8.01SC | Fall 2016 | Undergraduate

Classical Mechanics

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A cylinder of radius \(\displaystyle R\), mass \(\displaystyle M\), and moment of inertia \(\displaystyle I_{cm}\) about the axis passing through its center of mass starts from rest and moves down an inclined at an angle \(\displaystyle \phi\) from the horizontal. The center of mass of the cylinder has dropped a vertical distance \(\displaystyle h\) when it reaches the bottom of the incline. The cylinder rolls down the incline without slipping. Use the conservation of energy principle to calculate the speed of the center of mass of the cylinder when it reaches the bottom of the incline. Express your answer in terms of \(\displaystyle m\), \(\displaystyle h\), \(\displaystyle R\), \(\displaystyle g\), and \(\displaystyle I_{cm}\).

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Fixed Axis Rotation vs. Translation and Rotation

Here we contrast the expressions of the angular momentum and kinetic energy for a rigid object rotating about a fixed axis vs. a rigid object translating and rotating. For the case of fixed axis rotation, the object is pivoted about point S, left figure. Both rods are identical, length \(d\), mass \(m\), moment of inertia about an axis passing through the center of mass \(I_{cm}\), and about an axis passing through point \(S\) is \(I_S\). The plane of rotation is contained in he plane of the screen. The position vector of the center of mass measured with respect to \(S\) is \(\vec{r}_{cm}\).

Angular Momentum

\(\vec{L}_S = I_S\vec{\omega}\)

Kinetic Energy

\(\displaystyle K = \frac{1}{2}\;I_S\omega^2\)

Angular Momentum

\(\vec{L}_S = \vec{r}_s \times m\vec{v}_{cm} + I_{cm}\vec{\omega}\)

Kinetic Energy

\(\displaystyle K = \frac{1}{2}mv_{cm}^2+\frac{1}{2}\;I_{cm}\omega^2\)

Note:

In previous questions you have shown that if an object rotates about a fixed axis passing through point \(S\) and perpendicular to the plane of rotation, the angular momentum about point \(S\) is also given by: \(\vec{L}_S = m\vec{r}_{cm} \times \vec{v}_{cm} + I_{cm} \vec{\omega} \)

This is true because the center of mass is moving in a circle of radius \(d/2\) with center at point \(S\) and with the same \(\vec{\omega}\) as the object. As a result, the velocity of the center of mass has a magnitude \(\omega d/2\) therefore \(\vec{L}_S\) becomes \(\vec{L}_S = m\frac{d^2}{4} \vec{\omega} + I_{cm} \vec{\omega}\). Using the parallel axis theorem \(\vec{L}_S = I_S\vec{\omega}\).

You also showed that the kinetic energy for the object rotating about a fixed axis passing through point \(S\) is expressed as \(\displaystyle K = \frac{1}{2}mv_{cm}^2+\frac{1}{2}\;I_{cm}\omega^2\).

For the same reason as before, this is true because the speed of the center of mass is \(\omega d/2\) and the parallel axis theorem.

To avoid confusion, we advise you to always write \(\vec{L}_S = I_S\vec{\omega}\) and \(K = \frac{1}{2}I_S\omega^2\) for the case of fixed axis rotation.

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Problem Set 12 contains the following problems:

  1. Bicycle Wheel
  2. Rolling Without Slipping
  3. Cubical Block Collision with Low Ridge
  4. Yo-yo Rolling on Inclined Plane
  5. Billiards

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