# Rotation of Rigid Bodies: Rotating stick with disc on top

• Exath
In summary, the problem involves a thin cylindrical rod with a cylindrical disc attached to one end, both with given dimensions and masses. The arrangement can fall and rotate around a frictionless axis. The task is to calculate the kinetic energy, angular velocity, and center of mass velocity when the arrangement has rotated 90 degrees and is momentarily horizontal. Using energy principles and the parallel axis theorem, the potential energy for both the rod and disc can be calculated, and the inertia for the disc can be found using the formula Icm + Md^2. With these values, the angular velocity and kinetic energy can be determined.
Exath

## Homework Statement

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A thin cylindrical rod with the length of L = 24.0 cm and the mass m = 1.20 kg has a cylindrical disc attached to the other end as shown by the figure. The cylindrical disc has the radius R = 8.00 cm and the mass M 2.00 kg. The arrangement is originally straight up vertical), but it can fall and rotate around a frictionless axis. What is the arrangement's kinetic energy when it has rotated 90 degrees and is momentarily horizontal? What is the angular velocity and what is the velocity of the discs center of mass in this location? Compare this with what the velocity of the disc would have if the disc passed the line A if you had only dropped the disc straight down (without the rod).

Rod:
m = 1.2kg
L = 0.24m

Disc:
R = 0.08m
M = 2kg

## Homework Equations

Inertia of the rod = 1/3*mL^2

Energy equation: U1 + K1 = U2 + K2

Kinetic energy of rotating rigid body: K = 1/2*mv^2 + 1/2*Iw^2 (w = angular velocity, I = Inertia)

## The Attempt at a Solution

Pretty much stuck at the first part: Calculating the kinetic energy of the system when it reaches line A.

Started with the energy principles: U1 + K1 = U2 + K2. No kinetic energy at the start, K1 = 0, No potential energy at the end when it reaches line A because it momentarily stops according to the text: U2 = 0, This gives U1 = K2 so we only need to find the potential energy of the system. The potential energy of the disc is M*g*(L + R) (counting from center of mass). Not really sure how potential energy works for a rod rotating about an axis on one of its ends though.

The kinetic energy for a rotating rigid body is K = 1/2*mv^2 + 1/2*Iw^2, this becomes K2 = 1/2*Iw^2 (remove 1/2*mv^2 because there is no movement for this rigid body, only rotation). Here, I is the inertia of the entire system, which is the inertia of the rod (1/3*mL^2) plus the discs inertia. Now this is another problem. There is no formula for inertia for a disc spinning about an axis on its "edge". Maybe we could derive one with calculus, but that would be taking it too far... At one point i considered the disc as a single point meaning we could just use the general inertia formula m*r^2 for it, but that would also be going too far because a radius was given, it's clearly not meant to be just a point.

So we have U1 = K2 => (Potential rod + potential disc) = 1/2*(Inertia rod + Inertia disc)*w^2

with this i could solve the angular velocity w and then easily find the kinetic energy K2, but i have no idea what the potential energy for the rod is and inertia for the disc...

any ideas?

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Exath said:
Started with the energy principles: U1 + K1 = U2 + K2. No kinetic energy at the start, K1 = 0, No potential energy at the end when it reaches line A because it momentarily stops according to the text: U2 = 0, This gives U1 = K2 so we only need to find the potential energy of the system. The potential energy of the disc is M*g*(L + R) (counting from center of mass).
So far, so good.

Exath said:
Not really sure how potential energy works for a rod rotating about an axis on one of its ends though.
Just like the disk. What happens to the center of mass of the rod?

Exath said:
There is no formula for inertia for a disc spinning about an axis on its "edge".
Look up the parallel axis theorem.

It also falls like the disk, but counting from the center of mass it would be m*g*(L/2)?

Exath said:
It also falls like the disk, but counting from the center of mass it would be m*g*(L/2)?
Exactly.

Ok, so what's left is the inertia for the disk, how would i go about finding that?

Exath said:
Ok, so what's left is the inertia for the disk, how would i go about finding that?
See my second post in this thread.

Ah, the inertia should be Icm + Md^2 (d = distance between new axis and center of mass) = 1/2*MR^2 + M*(R + L)^2

Exath said:
Ah, the inertia should be Icm + Md^2 (d = distance between new axis and center of mass) = 1/2*MR^2 + M*(R + L)^2
Excellent!

Exath
Doc Al said:
Excellent!
Thanks for the help i got the rest from there :D

## 1. What is the difference between rotation and revolution?

Rotation refers to the movement of an object around its own axis, while revolution refers to the movement of an object around a fixed point or axis.

## 2. How is rotational motion different from translational motion?

Rotational motion involves movement around an axis, while translational motion involves movement in a straight line. In rotational motion, all points on an object move in a circular path, while in translational motion, all points move in the same direction.

## 3. What is the moment of inertia and how does it affect rotational motion?

The moment of inertia is a measure of an object's resistance to rotational motion. It depends on the mass and distribution of mass of the object. A larger moment of inertia means the object will be more difficult to rotate.

## 4. How does the location of the axis of rotation affect the motion of a rotating stick with a disc on top?

The location of the axis of rotation affects the rotational motion of the stick and disc. If the axis of rotation is at the center of mass of the system, the stick and disc will rotate together. If the axis of rotation is not at the center of mass, the stick and disc will experience different rotational speeds and directions.

## 5. What is the conservation of angular momentum and how does it apply to the rotation of rigid bodies?

The conservation of angular momentum states that the total angular momentum of a system remains constant in the absence of external torques. This means that if a rotating rigid body starts to spin faster, another part of the system must spin slower to maintain the same total angular momentum.

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