A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
Angular acceleration = 25 rad/s², Linear acceleration = 10 m/s²
Explanation
This question applies Newton's second law for rotational motion (τ = Iα) to a hollow cylinder. The key concept from the textbook is that for a thin ring/hollow cylinder rotating about its axis, the moment of inertia is I = MR². The torque is calculated as τ = F × R, and since there is no slipping, the linear acceleration equals the tangential acceleration (a = Rα). This follows the same methodology as Example 6.12 in the textbook.
Solution Steps
Step 1: Given values - Mass (M) = 3 kg, Radius (R) = 40 cm = 0.4 m, Force (F) = 30 N
Step 2: Moment of inertia - For a hollow cylinder, I = MR² = 3 × (0.4)² = = 0.48 kg m²
Step 3: Torque calculation - τ = F × R = = 12 N m
Step 4: Angular acceleration - Using τ = Iα, we get α = τ/I = 12/0.48 = 25 rad/s²
Step 5: Linear acceleration - Since there is no slipping, a = Rα = = 10 m/s²