Rotational Kinetic Energy of a rod

In summary, the conversation discusses a wooden block on a frictionless surface being hit by a bullet and the resulting energy conversions. The final kinetic energy equation is corrected to Kf = (1/2)(m + M)v^2 and the question is posed as to how to find the fraction of original kinetic energy converted into internal energy.
  • #1
klopez
22
0
1.A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

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What fraction of the original kinetic energy is converted into internal energy in the collision?



Here is my attempt:

Ki = (1/2)mv^2

kf = (m+M) L^2

Now I'm a bit confused what they mean by fraction, so I initial thought I had to divide Kf/Ki , which is:

2(m + M)L^2 / (mv^2)

but WebAssign marked it wrong. I have one more try and I want to try the other way, Ki/Kf but I don't want to get it wrong.

Can anyone help me and tell me if I'm doing this right or not. Thanks

Kevin
 
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  • #2
Your final kinetic energy equation is wrong. First off, it doesn't have the right units. It looks like you just used I. The final energy should be 1/2 I w^2, where the equation v = Lw holds.
 
  • #3
Oh okay I see how I got the final kinetic energy wrong. I did only use I. I fixed it, and now its:

Kf = (1/2)(m + M)v^2

Can anyone confirm this?

And if that's correct, how do I find the fraction?
 

FAQ: Rotational Kinetic Energy of a rod

1. What is rotational kinetic energy?

Rotational kinetic energy is a form of energy that an object possesses due to its rotation around an axis. It is dependent on the mass of the object, its angular velocity, and the moment of inertia.

2. How is rotational kinetic energy different from linear kinetic energy?

Rotational kinetic energy is different from linear kinetic energy because it is associated with an object's rotational motion, while linear kinetic energy is associated with an object's linear motion. This means that rotational kinetic energy is dependent on an object's angular velocity, while linear kinetic energy is dependent on an object's linear velocity.

3. How is rotational kinetic energy calculated?

Rotational kinetic energy is calculated using the formula K = 1/2 Iω^2, where K is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.

4. What factors affect the rotational kinetic energy of a rod?

The rotational kinetic energy of a rod is affected by its mass, moment of inertia, and angular velocity. Additionally, the shape and distribution of mass along the rod can also affect its rotational kinetic energy.

5. How is rotational kinetic energy used in real-world applications?

Rotational kinetic energy is used in many real-world applications, such as in the design and operation of machines and vehicles that involve rotational motion. It is also used in sports, such as in the spinning of a discus or the rotation of a figure skater. Additionally, rotational kinetic energy is important in understanding the behavior and properties of celestial bodies, such as planets and stars.

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