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A solid ball of mass M = 2.0 kg and radius R = 0.25 m starts from rest at a height h = 3.0 m above the bottom of the path. It rolls without slipping down the left side of the path. The right side of the path is frictionless. Moment of inertia of a hollow sphere is � = ! ! ��!. v What is the linear speed

Answer :

lublana

Answer:

6.48 m/s

Explanation:

We are given that

Mass,M=2 kg

Radius,R=0.25 m

Height,h=3 m

Moment of inertia of solid sphere=[tex]I=\frac{2}{5}MR^2[/tex]

We have to find the linear speed.

[tex]\omega=\frac{v}{R}[/tex]

By law of conservation of energy

[tex]mgh=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2[/tex]

[tex]mgh=\frac{1}{2}I(\frac{v}{R})^2+\frac{1}{2}mv^2=\frac{1}{2}v^2(\frac{I}{R^2}+m)[/tex]

Where [tex]g=9.8m/s^2[/tex]

Substitute the values

[tex]2\times 9.8\times 3=\frac{1}{2}(\frac{2}{5R^2}MR^2+M)=\frac{7}{10}Mv^2=\frac{7}{10}(2)v^2[/tex]

[tex]v^2=\frac{2\times 9.8\times 3\times 10}{7\times 2}[/tex]

[tex]v=\sqrt{\frac{2\times 9.8\times 3\times 5}{7}}[/tex]

[tex]v=6.48m/s[/tex]

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