To solve this problem, we need to determine the time required for the reel to reach an angular velocity of \(5 \, \text{rad/s}\) when subjected to a force \(F = 400 \, \text{N}\). The reel has a mass of \(50 \, \text{kg}\), a radius of gyration \(k_G = 2 \, \text{m}\), and the coefficient of kinetic friction between the reel and the ground is \(\mu_k = 0.2\). ### Step-by-Step Solution: 1. **Determine the Moment of Inertia:** The moment of inertia \(I\) of the reel about its center of mass \(G\) is given by: \[ I = m k_G^2 \] where \(m = 50 \, \text{kg}\) and \(k_G = 2 \, \text{m}\). \[ I = 50 \times 2^2 = 50 \times 4 = 200 \, \text{kg} \cdot \text{m}^2 \] 2. **Calculate the Torque:** The torque \(\tau\) applied by the force \(F\) is: \[ \tau = F \times r \] where \(r = 3 \, \text{m}\) (the radius at which the force is applied). \[ \tau = 400 \times 3 = 1200 \, \text{N} \cdot \text{m} \] 3. **Determine the Frictional Force:** The normal force \(N\) is equal to the weight of the reel: \[ N = mg = 50 \times 9.8 = 490 \, \text{N} \] The frictional force \(f_k\) is: \[ f_k = \mu_k N = 0.2 \times 490 = 98 \, \text{N} \] 4. **Calculate the Frictional Torque:** The frictional torque \(\tau_f\) is: \[ \tau_f = f_k \times r_f \] where \(r_f = 2 \, \text{m}\) (the radius at which the friction acts). \[ \tau_f = 98 \times 2 = 196 \, \text{N} \cdot \text{m} \] 5. **Net Torque:** The net torque \(\tau_{net}\) is: \[ \tau_{net} = \tau - \tau_f = 1200 - 196 = 1004 \, \text{N} \cdot \text{m} \] 6. **Angular Acceleration:** The angular acceleration \(\alpha\) is given by: \[ \alpha = \frac{\tau_{net}}{I} = \frac{1004}{200} = 5.02 \, \text{rad/s}^2 \] 7. **Time to Reach Angular Velocity:** Using the kinematic equation for angular motion: \[ \omega = \omega_0 + \alpha t \] where \(\omega = 5 \, \text{rad/s}\), \(\omega_0 = 0 \, \text{rad/s}\) (initial angular velocity), and \(\alpha = 5.02 \, \text{rad/s}^2\). Solving for \(t\): \[ 5 = 0 + 5.02 t \] \[ t = \frac{5}{5.02} \approx 0.996 \, \text{s} \] So, the time required for the reel to obtain an angular velocity of \(5 \, \text{rad/s}\) is approximately \(0.996 \, \text{s}\).

To solve this problem, we need to determine the angular velocity of the rod at the instant \(\theta = 60^\circ\). We will use the principles of work and energy to solve this. ### Given: - Length of the rod, \( L = 2 \, \text{m} \) - Mass of the rod, \( m = 30 \, \text{kg} \) - Force applied, \( F = 40 \, \text{N} \) - Initial angle, \( \theta = 0^\circ \) - Final angle, \( \theta = 60^\circ \) - Neglect friction and the masses of blocks A and B. ### Steps to solve: 1. **Determine the work done by the force \( F \):** The work done by the force \( F \) as the rod moves from \(\theta = 0^\circ\) to \(\theta = 60^\circ\) can be calculated by considering the horizontal displacement of point B. When \(\theta = 0^\circ\), point B is at the origin. When \(\theta = 60^\circ\), the horizontal displacement \( x_B \) of point B is: \[ x_B = L \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{m} \] The work done by the force \( F \) is: \[ W = F \cdot x_B = 40 \, \text{N} \times \sqrt{3} \, \text{m} = 40\sqrt{3} \, \text{J} \] 2. **Determine the change in potential energy:** The center of mass of the rod moves vertically as the rod rotates. Initially, when \(\theta = 0^\circ\), the center of mass is at a height of \( L/2 \) from the bottom. When \(\theta = 60^\circ\), the vertical height \( h \) of the center of mass is: \[ h = \frac{L}{2} \cos(60^\circ) = \frac{2}{2} \times \frac{1}{2} = \frac{1}{2} \, \text{m} \] The change in potential energy \( \Delta PE \) is: \[ \Delta PE = m g \Delta h = 30 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \left( \frac{1}{2} - 1 \right) \, \text{m} = 30 \times 9.81 \times \left( -\frac{1}{2} \right) = -147.15 \, \text{J} \] 3. **Apply the work-energy principle:** The work done by the force \( F \) is equal to the change in kinetic energy plus the change in potential energy: \[ W = \Delta KE + \Delta PE \] Since the rod starts from rest, the initial kinetic energy is zero. The final kinetic energy is: \[ \Delta KE = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia of the rod about the center of mass, and \( \omega \) is the angular velocity. For a rod rotating about its center, the moment of inertia \( I \) is: \[ I = \frac{1}{12} m L^2 = \frac{1}{12} \times 30 \, \text{kg} \times (2 \, \text{m})^2 = 10 \, \text{kg} \cdot \text{m}^2 \] Therefore: \[ 40\sqrt{3} = \frac{1}{2} \times 10 \times \omega^2 - 147.15 \] Solving for \(\omega\): \[ 40\sqrt{3} + 147.15 = 5 \omega^2 \] \[ 69.28 + 147.15 = 5 \omega^2 \] \[ 216.43 = 5 \omega^2 \] \[ \omega^2 = \frac{216.43}{5} = 43.286 \] \[ \omega = \

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

Hence calculate the coefficient of friction for this situation.

A 5 kg block attached to a string of length 5 m is pulled along an arc of a circle byapplying a constant force of 200 N as shown in the figure. Find the work done by theapplied force. [Use sin 530 = 45

A fisherman reels in his fishing line onto a spool with a diameter of 0.100 meters. If he rotates the spool 3.5 times, how much fishing line does he reel in, in meters?