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aksik [14]
3 years ago
11

A carousel has a radius of 1.70 m and a moment of inertia of 110 kg · m2. A girl of mass 44.0 kg is standing at the edge of the

carousel, which is rotating with an angular speed of 3.40 rad/s. Now the girl walks toward the center of the carousel and stops at a certain distance from the center d. The angular speed of the carousel is now 5.4 rad/s. How far from the center, in meters, did the girl stop?
Physics
1 answer:
bonufazy [111]3 years ago
8 0

Answer:

Explanation:

Initial moment of inertia of the carousel + girl

I₁ = 110 + 44 x 1.7²

= 110 + 127.16

= 237.16 kgm².

final moment of inertia of carousel + girl

I₂ = 110 + 44 x d²

applying law of conservation of angular momentum

I₁ ω₁ = I₂ω₂

ω₁ and ω₂ are angular velocities of the carousel before and after .

237.16 x 3.4 = (110 + 44 x d²)x 5.4

806.34 = 594 + 237.6 d²

237.6 d² = 212.34

d²= .8936

d = .9453 m

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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

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