Quiz 06 (10 pts) 1) The two rotating systems shown in the figure4) A wrench is a
ID: 1795886 • Letter: Q
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Quiz 06 (10 pts) 1) The two rotating systems shown in the figure4) A wrench is acting on a nut trying to turn it. The differ only in that the two identical movable masses length of the wrench lies directly to the east of the are positioned at different distances from the axis of nut. A force 150.0 N acts on the wrench at a position rotation. If you release the hanging blocks imulaneously from rest, and if the ropes do not 30.0° north of east. What is the magnitude of the slip, which block lands first? A) The block at the left lands first B) The block at the right lands first. / C) Both blocks land at the same time D) Not enough information 15.0 cm from the center of the nut in a direction torque about the center of the nut? A) 22.5 Nm B) 11.3 Nm C) 19.5 Nm 1) 2250 Nm : E) 1949 Nm 5) A solid, uniform, sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an 2) A tire is rolling akong a road, without slipping. nclined plane of height 7.0 m. What is the angular with a velocity v. A piece of tape is attached to the tire. When the tape is opposite the road (at the top of the tire), its velocity with respect to the road is velocity of the sphere at the bottom of the inclined plane? A) 5.8 rad/s B)9.9 rads C) 11 rad/s D) 7.0 rad/s A) 2v : B) v; C)1.5v: D) zero. E) The velocity depends on the radius of the tire. Possibly Useful Equations )A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly frorn 22.0 rads to 13 5 rads. what is the magru tude of the angular acceleration of the wheel? )0.616 rad/s; B)5.45 rad's : C) 111 rad/s 2-1+ 2a (0-80) 1, = mgdy. Krrans =mv, Krot =102 D) 22.5 rad's;E)10.9 rad's sphere MRExplanation / Answer
a) Both the block will land at the same time
b) velocity of tape upward = V
velocity with respect to road = V-0= V
c) 13.5^2= 22^2 +2 a *13.8
angular acceleration a = 10.9 rad/s^2
d) Torque = rF sin 30 = 0.15*150*sin 30 = 11.25
e) Energy = mgh = 2*7*9.81 = 137.34
I = 2/5 Mr^2 = 2.312
137.34= (2.312*angular^2)/2 + (2*angular^2*1.7^2)/2
angular = 5.8 rad/s
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