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.

Question

Himanshu Rathour · Answer

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≈48 o

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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.

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.

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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.

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.

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