A uniform sphere of radius Rand mass m is placed on an inclined plane which makes an angle `45^@ ` to the horizontal. For which of the following value of coefficient of friction, the sphere rolls without slipping,

Options

A. `3/7`
B. `1/2`
C. `5/8`
D. `1/7`

Answer

A, B, C

Explanation

Given,
Radius of a uniform sphere = R
Mass of sphere = m
Angle of inclination with horizontal,` theta = 45 ^(@)`
The free body diagram is shown in figure,

The equation for coefficient of friction is given by,
`mu = (f)/(N) " "...(1)`
From figure, according to newton's second law of motion,
`mg sin theta -f = ma`
`f = mg sin theta - ma " "...(2)`
The linear acceleration of solid sphere is given by,
`a = (g sin theta )/(1 + (I)/(M R^(2)))`
`a = (g sin theta )/(1+ (2/5 MR ^(2))/(MR ^(2)))" "(because I = 2/5 MR ^(2))`
`= (g sin theta )/( 1 + (2)/(5))`
`a = (5 g sin theta )/(7)`
Subtituting a value in equatin (2)
`f - mg sin theta -m (5 g sin theta )/(7)`
`=(2 mg sin theta)/(7) " "...(3)`
Also,
`N= mg cos theta" "...(4)`
Substituting equation (3) and (4) in equation (1)
`mu = ((2 mg sin theta )/(7))/(mg cos theta )`
` = 2/7 tan theta`
` = 2/7 tan 45`
`= 2/7`
To avoid slipping,
`mu ge 2/7`
From the given options, `3/7 , 1/2 and 5/8 are ge 2/7`