Question

please draw a picture

Q.3 A solid disk with a mass of 36 kg and a radius of 0.7 m is spinning around an axis through its center, it rotates with an angular speed of 3 full turns per second. You drop a 12 kg mass onto the disk at the edge, it sticks to the disk, reducing the angular speed to ws. Next you apply a 30 N force tangentially at the edge of the disk to slow it down, until it stops rotating Calculate the angular speed w, after the 12 kg is dropped onto the disk. Calculate the torque due to the 30 N force. Calculate how much time it takes to stop the disk with the tangential force.

Answer 1

Solutioni" - Axis of Rotation fig: Solid Disc A solid Disc with a mass of 36 kg and Radius of 0.7 m is spinning around an a xs through its centre. It rotates with an angular speed of 3 full turns per second. 1 full turn = 24 radian. : 3 full turn = 6a Radian. W; = 60 Radian/ sec. Drop a 12 Kg mass onto Disk at the edge , it sticks to the DISK: From Conservation of Angular Momentum, I j w i = If wf MR(Wi) = MRP+ MR2 Jw4 12 x (36 Kg) (0.7m 12 x 6x = T5 x(36kg) (0.7 m)% 6x + 12Kgx (0.7m) yurt 52:92R = 172. 133 wf . * long in Wt = 0.965 rad/s.

• NOW, the Apply BON Disk to slow force tangentially it down until at it the edge of stops rotating : Torque due to the son force = fx 8 y = 30N (0.7 m) . Y = 21 Nom. Now, we know, : Torque - (Moment of Inertia ) x (Angular Accelaration) 21 = ( MRT + MR2 ) x (*) 21 = 1 * 36 * (0.75"+ 12 (07)2] a 21 = 1417 & i d = 1.428 rad / S2. .

Time it takes to stop the Disk with tangential force is wf = wf wit-at .. (1) Here, wf = final Angular 'Speed s o Wil = Initial Angular speed = 0.965 rad/s. a = Angular Accenat Accelaration - 1.428 rad/s2. . from equin in 0 = i 0.965 - (1.428)t t = 0.675 sec. Hence, (1) Angular speed Wf after 12 kg is dropped onto the Disk is 0.965 rad/s. w Torque due to the Bon Force is 21 N-m. vii) time which takes to stop the Disk with tangential disk is 0.675 sec.