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Step-by-Step Solution
Step 1: Identify forces and motion
A solid cylinder of mass $m$ and radius $R$ is being pulled horizontally by a force $F$. Because it rolls without slipping, it undergoes both translational and rotational motion.
Step 2: Set up the translational equation
Let the linear acceleration of the cylinder be $a$. For the cylinder to accelerate horizontally, the net horizontal force equals the mass times acceleration:
$F - f = m a,$
where $f$ is the frictional force at the point of contact with the ground.
Step 3: Set up the rotational equation
Because the cylinder is rolling without slipping, the frictional force $f$ provides the torque about the center of the cylinder. The moment of inertia $I$ of a solid cylinder about its central axis is:
$I = \frac{1}{2} m R^2.$
The torque $\tau$ about the center due to friction is:
$\tau = f \times R.$
This torque produces an angular acceleration $\alpha$, so using the rotational form of Newton’s second law:
$\tau = I \alpha \quad \Longrightarrow \quad f \, R = \frac{1}{2} m R^2 \alpha.$
Hence,
$f = \frac{1}{2} m R \alpha.$
Step 4: Relate linear and angular accelerations
For rolling without slipping, the linear acceleration $a$ is related to the angular acceleration $\alpha$ by:
$a = \alpha R.$
So, $\alpha = \frac{a}{R}.$
Substitute $\alpha = \frac{a}{R}$ into $f = \frac{1}{2} m R \alpha$:
$f = \frac{1}{2} m R \left(\frac{a}{R}\right) = \frac{1}{2} m a.$
Step 5: Combine equations to find $a$
From the translational equation $F - f = m a$, substitute $f = \frac{1}{2} m a$:
$F - \frac{1}{2} m a = m a.$
Hence,
$F = \frac{3}{2} m a,$
giving
$a = \frac{2F}{3m}.$
Step 6: Determine the angular acceleration $\alpha$
Using $a = \alpha R$, we get:
$\alpha = \frac{a}{R} = \frac{\frac{2F}{3m}}{R} = \frac{2F}{3mR}.$
This is the required angular acceleration for the rolling cylinder.
Answer:
The angular acceleration of the cylinder is
$$ \frac{2F}{3mR}. $$
