A solid, uniform sphere of mass 2.0 kg and radius 1.8 m rolls from rest without slipping down an inclined plane of height 7.5 m. What is the

Question

A solid, uniform sphere of mass 2.0 kg and radius 1.8 m rolls from rest without slipping down an inclined plane of height 7.5 m. What is the angular velocity of the sphere at the bottom of the inclined plane

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Latifah 5 years 2021-08-05T21:43:52+00:00 1 Answers 26 views 0

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  1. Philomena
    0
    2021-08-05T21:45:32+00:00
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    Answer:

    w^2=5.5rads/s

    Explanation:

    From the question we are told that:

    Mass m=2.0kg

    Radius r=1.8m

    Height h=7.5m

    Generally the equation for Potential energy is mathematically given by

    Potential energy=Kinetic energy+Rotational energy

     mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2

    Since there is no slipping

     v=rw

    Therefore

     mgh=\frac{1}{2}mr^2w^2+\frac{1}{2}Iw^2

    Where

          I=\frac{1}{2}mr^2

          l=3.24m

     2*9.81*7.5=\frac{1}{2}(2)(1.8)^2w^2+\frac{1}{2}(3.24)w^2\\\\

     147.15=3.24w^2+1.62w^2

     w^2=\frac{147.15}{4.86}

     w^2=\sqrt{\frac{147.15}{4.86}}

     w^2=5.5rads/s

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