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zlopas [31]
3 years ago
5

Two disks of identical mass but different radii (r and 2r) are spinning on frictionless bearings at the same angular speed ?0, b

ut in opposite directions. The two disks are brought slowly together. The resulting frictional force between the surfaces eventually brings them to a common angular velocity.a) what is there final angular speed?b) what is the change in the rotational kinetic energy?
Physics
1 answer:
cestrela7 [59]3 years ago
8 0

Answer:

Explanation:

Moment of inertia of a disc = 1/2 M R²

Since mass is same for both and radius are r and 2r, their moment of inertia can be in the ratio of 1: 4 . Let them be I and 4I . Angular speed are ω₀ and   - ω₀ .

We shall apply law of conservation of angular momentum .

initial total angular momentum

I x ω₀ - 4I x ω₀ = - 3Iω₀

Let final common angular momentum be ω

total final angular momentum = ( I + 4I ) ω

Applying law of conservation of angular momentum

( I + 4I ) ω =  - 3Iω₀

ω = - 3 / 5 ω₀ .

b )

Initial total rotational K E

= 1/2 I ω₀² + 1/2 4I ω₀²

= 1/2 x5I ω₀²

Final total rotational K E

= 1/2 ( I + 4I ) ( - 3 / 5 ω₀ )²

= 1/2 x 9 / 5 I ω₀²

= 9 / 10I ω₀²

change in rotational kinetic energy = 9 / 10I ω₀² - 1/2 x5I ω₀²

(9/10 - 5/2) xI ω₀²

=( .9 - 2.5 )I ω₀²

= - 1.6 I ω₀² Ans

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Answer:

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Explanation:

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And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

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With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

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