Question

A coin kept on horizontal rotating disc has its
Centre at a distance of o.1'm from the axis
of rotation of the disc.. If the coefficient of friction
between the coin and the disc is 0.25 Find the
maximum angular Speed of disc at which the coin
would be aboutt to
slip off​

Answer 1

ANSWER

N=mg

fs=mgr

=

4

mg

f

s

=mrw

2

4

mg

=(m)(0.1)(w

2

)

w

2

=

0.4

g

w=

4

100

w

2

=5rads/sec

Answer 2

Answer:

N = mg

fs = mgr

   = $\frac{mg}{4\\}$

fs = $mrw^{2}$

$\frac{mg}{4}$ = (mg)(0.1)($w^{2}$)

$w^{2}$ = $\frac{g}{0.4}$

w =$\frac{100}{4}$

$w^{2}$= 5 rads/sec