hurricane modeled as a uniform cylinder. The equation used was KE = (1/2)*I*w^2, with I=(1/2)MR^2 for a uniform cylinder. However, the given density of air (1.3 kg/m^3) was mistaken for mass, and the height (5 km) was neglected. Once these were corrected, the estimated energy stored in the hurricane's rotational motion was found to be 4E17 J.
  • #1
sinclair18
4
0
Rotational Motion Question -- please help

Homework Statement



Estimate the energy stored in the rotational motion of a hurricane. Model the hurricane as a uniform cylinder 300 km and 5 km high, made of air whose mass is 1.3 kg/m^3. Estimate the outer edge of the hurricane to move at a speed of 200 km/h.



Homework Equations



KE = (1/2)*I*w^2

v= wR

For a uniform cylinder, I=(1/2)MR^2 (but it's hollow so would that make a difference, even though you're only given one radius)

The Attempt at a Solution


So here's how I started out:

KE = (1/2)*I*w^2

v= wR

--> Therefore KE = (1/2)[(1/2)MR^2][v/R]^2

The answer is 4E17 J, but I can't seem to get that. Can someone please tell me where I'm going wrong? I've been doing this problem for so long and I'm just not getting what I keep doing wrong. I feel like it might have something to do with the mass I'm using (M=1.3 kg/m^3) or the fact that I'm neglecting the height (h=5km)? It's not supposed to be a difficult problem...please help!
 
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  • #2
sinclair18 said:

Homework Statement



Estimate the energy stored in the rotational motion of a hurricane. Model the hurricane as a uniform cylinder 300 km and 5 km high, made of air whose mass is 1.3 kg/m^3. Estimate the outer edge of the hurricane to move at a speed of 200 km/h.

Homework Equations



KE = (1/2)*I*w^2

v= wR

For a uniform cylinder, I=(1/2)MR^2 (but it's hollow so would that make a difference, even though you're only given one radius)

The Attempt at a Solution


So here's how I started out:

KE = (1/2)*I*w^2

v= wR

--> Therefore KE = (1/2)[(1/2)MR^2][v/R]^2

The answer is 4E17 J, but I can't seem to get that. Can someone please tell me where I'm going wrong? I've been doing this problem for so long and I'm just not getting what I keep doing wrong. I feel like it might have something to do with the mass I'm using (M=1.3 kg/m^3) or the fact that I'm neglecting the height (h=5km)? It's not supposed to be a difficult problem...please help!

What you're given is not a "mass" of air - it's the density of air. Units are mass per unit volume.

What's the volume of a cylinder? Hence, what is its mass? Work in symbols throughout (try to use LaTex, if possible), as there's less chance of error.

You definitely need the height that's given. Also, I'm assuming that 300km is the diameter, not the radius?
 
  • #3
Yeah I just figured it out actually thanks soooo much!
 
  • #4
sinclair18 said:
Yeah I just figured it out actually thanks soooo much!

No problem. Glad to help (if I did).
 
  • #5


First, it's important to note that the given dimensions for the hurricane are not sufficient to accurately estimate the energy stored in its rotational motion. In reality, the hurricane is a complex system with varying wind speeds and air densities throughout its structure. Therefore, the estimate provided may not be entirely accurate.

That being said, let's work through the problem with the given information. You are correct in using the formula KE = (1/2)*I*w^2, where I is the moment of inertia and w is the angular velocity.

To find the moment of inertia for a cylinder, we use the formula I = (1/2)MR^2. Since we are given the mass per unit volume, we can calculate the mass of the cylinder by multiplying the volume by the density. The volume of a cylinder is given by V = πr^2h, where r is the radius and h is the height. However, since we are only given one radius, we will assume that the cylinder is solid and use the formula V = πr^2h to calculate the volume. Therefore, the mass of the cylinder is:

M = (1.3 kg/m^3)(π)(300 km)^2(5 km) = 6.54 x 10^14 kg

Now we can plug in the values for I and w into the formula for kinetic energy:

KE = (1/2)*(1/2)(6.54 x 10^14 kg)(300 km)^2*(200 km/h)^2

= 4.91 x 10^22 J

This is a very large amount of energy, but keep in mind that this is just an estimate and the actual value may be different due to the simplifications we made in our calculations. Also, the height of the hurricane may have some impact on the energy stored in its rotational motion, but it is likely to be negligible compared to the other factors.

In conclusion, the key to solving this problem is to use the correct formula for moment of inertia and to be careful with units. I hope this helps!
 

1. What is rotational motion?

Rotational motion is the movement of an object around a fixed point or axis. This type of motion can be seen in objects such as wheels, planets, and spinning tops.

2. What is the difference between rotational motion and linear motion?

The main difference between rotational motion and linear motion is the type of path the object takes. In rotational motion, the object moves in a circular or curved path around a fixed point, while in linear motion, the object moves in a straight line.

3. How is rotational motion measured?

Rotational motion is measured using units such as radians, revolutions, and degrees. These units are used to measure the angle of rotation of the object around the axis or fixed point.

4. What factors affect the rotational motion of an object?

The rotational motion of an object is affected by factors such as its mass, shape, and the force applied to it. The distribution of mass in an object also plays a role in determining its rotational motion.

5. How is rotational motion related to angular velocity and acceleration?

Angular velocity and acceleration are measures of how fast an object is rotating and how quickly its rotational speed changes, respectively. The greater the angular velocity or acceleration, the greater the rotational motion of the object.

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