Question
Question: A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis th...
A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis through speed 2rads−1. the radius of the disc is50cm. Find the minimum coefficient of friction between disc and coin so that the coin does not slip (g=10ms−2)
A. 0.1 B. 0.2 C. 0.3 D. 0.4
Solution
Friction is the resisting force between the two surfaces that are sliding or slipping across each other which always work in the opposite direction in the object is moving. Use the given terms and its correlation to find out the minimum coefficient of the friction.
Complete step by step answer:
Angular Speed, ω=2 radsec−1
Radius, r=50 cm = 10050m
g=10ms−2
Mass is =m
The coefficient of the friction is μ
Also Friction,
f=mω2r .......(1) f=μN f=μmg ........(2)
Equate the right hand side of the equations (1) and (2)
mω2r=μmg
Take mass “m” common from both the sides of the equation and remove it.
ω2r=μg
Now, place the values of the known terms.
(2)2×10050=μ×10
Make the coefficient of the friction, μ the subject –
μ=100×104×50 μ=1000200 μ=0.2
Therefore, the required answer is - the minimum coefficient of friction between disc and coin so that the coin does not slip is 0.2.
Hence, from the given multiple choices – the option B is the correct answer.
Note: Since the friction and load are measured in units of force, they cancel each other as a result the unit of coefficient of friction (μ=LF ) is dimensionless. Basically there are two types of coefficient of friction.
-Static coefficient friction – It is applied to the objects which are motionless.
-Kinetic coefficient friction – It is applied to the objects which are in motion.