Consider the diagram below. Calculate the value of F2. Calculate the value ol th
ID: 2013556 • Letter: C
Question
Consider the diagram below. Calculate the value of F2. Calculate the value ol the distance X. Consider the diagram below. Calculate tho tension. T, in the wire. Calculate the magnitude and direction ol the vertical force the wall exerts on the hinge. A solid sphere is attached to the shaft of an electric motor. The sphere's mass is 3.00 kg and its radius is 0.150 meters. The electric motor increases the sphere's rotation from 400 rovolution/second to 20.0 revolutions/second in 800 seconds. Calculate the sphere's angular acceleration. Calculate the torque applied to the sphere. Calculate the disk's angular displacement during the time its rotation increased from 4.00 revolutions/second to 20.0 revolutions/second. Calculate the work done by the electric motor during the time the sphere's rotation increased from 4.00 revolutions/second to 20.0 revolutions/second.Explanation / Answer
assignment 1a) Use sum of vertical forces. We have 55.0 N and F2 up, while we have the weights of the bar and the weight itself pulling down. Assuming a stationary system, the 4 forces must add to 0. So we have 55.0N +F2 -5.00(9.8) -8.00(9.8) =0. Notice that we are gives MASSES, not weights, so we have to multiply the 5.00 and 8.00 by g (which I'm using as 9.8, if your teacher gave you a different number, use that instead and recalculate). Also note that the 55.0 and F2 are + as they go up, while the two weights go down, making them -. So F2 =-55.0 +9.8(5.00+8.00) = 72.4N. assignment 1b) For this, we use the fact that the sum of the torques =0. Now sice the 'axis point' (the point we measure the lever arms) is arbitrary, I'm going to choose the center of mass for the beam. By doing so, the beam itself produces no torque, the weight produces a torque of 5.00(9.8)(0.100) clockwise, F1 has a torque of 55.0X, also clockwise, while the torque of F2 is 72.4(0.100+0.120) counter-clockwise. Putting this all together (using - for clockwise), we have 72.4(0.220) -55.0X -5.00(9.8)(0.100) =0-->-55.0X = 4.90-72.4(0.220) =-11.028---> X=0.20050909.., which rounds to 0.201. not sure about assignment 2, will skip assignment 3a) Not sure on the greek letters, but the angular acceleration = change in angular veloctiy/time, that is (final angular velocity - starting angular velocity)/time = (20.0rev/s -4.00rev/s)/8.00s =2.00 rev/s^2 b)torque=Fr=mar=mr^2(angular acceleration)=3.00kg(0.150m)^2(2.00rev/s^2)(2pi rad/rev)=0.848Nm. c) use the angular equivalent of vf^2 = vo^2 +2ad (so change d to theta, v to the greek letter for angular velocity, and likewise for a). From part a, the acceleration is 2.00 rev/s^2, and we're told the velocity goes from 4rev/s to 20rev/s, so (20.0rev/s)^2 =(4.00rev/s)^2 +2(2.00rev/s^2)theta. 400rev^2/s^2 -16rev^2/s^2 =4rev/s^2theta. 96.0rev=theta. d) W=Fd=Fr(theta)=(torque)(theta). From b) and c), W=0.848Nm*96.0rev(2pi rad/rev). W=512J