A solid sphere rolls down incline without slipping — find acceleration

Question

A solid sphere rolls down an inclined plane (angle $\theta$ ) without slipping. Find the linear acceleration of the sphere.

Solution — Step by Step

Three forces act on the rolling sphere:

Weight $mg$ — acting downward at the centre of mass
Normal force $N$ — perpendicular to the incline surface
Static friction $f$ — acting up the incline at the contact point (it’s static friction because the sphere rolls without slipping — no relative sliding at contact)

Static friction provides the torque that causes angular acceleration.

Along the incline (taking down as positive):

mg\sin\theta - f = ma \quad \cdots (1)

where $a$ is the linear acceleration of the centre of mass.

Static friction $f$ acts at the contact point (distance $R$ from centre) and provides torque about the centre:

\tau = fR = I\alpha

For a solid sphere: moment of inertia $I = \frac{2}{5}mR^2$

fR = \frac{2}{5}mR^2 \cdot \alpha \quad \cdots (2)

For rolling without slipping: $a = \alpha R$ , so $\alpha = a/R$ .

Substituting into equation (2):

f = \frac{2}{5}mR \cdot \frac{a}{R} = \frac{2}{5}ma

Now substituting $f = \frac{2}{5}ma$ into equation (1):

mg\sin\theta - \frac{2}{5}ma = ma

mg\sin\theta = ma\left(1 + \frac{2}{5}\right) = ma \cdot \frac{7}{5}

\boxed{a = \frac{5g\sin\theta}{7}}

Why This Works

The $\frac{5}{7}$ factor arises because some of the gravitational potential energy goes into rotational kinetic energy (spinning) rather than all into translational kinetic energy. For a sliding object (no rotation), $a = g\sin\theta$ . For a rolling solid sphere, only 5/7 of that acceleration is achieved — the sphere rolls more slowly than it would slide.

The general formula for any rolling body is:

a = \frac{g\sin\theta}{1 + I/(mR^2)}

For a solid sphere: $I = \frac{2}{5}mR^2$ , so $I/(mR^2) = 2/5$ , giving $a = g\sin\theta/(1 + 2/5) = 5g\sin\theta/7$ .

Alternative Method

Using energy conservation (faster but gives no friction info):

At the bottom, all gravitational PE has converted to KE:

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{2}{5}mR^2\cdot\frac{v^2}{R^2} = \frac{7}{10}mv^2

v^2 = \frac{10gh}{7}

Using $v^2 = 2aL$ where $L = h/\sin\theta$ (length of incline):

a = \frac{v^2}{2L} = \frac{10gh/7}{2h/\sin\theta} = \frac{5g\sin\theta}{7}

Same result, achieved without solving the dynamics equations.

JEE Main and Advanced regularly test rolling motion with different objects: solid sphere ( $I = 2mR^2/5$ , $a = 5g\sin\theta/7$ ), hollow sphere ( $I = 2mR^2/3$ , $a = 3g\sin\theta/5$ ), solid cylinder ( $I = mR^2/2$ , $a = 2g\sin\theta/3$ ), hollow cylinder ( $I = mR^2$ , $a = g\sin\theta/2$ ). Memorise which rolls fastest (solid sphere, smallest $I$ ) and slowest (hollow cylinder, largest $I$ ).

Common Mistake

Students often assume static friction acts down the incline (thinking it opposes the rolling motion). It actually acts up the incline. Here’s why: without friction, the sphere would slide down and translate without rotating. Friction at the contact point must provide an upward torque to generate the clockwise rotation needed for rolling. So friction acts upward along the incline — opposing translation but enabling rotation.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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