91Ó°ÊÓ

To find the tension, it is necessary to apply Newton's Second Law in the rotational form. This gives: \[ \tau = I \alpha \], where \( \tau \) is the torque, \( I \) is the moment of inertia of the cylinder, and \( \alpha \) is the angular acceleration. The torque created is due to the tension in the string, so we replace \( \tau \) with \( T \cdot R \) (T being the tension we want to find and R the radius). The moment of inertia for a cylinder is \( \frac{1}{2} MR^2 \) and the relationship between angular and linear acceleration is \( \alpha = \frac{a}{R} \). Replacing all these results in: \( TR = \frac{1}{2} MR^2 \cdot \frac{a}{R} \), which simplifies to \( T = \frac{1}{2} M a \). To find 'a', we need to consider the forces in the vertical direction: \( Mg - T = Ma \), as the tension is pulling upwards and the weight, \( Mg \), downwards. By substitifying \( T \) from our previous calculation, we find: \( Mg - \frac{1}{2} M a = Ma \) which leads to \( a = \frac{2}{3}g \) and \( T = \frac{1}{2} M a = \frac{1}{3} M g \). Therefore, the tension in the string is one-third the weight of the cylinder.

The magnitude of the acceleration of the center of gravity has already been found while finding the tension. It is \( a = \frac{2}{3} g \). This is the acceleration the cylinder would have if we dropped it straight down, which is less than simply dropping it because in this case the force is not only defeating gravity, but also giving the cylinder rotational kinetic energy.

We can find the speed of the center of gravity after the cylinder has descended through distance 'h' using the kinematic equation \( v^2 = u^2 + 2as \), where 'v' is the final velocity we want to find, 'u' is the initial velocity (which in this case is 0 as the cylinder is released from rest), 'a' is the acceleration, and 's' is the displacement (or distance descended). By direct substituition, we get \( v^2 = 0 + 2 \cdot (\frac{2}{3} g) \cdot h \) which simplifies to \( v^2 = \frac{4}{3} \cdot 2gh = \frac{8}{3} gh \). Therefore, \( v = \sqrt{\frac{8}{3} gh} = \sqrt{\frac{4}{3}} \sqrt{2gh} = (\frac{4}{3} g h )^{1 / 2} \). Therefore, the speed of the center of gravity is \((4gh / 3)^{1/2}\) after the cylinder has descended through distance 'h'.

The energy approach involves equating the initial and final energies. Initially, the cylinder has potential energy (\( Mgh \)) and no kinetic energy (as it isn't moving). Finally, it has no potential energy (as it is on the ground) and kinetic energy split into two parts - translational (\( \frac{1}{2} M v^2 \)) and rotational (\( \frac{1}{2} I \omega^2 \), where \( \omega = \frac{v}{R} \) is the angular speed). Setting these equal gives: \( Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \) which simplifies to: \( Mgh = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2 \). Solving for 'v' gives: \( v = \sqrt{\frac{4}{3}} \sqrt{2gh} = (\frac{4}{3} g h )^{1 / 2} \). As you can see, this matches the result from the previous part, confirming the solution.