Câu hỏi 22

Gia tốc của con lắc đơn:

\(\begin{array}{*{20}{l}}
\begin{gathered}
a = \sqrt {a_t^2 + a_n^2} \hfill \\
\,\, = \sqrt {{g^2}.{{\sin }^2}\alpha + 4{g^2}.{{\left( {\cos \alpha - \cos {\alpha _0}} \right)}^2}} \hfill \\
\end{gathered} \\
\begin{gathered}
{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = g\sqrt {{{\sin }^2}\alpha + 4{{\left( {\cos \alpha - \cos 60} \right)}^2}} \hfill \\
\,\, = 10.\sqrt {{{\sin }^2}\alpha + 4\left( {{{\cos }^2}\alpha - \cos \alpha + {{0,5}^2}} \right)} \hfill \\
\end{gathered} \\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 10\sqrt {{{\sin }^2}\alpha + 4{{\cos }^2}\alpha - 4\cos \alpha + 1} } \\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 10\sqrt {{{\sin }^2}\alpha + {{\cos }^2}\alpha + 3{{\cos }^2}\alpha - 4\cos \alpha + 1} } \\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 10\sqrt {3{{\cos }^2}\alpha - 4\cos \alpha + 2} }
\end{array}\)

Ta có: \(3{\cos ^2}\alpha  - 4\cos \alpha  + 2 = {\left( {\sqrt 3 \cos \alpha  - \dfrac{2}{{\sqrt 3 }}} \right)^2} + \dfrac{2}{3} \ge \dfrac{2}{3}\)

\( \Rightarrow {a_{\min }} \Leftrightarrow {\left( {3{{\cos }^2}\alpha  - 4\cos \alpha  + 2} \right)_{\min }} = \dfrac{2}{3}\)

Dấu “=” xảy ra khi \(\sqrt 3 \cos \alpha  = \dfrac{2}{{\sqrt 3 }} \Rightarrow \cos \alpha  = \dfrac{2}{3}\)

Khi đó lực căng dây có độ lớn:

\(T = mg\left( {3.\cos \alpha  - 2.\cos {\alpha _0}} \right) = 0,1.10.\left( {3.\dfrac{2}{3} - 2.\cos 60} \right) = 1N\)

Chọn D.