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)
Question
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)
Solution
To solve this problem, we need to use the principles of rotational motion and Newton's second law.
Step 1: Identify the forces acting on the cylinder. The horizontal force is 40 N and the gravitational force is mass times gravity, or 5 kg * 9.8 m/s^2 = 49 N.
Step 2: Calculate the torque. Torque (τ) is the force times the radius, or 40 N * 0.5 m = 20 Nm.
Step 3: Use Newton's second law for rotation, which states that torque is equal to the moment of inertia times the angular acceleration. The moment of inertia (I) for a hollow cylinder is given by the formula I = mr^2, or 5 kg * (0.5 m)^2 = 1.25 kgm^2.
Step 4: Solve for the angular acceleration (α). Rearranging the formula from step 3 gives α = τ/I, or 20 Nm / 1.25 kg*m^2 = 16 rad/s^2.
So, the angular acceleration of the cylinder is 16 rad/s^2.
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