Question

Suppose that the metal cylinder in the last problem has a mass of 0.10 kg and the coefficient of static friction between the surface and the cylinder is 0.12. If the cylinder is 0.20 m from the center of the turntable, what is the maximum speed that the cylinder can move along its circular path without slipping off the turntable?

Answer 1

Answer:

v = 0.485 V

Explanation:

Let the Centripetal force be F

$F = \frac{mv^{2} }{r}$................(1)

mass, m = 0.10 kg

radius, r = 0.20 m

speed, v = ?

$F = \mu mg$..................(2)

$\frac{mv^{2} }{r} = \mu mg$

$v^{2} = \mu rg$

$v = \sqrt{\mu rg}$

$v = \sqrt{0.12 * 0.2 * 9.8}$

$v = 0.485 V$

Answer 2

Answer:

The speed maximum speed is $0.49ms^{-1}$

Explanation:

The centrifugal force always acts on the cylinder and move away the rotating platform from the  rotational axis. so the centripetal force provide by the frictional force:

Therefore

$\frac{mv^{2} }{r} =u_{s} mg$        

coefficient of static friction: $u_{s} =0.12$

mass of the cylinder: $m=0.10kg$\

distance of the cylinder from the turntable: $r=0.20m$

$\frac{mv^{2} }{r} =u_{s} mg$

cross multiply to find v

$v^{2} =u_{s} rg\\v=\sqrt{u_{s}rg } \\v=\sqrt{0.12*0.20*9.80}\\ v=0.49m/s$