Question

Fill in the Blank : A uniform disc of mass m and radius R is rolling up a rough inclined plane which makes an angle of$30^{\circ}$ with the horizontal. If the coefficients of static and kinetic friction are each equal to$\mu$ and the only forces acting and the are gravitational and frictional, then the magnitude of the frictional force acting on the disc is . . . . . . and its direction is . . . . . . (write up or down) the inclined plane.

Answer 1

Answer:

Explanation:

Force equation = mgsin30 -f  = ma

Torque equation = fR = 1/2mR²(a/R) = f = ma/2

thus, mg/2-ma/2 = ma

a = g/3 = f = mg/6

Therefore, : A uniform disc of mass m and radius R is rolling up a rough inclined plane which makes an angle of30^{\circ} with the horizontal. If the coefficients of static and kinetic friction are each equal to\mu and the only forces acting and the are gravitational and frictional, then the magnitude of the frictional force acting on the disc is six and its direction is upwards on the inclined plane.