Question

A uniform solid cylinder of mass $m$ and radius $R$ is pulled along a horizontal smooth road by a horizontal force $F$ applied at its center of mass. If the cylinder rolls without slipping, the angular acceleration $\alpha$ of the cylinder is:

• A. $\frac{F}{2mR}$
• B. $\frac{3F}{2mR}$
• C. $\frac{2F}{3mR}$
• D. $\frac{F}{3mR}$

Hint:

For any symmetric rolling body pulled at its center of mass, the linear acceleration is $a = \frac{F}{m + I/R^2}$.
For a solid cylinder, $I/R^2 = m/2$, so $a = \frac{F}{1.5m} = \frac{2F}{3m}$.
Using $\alpha = a/R$ immediately gives $\alpha = \frac{2F}{3mR}$.

Answer 1

The Correct Option is C

Step 1: Use energy instead of separate force and torque equations.
For rolling without slipping, the cylinder's total kinetic energy combines translation and rotation, \[ KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] with $I=\frac{1}{2}mR^2$ and $\omega=\frac{v}{R}$ for pure rolling.
Step 2: Simplify the kinetic energy. \[ KE = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\frac{v^2}{R^2} = \frac{3}{4}mv^2 \]
Step 3: Apply the power relation.
The power delivered by $F$ equals the rate of increase of this kinetic energy, $Fv = \frac{d}{dt}\left(\frac{3}{4}mv^2\right) = \frac{3}{2}mva$, so $F=\frac{3}{2}ma$, giving $a=\frac{2F}{3m}$.
Step 4: Convert to angular acceleration.
Using $a=R\alpha$, \[ \boxed{\alpha = \frac{2F}{3mR}} \]