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Consider the following situation:

A sphere with radius $r$ rolls down a plank which forms an angle $\theta$ with the horizontal axis. The starting hight of the centre of mass is $h$. We assume that the sphere rolls without sliding.

Clearly, the initial energy is $E=mgh$.

Now we calculate the energy at the lowest point (where the potential energy is $0$). We consider the movement of the sphere like a rotation around its contact point with the ground. The second axiom of rotation tells us: $$ \sum \overset{\to}\tau=\overset{\to}\alpha\cdot I $$ We can see that $\left|\sum \overset{\to}\tau\right|=rmg\sin\theta$ and because the sphere rolls without sliding, we have $a=-\alpha r$. Thus we get $|a|=\frac{mr^2}{I}g\sin\theta$. The linear velocity at the lowest point is thus given by: $$ \frac{h}{\sin\theta}=\frac{a}{2}t_{end}^2\implies t_{end}=\sqrt{\frac{2h}{a\sin\theta}}\implies\\ v_{end}=at_{end}=a\sqrt{\frac{2h}{a\sin\theta}}=\sqrt{\frac{2ha}{\sin\theta}} $$ The energy at the lowest point is the sum of the linear kinetic energy and rotational energy. Again because there is no sliding, we have $v_{end}=-\omega_{end} r$ and thus: $$ E=\frac{mv_{end}^2}{2}+\frac{I\omega_{end}^2}{2}=\frac{mv_{end}^2}{2}\left(1+\frac{I}{mr^2}\right)=\frac{2mha}{2\sin\theta}\left(1+\frac{I}{mr^2}\right)=\frac{m^2r^2gh}{I}\left(1+\frac{I}{mr^2}\right)=\\ mgh+\frac{m^2r^2gh}{I} $$ So it is greater than before. But I don't see why the energy shouldn't be conserved, so where did I go wrong?

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You are making mistake in assuming the moment of inertia of a (solid) sphere to be same about a tangent to its surface (rotation around its contact point with ground) which is lets say $I_1=\frac{7}{5}mr^2$and moment of inertia about its axis (diameter) which is lets say $I_2=\frac{2}{5}mr^2$.

With these values your final equation becomes $$E_{end}=\frac{m^2r^2gh}{I_1}(1+\frac{I_2}{mr^2})=(\frac{5}{7}+\frac{2}{7})mgh=mgh$$

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