91Ó°ÊÓ

First, we'll determine the acceleration of the bucket as it falls. Using Newton's second law, we know that the net force acting on the bucket is its weight minus the tension in the rope. The equation for net force is:\[ F_{ ext{net}} = m_{ ext{bucket}} imes g - T \]where \( m_{\text{bucket}} = 15.0 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity). The net force is also equal to the mass of the bucket times its acceleration (\( a \)), so:\[ m_{\text{bucket}} imes a = m_{\text{bucket}} imes g - T \]But, since the rope is wrapped around a cylinder, the tension produces a torque which accelerates the cylinder. The equation for torque \( \tau \) is:\[ \tau = I \times \alpha \]where \( I \) is the moment of inertia \( = \frac{1}{2} m_{\text{cylinder}} R^2 \), \( R \) is the radius of the cylinder, and \( \alpha \) is the angular acceleration equal to \( a/R \). Setting up equations for linear and rotational dynamics together gives us:\[ m_{\text{bucket}} g - T = m_{\text{bucket}} a \] (1)\[ T R = \frac{1}{2} m_{\text{cylinder}} R^2 \frac{a}{R} \] (2)Solving these simultaneously for \( a \):From equation (2), we get:\[ T = \frac{1}{2} m_{\text{cylinder}} a \]Substituting for \( T \) in equation (1):\[ m_{\text{bucket}} g - \frac{1}{2} m_{\text{cylinder}} a = m_{\text{bucket}} a \]\[ a = \frac{m_{\text{bucket}} g}{m_{\text{bucket}} + \frac{1}{2} m_{\text{cylinder}}} \]Plug in values: \( m_{\text{bucket}} = 15.0 \, \text{kg}, m_{\text{cylinder}} = 12.0 \, \text{kg}, g = 9.8 \, \text{m/s}^2 \)\[ a = \frac{15.0 \times 9.8}{15.0 + \frac{1}{2} \times 12.0} \approx 6.54 \, \text{m/s}^2 \]

Using the acceleration found in Step 1, substitute \( a \) back into the relation for tension T:\[ T = \frac{1}{2} m_{\text{cylinder}} a \]Substituting in known values:\[ T = \frac{1}{2} \times 12.0 \times 6.54 \approx 39.24 \, \text{N} \]

We use the kinematic equation to find the final speed \( v \) with initial speed \( u = 0 \):\[ v^2 = u^2 + 2as \]where \( s = 10.0 \, \text{m} \), \( a = 6.54 \, \text{m/s}^2 \):\[ v^2 = 2 \times 6.54 \times 10.0 \]\[ v = \sqrt{130.8} \approx 11.44 \, \text{m/s} \]

Using the equation \( s = ut + \frac{1}{2}at^2 \), solve for \( t \):\[ 10.0 = 0 \times t + \frac{1}{2} \times 6.54 \times t^2 \]\[ t^2 = \frac{20.0}{6.54} \approx 3.06 \]\[ t = \sqrt{3.06} \approx 1.75 \, \text{s} \]

The cylinder exerts force equal to the net torque it experiences. Torque is caused by the tension in the rope:\[ \tau = T \times R = \frac{1}{2} m_{\text{cylinder}} a \times R \]The force exerted by the axle in the radial direction is equal to the tension in the rope (as friction is zero and no external radial forces act on the system), therefore:\[ F_{\text{axle}} = \sqrt{(T^2 + (m_{\text{cylinder}} g)^2)} = \sqrt{(39.24)^2 + (12.0 \times 9.8)^2} \approx 167.0 \, \text{N} \]