As shown in Fig. E10.20, a string is wrapped several times around the rim of a small hoop with radius 0.0800 m and mass 0.180 kg. The free end of the string is pulled upward in just the right way so that the hoop does not move vertically as the string unwinds. (a) Find the tension in the string as the string unwinds. (b) Find the angular acceleration of the hoop as the string unwinds. (c) Find the upward acceleration of the hand that pulls on the free end of the string. (d) How would your answers be different if the hoop were replaced by a solid disk of the same mass and radius?

Solution 78P Problem (a) Step 1: Radius of the hoop R = 0.08 m Mass of the hoop M = 0.18 Kg To find the tension in the string Step 2: Tension T = Mg -----(1) 2 g = 9.8 m/s (acceleration due to gravity) T = 0.18 x 9.8 T = 1.76 N Tension in the string is 1.76 N Problem (b) Step 1: To find the angular acceleration of the hoop Total Torque produced = I-----(2) Total torque = force x perpendicular distance, here the force is nothing but the tension produced in the string. And perpendicular distance is radius of the hoop R. = TR -----(3) Moment of inertia of the hoop I = MR ----(4) Step 2: From equations (1) to (4), we can find angular acceleration value = TR I = Mg.R MR 2 g = R -----(5) = 0.08 = 122.5 rad/m 2 The angular acceleration of the hoop as the string unwinds is 122.5 rad/m 2 Problem (c) Step 1: To find the upward acceleration Upward acceleration is tangential acceleration (a). a = R----(6) From equation (5) a = R. g R a = g 2 a = 9.8 m/s 2 Upward acceleration is 9.8 m/s Problem (d) Step 1: The change in T, and a if the hoop is replaced by a solid disk of same mass and radius. 1 2 Moment of inertia of the disk I = MR 2 Tension T won’t be changed because T is independent of moment of inertia. Step 2: Change in Angular acceleration ( ) = TR I = T.R MR2 2 = 2 TR MR 2 = 2 The value of angular acceleration is doubled