Question

A uniform disc of mass m and radius R is rolling down a rough inclined plane which makes an angle 30° with the horizontal. If the coefficients of static and kinetic friction are each equal to m and the only force acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is –

• A. (mg/3) upwards
• B. (mg/3) downwards
• C. (mg/6) upwards
• D. (mg/6) downwards

Answer 1

Answer: (C)

(mg/6) upwards

Answer 2

mg sinq – f = ma ....(1)

Torque w.r.t. centre of disc,

f.R = Ia Þ f.R. = $\frac { 1 } { 2 }$mR². Þ a = $\frac { 2 \mathrm { f } } { \mathrm { m } }$ ....(2)

\ From eqⁿ (1) = eqⁿ(2)

mg sinq – f = 2f Þ f

= = $\frac { \mathrm { mg } } { 6 }$