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a cylinder of m=0.53 kg and radiuS 0.32 m can rotate around an axis through its...

a cylinder of m=0.53 kg and radiuS 0.32 m can rotate around an axis through its center. the cylinder moment of inertial is given by MR^2 /2. the cylinder is initially at rest. A force of 0.70 N is then applied tangentially at the edge of the cylinder. how many radians will the cylinder rotate through during the first 2.5 seconds that the force is applied
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\!\!\!\!\!\!\!\!\!$Moment of inertia of the cylinder, $I = \frac{MR^2}{2} = \frac{0.53\times (0.32)^2}{2}=0.027 $kgm$^2\\\\ $Torque on the cylinder due to the tangential force, $\\\\\tau = RF = 0.32\times 0.7 = 0.224$Nm$\\\\ $Torque is related to moment of inertia as, $\tau = I\alpha\\\\ $Angular acceleration of the cylinder, $\alpha = \frac{\tau}{I} =\frac{0.224}{0.027}=8.3$rad/s$^2\\\\ $Initial angular speed of the cylinder, $\omega_i = 0\\\\ $Angular displacement of the cylinder in the first 2.5s, $\\\\ \Delta \theta = \omega_i t + \frac{1}{2}\alpha t^2= \frac{1}{2}\times 8.3\times (2.5)^2=25.94$radians

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