A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is :

\frac{F}{2 m R}

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\frac{F}{3 m R}

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\frac{2 F}{3 m R}

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\frac{3 F}{2 m R}

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Solution

The correct option is D $\frac{2 F}{3 m R}$
$\begin{matrix} By Translation force equation - \begin{matrix} F - f = m a F - f = m r α - (1 : 0 = r α) \end{matrix} \end{matrix}$
$\begin{matrix} By Rotational motion equation - f t r = \frac{m r^{2}}{2} \cdot α (sin u I = \frac{m r^{2}}{2}) - (2) \end{matrix}$
$\begin{matrix} From eq 1 and (2) we get α = \frac{2 F}{3 m r} Hence option (C) is correct. \end{matrix}$

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A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is :

The correct option is D 2F3mR By Translation force equation - F−f=maF−f=mrα−(1:0=rα) By Rotational motion equation - ftr=mr22⋅α(sinuI=mr22)−(2) From eq 1 and (2) we get α=2F3mr Hence option ( C ) is correct.