PHYS 5204

  1. 3—I think that choice 2 could also be right, but it is described a little confusing for me. As you can see in the pic, the velocity vector V is tangent to a point on the circumference which is also parallel to the plane of rotation. If choice 2 had the word “tangent” in the description I would feel more confident, otherwise I’m not sure it describes a point on the circumference which actually draws out a curved path, not a straight path that the velocity vector does
  2. 3—action and reaction forces act on both bodies
  3. 2—steeper is faster
  4. 1—speed is based on height: Potential Energy is converted to Kinetic Energy, it’s just that the steeper plane converts it faster
  5. 2—Higher torque results in higher angular acceleration and velocity if turning the same load
  6. 2—Right Hand Rule indicates the Angular velocity vector is to the left and rotation is counter clockwise
  1. 1—Right Hand Rule indicates the Angular acceleration vector is slowing the disk down
  1. 5—Only force acting on the system is the gravitational force pulling on m2 and that force is m2g
  2. 3—That same force is being applied to the entire mass of the system (m1+m2)
  3. 7—simple division of m2g/(m1+m2)
  4. 4—Tension is force acting on m1 or m1a or m1m2g/(m1+m2)
  5. 1—Tension on block A is not working against any force (like friction or gravity) so therefore no work is being done
  6. 5—KE1=m1v2/2 and v2=2ad so KE1=m1ad= m1m2gd/(m1+m2)
  7. 4—All of the Potential Energy in Block B is converted to Kinetic Energy in Blocks A and B
  8. 1—Rotational inertia of a solid cylinder (since the mass is evenly distributed) is I=mr2/2, with m=250kg and r=d/2=.30/2=.15mI=250(.15)2/2=2.81kg-m2
  9. 3—Torque=Rotational Inertia x Angular accelerationτ=Iα so,
    α=τ/I=120 N-m/2.81 Kg-m2=42.7 rad/s2
  10. 2—Angular acceleration is the change in Angular velocity over the change in timeα=Δω/Δt and Δω=ωfinal-ωinitial Since ωinitial=0 rad/s, we can easily solve for Δt=Δω/α=
    3000rpm(1min/60s)(2π rad/1 rev)/42.7 rad/s2=7.35 sec
  11. 3—Power(in kW) from a motor=Torque(N-m)(2π)(# rpm)/60,000=(120)(2π)(3000)/60,000=
    37.68 kW=37680 W and 1 hp=746 W so the motor generates 37680/746=50.5 hp
  12. 4—All of the rotational Kinetic Energy of the motor will be converted to Heat. KErot=Iω2/2
    =(2.81 kg-m2)((3000 rpm)(1min/60s)(2π rad/1 rev))2/2=138,527 Joules
  13. 2—The Force of friction is related to the weight of the armature on the bearings. Recall W=mg=(250kg)(9.8m/s2)=2450N. This corresponds to a friction force of Ffric=Wη (where η is the coefficient of friction=0.008) so Ffric=(2450N)(0.008)=19.6N. Recall that τ=Fr where the Force (F) is applied at a distance (r) from the center of the shaft to create Torque (τ). Since we already know the Friction force = 19.6N, what is the equivalent torque due to friction on the bearings? The torque is τ=Fr=(19.6N)(.016m)=.314N-m
  1. 4—The torque of friction creates a small angular deceleration α where τ=Iα from Prob 16 above. Solving for α α=τ/I=.314N-m/2.81 kg-m2=-0.112rad/s2. We can use this to calculate the time for the angular velocity to decrease from 3000 rpm to a complete stop with the relationship from Prob 17: α=Δω/Δt and Δω=ωfinal-ωinitial Since ωinitial=3000 rpm and ωfinal=0 rpm, we can easily solve for Δt Δt=Δω/α=(0-(3000rpm)(1min/60s)(2π rad/1 rev))/(-0.112rad/s2)=2810 sec
  1. 4—output torque=(input torque)(gear reduction ratio)(120N-m)(3/1)=360N-m