~~~~~NEED HELP ASAP~~~~~ A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a ra

Question

~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)

a.) Determine the linear regression

b.) At this given angular velocity, what is the rotational kinetic energy?

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Thái Dương 7 months 2021-07-15T01:41:22+00:00 1 Answers 0 views 0

Answers ( )

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    2021-07-15T01:42:49+00:00

    Answer:

    Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

    Explanation:

    a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

    a= ω^2*r

    ω= 300rev/min

    convert into rev/s

    300/60= 5rev/s

    a= 18.75m/s^2

    b) use Krot= 1/2 Iω^2

    plug in gives

    1/2(30*2.25)(25)= 843.75 J

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