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Grade: 12th pass
        
An insect is released from rest at the top of the smooth bowling ball such that it slides over the ball. Prove that it will loose it's footing with the ball at an angle of about 48° with the vertical.
one year ago

Answers : (1)

Himanshu Rathour
119 Points
							
Insect when released will be in circular motion with centre at centre of ball
Let insect loses contact when it makes angle ø with vertical
Height descended=r(1-cosø)
By work energy theorem,
Work done by gravity=change in kinetic energy
1/2mv²=mgr(1-cosø)
v²=2gr(1-cosø)
Now,
mgsinø-N=mv²/r
mgsinø-N=m2gr(1-cosø)/r
When insect loses contactc N=0
mgsinø=2mgr(1-cosø)/r
Solving it
ø≈48o
 
one year ago
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