Question

hey, can anyone solve this question it's urgent?
A disc revolves in a horizontal plane at a steady rate of 3 revolutions per second. a coin of mass 6g just remains on the disc if kept at a mean distance of 2 cm from the axis of rotation. what is the coefficient of friction between the coin and the disc? (g=10m/s²)​

Answer 1

Answer:

0.018

Explanation:

Given:

ω=3 rev/sec

where ω is the angular velocity

m=6 g

=0.006 kg

r=2 cm

=0.02 m

Solution:

Forces acting on the coin are

1. Its weight vertically downwards
2. Normal reaction acting vertically upwards
3. Frictional force (radially inwards)
4. Centripetal force (radially outwards)

We see that the coin is in rest, this means:

1. Forces in vertical direction are balanced
2. Forces in horizontal direction are balanced

According to 1.

mg=N

and

According to 2.

Fc=Ff

where

Fc=centripetal force

Ff=frictional force

Ff=Nμ

=mgμ       (since N=mg)

where μ=coefficient of friction

Fc=mω²r

Equating the two forces

mgμ=mω²r

gμ=ω²r

Plugging in the values

10μ=9*2*10⁻²

=18*10⁻3

=0.018

Hope this helps

Answer 2

Answer:

0.71

Explanation:

Here, the centrifugal force which is radially outwards will be balanced by friction force which is towards the centre .

The centrifugal force that acts on the coin is = mrw^2

Friction force = uN

= uMg = u*(6/1000)*10

= 0.06 (u)

w = 3*2π = 6*(22/7)

w^2 = 17424/49

Now, (6/1000)*(2/100)*(17424/49) = uN

(0.006)*(0.02)*(17424/49) = 0.06u

u = (0.1)*(0.02)*(17424/49)

= (0.002)*(17424/49)

= 711/1000

= 0.71