Question

an object is placed 5 feet from the center of a horizontally rotating platform. the coefficient of friction is 0.2. the object will begin to slide off when platform speed is nearest to

Answer 1

Gives

$\mu=0.2;R=5feet=1.524m$

$V=\sqrt{\mu R g}$

$V=\sqrt{0.2\times\times1.524\times9.8}$

$V=\sqrt{2.98704}$ m/sec

$V=1.73m/sec$