1. (40 POINTS) A cylinder and pulley have radii Rc = 0.150 m and Rp = 0.120 m, r

Question

1. (40 POINTS) A cylinder and pulley have radii Rc = 0.150 m and Rp = 0.120 m, respectively. Consider these two rotating objects to be uniform solid cylinders (i.e. disks) with masses Mc = 5.00 kg and Mp = 2.00 kg, respectively. The cylinder and pulley turn without friction about their stationary axles that pass through their centers. A light rope is wrapped around the cylinder. One side of the rope passes over the pulley and has a 4.00 kg uniform rectangular block suspended from its free right end. The other side has a 2.00 kg uniform rectangular -shaped load suspended from the rope?s left free end. The box and load have the same vertical thickness. The box is released from rest when its bottom surface is an initial height H = 6.00 m above the ground. The box descends and the load rises (ascends) as the rope turns with the cylinder and pulley. There is no slippage between rope and the cylinder and pulley surfaces. (a) (30 points) Find the common speed v of the box and load when the box has fallen 3.00 m. (b) (10 points) What is the ratio ?p/?c of the pulley?s angular velocity to the cylinder?s angular velocity while the box is descending? 2. Extra Credit (15 POINTS. No penalty if not done.) Referring to the previous problem, (a) (6 points) what is the magnitude a of the common acceleration of the box and load? (b) (9 points) what are the tension magnitudes (i) Tb in the section of string between the pulley and the box. (ii) Tp in the section of the string between the pulley and cylinder (iii)Tc in the section of the string between the cylinder and the load?

Explanation / Answer

as the box comes down the load goes up ; MB - box mass;ML = load mass

find out decreases in PE = (MB-ML)*x    (as box moves down by x load moves up by x)

this is equal to increase in KE of the system as energy is conserved for no friction

velocity of box and load are going to be the same V as they are connected by a taut rope.

also angular velocity = V/respective radius assuming no slipping over the rotational objects

hence KE = 0.5(I1*1^2 +I2*22 + (MB+ML)V^2)

1 = V/R1 ; 2 = V/R2

differentiate the energy equation to get force equation and then apply it to individual members as F=ma or = I

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