Question

dynamics

A force of P = 60 N is applied to the cable, which causes the 230-kg reel to turn since it is resting on the two rollers A and B of the dispenser, Neglect the mass of the rollers and the mass of the cable. Assume the radius of Gyration of the reel about its center axis remains constant at ko 0.6 m. (Figure 1) Part A Determine the angular velocity of the real measured clockwise, after it has made two revolutions starting from rest. Express your answer using three significant figures. Enter positive value if the angular velocity is clockwise and angular velocity is counterclockwise. 0 AED It vec ? rad/s Figure Submit 1 of 1 > RestA Provide Feedback OGOS

Answer 1

First of all calculate the work done due to load p when reel move two revolution.

Then calculate initial kinetic energy

Then calculate final kinetic energy

Then using work energy principle we can find the the angular velocity of Reel

Given data applied force = P = GON mass of reel = m = 23019 Radicis af gyration of reof. about cito= ko = 0.6m confral axis P-GON 0 im 777777TTIA 0.6m We have to find out angular velocity of reel measchod clockwise after it has made two revolution. Assumption : mass of rollars and cable neglected Reel is initially in rest condition. (W, =o)

Solution We know that for a revolution to o= ar radian i for & rovolution -o 0:2x27 = 47. radion P=60N NB W = mg= - 230X9.815 Total work by force p dile to: two revolution af reel O= 47 radian alo e = ox8= 458 0.75= 37m :: Work done by force p= Wp = PxP= 60x37 hp=1800J

Initial kinetic energy = Ti = of 2002= 0v final kinetic energy a = $x Io cu 2 To af xlm ke?)xcede? Tags $x[230x (0.632] x custom Ta = 41.4 co² (where Io = moment of Inertia about it centre = m ko? = 230 0.62- 89.8 kg-m2 from principal of work and energy Tit EU-2 = Ta 0 + 180 T = 41.402 :: Wo ? 1807 41-4 .; Wg = 3.6958 rad/s . Anguter velocity af reel = W2= 3.6958 rad /s (Ang)