Rotational Motion: Exam-Pattern Drill (10)

easy 2 min read

Question

A solid uniform disc of mass M=2kgM = 2\,\text{kg} and radius R=0.5mR = 0.5\,\text{m} rolls without slipping down a smooth incline of angle 3030^\circ. Find the linear acceleration of its centre of mass and the friction force acting on it. Take g=10m/s2g = 10\,\text{m/s}^2.

Solution — Step by Step

For rolling without slipping, both translation and rotation matter. Along the incline (taking down as positive):

Mgsinθf=MaMg\sin\theta - f = Ma

About the centre, friction provides torque:

fR=Iαf R = I\alpha

where I=12MR2I = \frac{1}{2}MR^2 for a solid disc.

No slipping means a=Rαa = R\alpha, so α=a/R\alpha = a/R. Substitute:

fR=12MR2aR    f=12Maf R = \frac{1}{2}MR^2 \cdot \frac{a}{R} \implies f = \frac{1}{2}Ma

Substitute ff back into the translational equation:

Mgsinθ12Ma=MaMg\sin\theta - \frac{1}{2}Ma = Ma

gsinθ=32a    a=2gsinθ3g\sin\theta = \frac{3}{2}a \implies a = \frac{2g\sin\theta}{3}

Plugging in: a=2×10×0.53=1033.33m/s2a = \frac{2 \times 10 \times 0.5}{3} = \frac{10}{3} \approx 3.33\,\text{m/s}^2.

f=12Ma=12×2×103=1033.33Nf = \frac{1}{2}Ma = \frac{1}{2} \times 2 \times \frac{10}{3} = \frac{10}{3} \approx 3.33\,\text{N}

Final: a3.33m/s2a \approx 3.33\,\text{m/s}^2 and f3.33Nf \approx 3.33\,\text{N} up the incline.

Why This Works

Rolling friction here is static, not kinetic — the contact point momentarily has zero velocity. That is why we cannot use f=μNf = \mu N blindly. Instead, friction adjusts itself so the no-slip condition is satisfied.

The factor 23\frac{2}{3} in the acceleration is specific to a solid disc. For a ring, you would get 12gsinθ\frac{1}{2}g\sin\theta, and for a solid sphere, 57gsinθ\frac{5}{7}g\sin\theta. Toppers memorise these three.

Alternative Method

Energy conservation gives speed but not acceleration directly. Using the instantaneous-axis-through-contact-point trick:

τcontact=Icontactα=(I+MR2)α\tau_{\text{contact}} = I_{\text{contact}}\alpha = (I + MR^2)\alpha

Gravity provides the torque, friction does not (it acts at the axis). Solve to get the same aa.

Common Mistake

Students assume friction must point down the incline because the disc is moving downhill. Actually, friction here points up the incline — it is what gives the disc the angular acceleration needed to keep rolling without slipping.

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