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 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:
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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 $\alpha = \frac{2F}{3mR}$.
For a solid cylinder, $I/R^2 = m/2$, so $a = \frac{F}{1.5m} = \frac{2F}{3m}$.
Using $\alpha = a/R$ immediately gives $\alpha = \frac{2F}{3mR}$.
Updated On: Jul 22, 2026
- $\frac{F}{2mR}$
- $\frac{3F}{2mR}$
- $\frac{2F}{3mR}$
- $\frac{F}{3mR}$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We need to find the angular acceleration of a uniform solid cylinder rolling without slipping on a horizontal surface when pulled by a horizontal force $F$ at its center of mass.
Step 2: Key Formula and Approach:
For pure rolling (rolling without slipping), the linear acceleration $a$ of the center of mass and the angular acceleration $\alpha$ are related by:
\[ a = R\alpha \] We will apply Newton's second law for linear motion and rotational motion about the center of mass.
Step 3: Detailed Explanation:
• Equations of motion:
Let $f_s$ be the static friction force acting backwards at the contact point.
Linear force equation:
\[ F - f_s = m a = m (R\alpha) \quad \text{--- (Equation 1)} \] Torque equation about the center of mass:
The only force creating torque is friction $f_s$:
\[ \tau = f_s R = I\alpha \] For a solid cylinder, the moment of inertia is $I = \frac{1}{2}mR^2$:
\[ f_s R = \left(\frac{1}{2}mR^2\right)\alpha \Rightarrow f_s = \frac{1}{2}mR\alpha \quad \text{--- (Equation 2)} \]
• Solve for $\alpha$:
Substitute $f_s$ from Equation 2 into Equation 1:
\[ F - \frac{1}{2}mR\alpha = mR\alpha \] \[ F = \frac{3}{2}mR\alpha \] \[ \alpha = \frac{2F}{3mR} \]
Step 4: Final Answer:
The angular acceleration of the cylinder is $\frac{2F}{3mR}$, which corresponds to Option (C).
We need to find the angular acceleration of a uniform solid cylinder rolling without slipping on a horizontal surface when pulled by a horizontal force $F$ at its center of mass.
Step 2: Key Formula and Approach:
For pure rolling (rolling without slipping), the linear acceleration $a$ of the center of mass and the angular acceleration $\alpha$ are related by:
\[ a = R\alpha \] We will apply Newton's second law for linear motion and rotational motion about the center of mass.
Step 3: Detailed Explanation:
• Equations of motion:
Let $f_s$ be the static friction force acting backwards at the contact point.
Linear force equation:
\[ F - f_s = m a = m (R\alpha) \quad \text{--- (Equation 1)} \] Torque equation about the center of mass:
The only force creating torque is friction $f_s$:
\[ \tau = f_s R = I\alpha \] For a solid cylinder, the moment of inertia is $I = \frac{1}{2}mR^2$:
\[ f_s R = \left(\frac{1}{2}mR^2\right)\alpha \Rightarrow f_s = \frac{1}{2}mR\alpha \quad \text{--- (Equation 2)} \]
• Solve for $\alpha$:
Substitute $f_s$ from Equation 2 into Equation 1:
\[ F - \frac{1}{2}mR\alpha = mR\alpha \] \[ F = \frac{3}{2}mR\alpha \] \[ \alpha = \frac{2F}{3mR} \]
Step 4: Final Answer:
The angular acceleration of the cylinder is $\frac{2F}{3mR}$, which corresponds to Option (C).
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