Questions: In the figure, a constant horizontal force ( vecFa p p ) of magnitude 24.6 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 16.1 kg, its radius is 0.254 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder? (a) Number Units (b) Number Units (c) Number ( hati+ ) Units

Solution Steps

• Given:
  • Force applied, \( F_{\text{app}} = 24.6 \, \text{N} \)
  • Mass of the cylinder, \( m = 16.1 \, \text{kg} \)
  • Radius of the cylinder, \( R = 0.254 \, \text{m} \)
• Required:
  • (a) Acceleration of the center of mass, \( a_{\text{cm}} \)
  • (b) Angular acceleration, \( \alpha \)
  • (c) Frictional force, \( f \)

• For a solid cylinder, the moment of inertia about its center is: \[ I = \frac{1}{2} m R^2 \] Substituting the given values: \[ I = \frac{1}{2} \times 16.1 \, \text{kg} \times (0.254 \, \text{m})^2 = 0.520 \, \text{kg} \cdot \text{m}^2 \]

• The net force acting on the cylinder is: \[ F_{\text{net}} = F_{\text{app}} - f \] According to Newton's second law: \[ F_{\text{net}} = m a_{\text{cm}} \] Therefore: \[ a_{\text{cm}} = \frac{F_{\text{app}} - f}{m} \]

• The torque due to the applied force is: \[ \tau = F_{\text{app}} R \] According to Newton's second law for rotation: \[ \tau = I \alpha \] Therefore: \[ \alpha = \frac{F_{\text{app}} R}{I} \] Substituting the values: \[ \alpha = \frac{24.6 \, \text{N} \times 0.254 \, \text{m}}{0.520 \, \text{kg} \cdot \text{m}^2} = 12.02 \, \text{rad/s}^2 \]

• For rolling without slipping: \[ a_{\text{cm}} = \alpha R \] Substituting the value of \( \alpha \): \[ a_{\text{cm}} = 12.02 \, \text{rad/s}^2 \times 0.254 \, \text{m} = 3.05 \, \text{m/s}^2 \]

• Using the relation \( a_{\text{cm}} = \frac{F_{\text{app}} - f}{m} \): \[ f = F_{\text{app}} - m a_{\text{cm}} \] Substituting the values: \[ f = 24.6 \, \text{N} - 16.1 \, \text{kg} \times 3.05 \, \text{m/s}^2 = 24.6 \, \text{N} - 49.1 \, \text{N} = -24.5 \, \text{N} \]

Final Answer