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**1. Homework Statement**

Rotating with angular velocity ##ω_0## solid homogeneous cylinder of radius ##r## placed without starting forward speed at the bottom of the inclined plane, forming an angle ##\alpha## with the horizontal plane, and starts to roll in up. Determine the time during which the cylinder reaches the highest position on the inclined plane.

**2. Homework Equations**

##m\dfrac{dv}{dt}=F-mg\sin\alpha##

##I\dfrac{d\omega}{dt}=Fr##

**3. The Attempt at a Solution**

From integrating of the above eqns, I get:

##\dfrac{\omega r}{2}-v=g\sin\alpha \cdot t-const##

Where the constant we can determine from the initial conditions, e.g.

##const=-\dfrac{mr\omega_0}{2}##

So, I get eqn

##\dfrac{\omega r}{2}-v=g\sin\alpha \cdot t-\dfrac{\omega_0 r}{2}##

I think that at the start of pure rotation, the cylinder has to stop, ie, it reaches the maximum height. How is it possible to argue?

Then ##v=-\omega r##, and ##\omega=0##

##0=g\sin\alpha \cdot t-\dfrac{\omega_0 r}{2}##

##t=\dfrac{\omega_0 r}{2g\sin\alpha}##

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