(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the maximum theta given that

h (initial height of block)

m (mass of block)

M (mass of rod)

L (length of rod)

2. Relevant equations

3. The attempt at a solution

I did it three different ways and none of them worked. What is wrong with these ideas?

Method 1: Potential Energy

The gravitational potential energy of the block and rod's center of mass must be equal before and after since initially and finally, the system has no kinetic energy

[tex]U_{blocki} + U_{rodi} = U_{blockf} + U_{rodf}[/tex]

[tex]mgh + Mg\frac{L}{2} = mg(L-L\cos\theta) + Mg(L-\frac{L}{2}\cos\theta)[/tex]

[tex]mh + \frac{ML}{2} = mL(1-\cos\theta) + \frac{ML}{2}(1+1-\cos\theta)[/tex]

[tex]mh + \frac{ML}{2} = mL(1-\cos\theta) + \frac{ML}{2} + \frac{ML}{2}(1-\cos\theta)[/tex]

[tex]mh = (mL + \frac{ML}{2})(1-\cos\theta)[/tex]

[tex]\cos\theta = 1 - \frac{mh}{mL + \frac{ML}{2}}[/tex]

Method 2: Torque

The velocity of the block at the bottom of the ramp is

[tex]v = \sqrt{2gh}[/tex]

The angular momentum of the system is the angular momentum of the block just before it hits (since the rod is at rest)

[tex]l = I\omega = (mL^2)(\frac{v}{L})[/tex]

[tex]l = mL\sqrt{2gh}[/tex]

Gravity provides torque which changes the angular momentum from that quantity down to zero, so

I found the torque of the block and the rod separately and added them together

[tex]\tau_{block} = r \times F = L(mg)\sin\theta[/tex]

[tex]\tau_{rod} = r \times F = \frac{L}{2}(Mg)\sin\theta[/tex]

[tex]mL\sqrt{2gh} = \int_0^{\theta_{max}} \tau d\theta[/tex]

[tex]mL\sqrt{2gh} = \int_0^{\theta_{max}} (Lmg+\frac{LMg}{2})\sin\theta d\theta[/tex]

[tex]m\sqrt{2gh} = (mg + \frac{Mg}{2})(1-\cos\theta_{max})[/tex]

[tex]\cos\theta_{max} = 1 - \frac{m\sqrt{2gh}}{mg + \frac{Mg}{2}}[/tex]

Why aren't they the same?

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# Homework Help: Block and Pendulum

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