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kow [346]
1 year ago
5

Problem 3A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined pla

ne of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane
Physics
1 answer:
frosja888 [35]1 year ago
4 0

Answer:

5.1 rad/s

Explanation:

Mechanical energy of the system is conserved since no external work is done on the sphere.

mgh = mv^2/2 + I\omega^2/2

Substituting v = ωr and I = 2 m r^2/5, we get,

=> mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2

=> mgh = m\omega^2r^2/2 + m\omega^2r^2/5

=> gh =\omega^2r^2/2+\omega^2r^2/5

=>  gh = 7\omega^2 r^2/10

=>  \omega r = (10gh/7)^{1/2}

=> \omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2} = 5.1 rad/s

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Given that force is applied at an angle of 30 degree below the horizontal

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