11.7 FE/EIT -Forces and motion for rope/pulley blocks on horizontal plane (related Hw 9.5) Block A slides along a fricti

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11 7 Fe Eit Forces And Motion For Rope Pulley Blocks On Horizontal Plane Related Hw 9 5 Block A Slides Along A Fricti 1 (55.59 KiB) Viewed 50 times

11.7 FE/EIT -Forces and motion for rope/pulley blocks on horizontal plane (related Hw 9.5) Block A slides along a frictionless horisontal surface. It connects to a block B by a rope that p over a massive pulley P. Description Mam of block A or It Earth's gravitational acceleration Pulley Ps radius Pulley P's moment of inertia about Horisontal measure of block A's position Tension in rope on upper-left of pulley Tesoon in rope on lower-right of pulley 9 R 1 xin T₂ TH Type Constant Constant Constant Constant Variable Uskonn Unknown Fill bank with the most accurate of Baseline case a. Massive rope) b. Pulley inertia I>0 Shown right are baseline results (from Hw 9.5) for block A sliding on a frictionless surface and connected to block B by a maseless rope that passes over a maseless frictionless pulley. Using e. Rotational friction in pulley T physical intuition (guess), fill in each blank. a. Rope is massive. Pulley is massless, frictionless b. Rope is mamless Pulley is massive, frictionless c. Rope and pulley are maalesa Pulley has friction. FBD of block A FBD of block B • Draw a free-body diagram for block A, block B. and pulley P. Assume the rope is mases- less and the pulley is frictionless. Case 1: r(t) Earth 7th To • Differentiate block A's position vector from C, to form A's velocity and acceleration. Knowing the rope is taut (not slack) and inextensible (not elastic), determine block B's velocity and acceleration. Knowing the rope rolls over pulley P, determine P's angular velocity and angular acceleration a PA- 2.¹. 2. Result: Wa ê gt² . • Form equations of motion. Solve for and the difference in rope tension across the pulley. mg R2 Result: ATension TB - TA I + 2m R² Case 2: 1 = 1m R² 2-19 mgl 1 + 2m R² • Determine for each given moment of inertia I. Solve for r(t) when r(0)-2 m, f(0) = 3. Result: 2-19 t+{g²² TA 2 Ta 2 7A S FBD of pulley P z(t) 9 mg -C₂