Question

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 coefficient of friction is \(\mu\) and only forces acting are gravitational and frictional, then the magnitude of the frictional force '\(f\)' is \(\left(\sin 30^{\circ }=\frac{1}{2}\right)\)

• A. \(\frac{mg}{6}\)
• B. \(6mg\)
• C. \(\frac{mg}{3}\)
• D. \(3mg\)

Hint:

Write the linear equation along the incline and the torque equation about the centre, with the rolling condition linking them.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The disc rolls without slipping, so \(a=R\alpha\). Friction acts up the incline and gives the torque about the centre.

Step 2: Equations of motion
Along the incline: \(mg\sin30^{\circ}-f=ma\) (retardation \(a\)).
About the centre: \(fR=I\alpha=\dfrac{1}{2}mR^{2}\cdot\dfrac{a}{R}\), so \(f=\dfrac{ma}{2}\).

Step 3: Calculation
\(\dfrac{mg}{2}=ma+f=2f+f=3f\)
So \(f=\dfrac{mg}{6}\).

Final Answer:
The frictional force is mg divided by 6. \[ \boxed{\text{(A) }\dfrac{mg}{6}} \]