A 240-N sphere 0.20 m in radius rolls without slipping 6.0 m down a ramp that is inclined at 37° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

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Answer 1 · 2018-01-22T08:53:27

#42.4 (rad)/s#

Total energy=potential energy of the sphere at the top of the inclined plane is #m×g×h = m×g×l sin 37 # (where,#l# is the length of the plane)

Given, #mg = 240N#, #l=6m#

So,total energy = #240×6×(3/5) N or 864 J#

When the sphere will be at the bottom its total energy will be potential energy i.e #1/2×m×v^2#,or #12v^2# ,where #v# is the linear velocity,

So,using law of conservation of energy we get, #12 v^2 = 864# or, #v = 8.84 m/s#

So,angular velocity i.e #omega =v/r=8.48/0.20 = 42.4 (rad)/s#