- #1

LCSphysicist

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- Homework Statement
- A stick is pivoted at the origin and is arranged to swing around in a

horizontal plane at constant angular speed ω. A bead of mass m slides

frictionlessly along the stick. Let r be the radial position of the bead.

Find the conserved quantity E given in Eq. (6.52). Explain why this

quantity is not the energy of the bead.

- Relevant Equations
- All below...

A stick is pivoted at the origin and is arranged to swing around in a

horizontal plane at constant angular speed ω. A bead of mass m slides

frictionlessly along the stick. Let r be the radial position of the bead.

Find the conserved quantity E given in Eq. (6.52). Explain why this

quantity is not the energy of the bead.

I came across with such question and i am a little confuse with the answer.

$$L = \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$

$$E = \frac{m(r'^{2} + (r\theta ')^{2})}{1}- \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$

$$E = \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$

But the answer should be... $$E = \frac{m(r'^{2} - (r\theta ')^{2})}{2}$$

horizontal plane at constant angular speed ω. A bead of mass m slides

frictionlessly along the stick. Let r be the radial position of the bead.

Find the conserved quantity E given in Eq. (6.52). Explain why this

quantity is not the energy of the bead.

I came across with such question and i am a little confuse with the answer.

$$L = \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$

$$E = \frac{m(r'^{2} + (r\theta ')^{2})}{1}- \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$

$$E = \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$

But the answer should be... $$E = \frac{m(r'^{2} - (r\theta ')^{2})}{2}$$