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Nadusha1986 [10]
3 years ago
7

Two disks are rotating about the same axis Disk A has a moment of inertia of 9.99 kg m2 and an angular velocity of 4.51 rad s Di

sk B is rotating with an angular velocity of 6.35 rad s The two disks are then linked together without the aid of any external torques so that they rotate as a single unit with an angular velocity of 4.24 rad s The axis of rotation for this unit is the same as that for the separate disks What is the moment of inertia of disk B
Physics
1 answer:
prisoha [69]3 years ago
3 0

Answer:

Explanation:

Conservation of angular momentum

If both disks have original angular velocity in the same direction, we would expect their final angular velocity to be

4.51 < ω < 6.35

Where 6.35 rad/s would occur if disk B had I = 0

and 4.51 rad/s would occur if disk B had I = ∞

As both disks have a final angular velocity less than their original, One disk must have changed direction of rotation during the collision. Else angular momentum is Not conserved.

If Disk A did not change direction

Disk A had an angular momentum <u>change</u> of

ΔL = I(ωf - ωi) = 9.99(4.24 - 4.51) = -2.6973 kg•m²/s

Disk B changed direction. It first had to reduce its original momentum to zero before spinning back up to its final momentum in the opposite direction. This total momentum change will be equal to that lost by Disk A

I(4.24 - (-6.35)) = 2.6973

I = 2.6973 / 10.59 = 0.254702... ≈ 0.255 kg•m²

If Disk A was the one to change direction

Its change in angular momentum was

ΔL = 9.99(4.51 + 4.24) = 87.4125 kg•m²/s

Disk B had to lose the same amount of angular momentum during the collision

87.4125 = I(6.35 - 4.24)

I = 41.42772 ≈ 41.4 kg•m²

I leave it to you to determine which is correct as we cannot tell from the information given in the question.

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A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

  • work done = force x distance x cos θ
  • since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

  • the work done by gravity = weight of carton x distance x cos θ
  • The weight of the carton = force exerted by the mass of the carton = m x g
  • the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

  • work done by the normal force = force x distance x cos θ
  • the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

  • The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

  • work done by the rope= force x distance x cos θ
  • the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

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Two tiny conducting spheres are identical and carry charges of -18.8 µC and +46.5 µC. They are separated by a distance of 2.47 c
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Answer:

F=-12896N, attractive.

Explanation:

For calculating this force we use the Coulomb Law:

F=\frac{kq_1q_1}{r^2}

Where k=9\times10^9Nm^2/C^{-2} is the Coulomb's constant, q_1 and q_2 the values of each charge and r the distance between them.

Since the Coulomb's constant as I wrote it is in S.I. we have to write all the magnitudes in that system of units, and substitute:

F=\frac{(9\times10^9Nm^2/C^{-2})(-18.8\times10^{-6}C)(46.5\times10^{-6}C)}{(0.0247m)^2}=-12896N

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