- #1

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In b). Since the force T is perpendicular to the trajectory of the mass m, T does not perform any work on m, therefore the translational mechanical energy is conserved, from which I deduce that the initial speed is equal to the final speed, moreover, the speed is constant.

Now, when analyzing the linear momentum, due to the force on m T, it can be deduced that when evaluating the integral of this force in time, it gives different from 0, but since the speed of the mass m remains constant as well as its mass, its linear momentum must remain constant, so the previous integral would be equal to 0 ... can you see this possible contradiction?

I am right?

\begin{equation*}

\int T(t)\,\mathrm{d}t=0?

\end{equation*}

Now, when analyzing the linear momentum, due to the force on m T, it can be deduced that when evaluating the integral of this force in time, it gives different from 0, but since the speed of the mass m remains constant as well as its mass, its linear momentum must remain constant, so the previous integral would be equal to 0 ... can you see this possible contradiction?

I am right?

\begin{equation*}

\int T(t)\,\mathrm{d}t=0?

\end{equation*}