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

(a, b, c) Given
radius of sphere = R
mass of sphere = m
and the angle of inclination in inclined plane, ` theta = 45^@`
The linear acceleration of solid sphere is given by

` a = (g sin theta)/(1 + (I)/(MR^2) ) = (g sin theta)/(1 + (2/5 MR^2)/(MR^2) ) " " [ because I = 2/5 MR^2]`
` a = 5/7 g sin theta`
Applying Newton's second law of motion for linear motion of the centre of mass.
` mg sin theta - f = ma`
`rArr f = mg sin theta - ma = mg sin theta - m 5/7 . g sin theta = 2/7 mg sin theta`
` therefore ` Coefficient of friction `(mu_s) = f/N = (2/7 mg sin theta)/(mg cos theta)`
` = 2/7 tan theta`
To prevent slipping, ` mu_s gt= 2/7 tan theta`
`mu_s gt= 2/7 " " [ because tan 45^@ = 1]`
Hence `3/7 , 1/2 ` and `5/8` are all greater than `2/7`