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  • A rod with linear mass density given, as -         <img alt="\lambda=\frac{\lambda_0 x^2}{l^2}" src="https://entrancecorner.oncodecogs.com/gif.late

A rod with linear mass density given, as -

        \lambda=\frac{\lambda_0 x^2}{l^2}

where, x is the distance from its left most end and ' l ' is the length of the rod. If the rod is placed on a smooth horizontal plane and a force of constant magnitude F is applied on its rightmost end then the angular acceleration of the rod will be.

Option: 1

\alpha=\frac{F}{\lambda_0 l^2}


Option: 2

\alpha=\frac{3F}{\lambda_0 l^2}


Option: 3

\alpha=\frac{5F}{\lambda_0 l^2}


Option: 4

\alpha=\frac{7F}{\lambda_0 l^2}


Answers (1)

best_answer

dI=(\lambda dx)x^2

\begin{aligned} &\begin{aligned} \int d I & =\int_0^l \frac{\lambda_0 x^4}{l^2} d x \\ I & =\frac{\lambda_0}{l^2} \times \frac{l^5}{5}=\frac{\lambda_0 l^3}{5} \end{aligned}\\ &\begin{aligned} \tau & =I \alpha \\ l F & =\frac{\lambda_0 l^3}{5} \alpha \\ \alpha & =\frac{5 F}{\lambda_0 l^2} \end{aligned} \end{aligned}

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shivangi.bhatnagar

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