A uniform disc of mass 100 g has a diameter of 10 cm. Calculate the total kinetic energy of the disc when rolling along a horizontal table with a velocity of $20\ \text{cm s}^{-1}$. (Take the surface of the table as reference.)

A uniform disc of mass 100 g has a diameter of 10 cm. Calculate the total kinetic energy of the disc when roll

V. Numerical Problems · Q1

20\ \text{cm s}^{-1}

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Concept understanding — Rolling Motion on an Inclined Plane

For a round object of mass $m$ and radius $R$ rolling without slipping down an incline of angle $\theta$ , two forces act along the slope: the driving component of gravity, $mg\sin\theta$ , and an opposing static frictional force $f$ (the perpendicular component $mg\cos\theta$ is simply balanced by the normal force $N$ ).

Setting up the two equations of motion. For translation along the incline: $mg\sin\theta-f=ma$ . For rotation about the object's own center (about which $mg\sin\theta$ produces no torque, since it passes through the center; only friction does, via $\tau=Rf$ ): $Rf=I\alpha$ . Using $a=R\alpha$ and $I=mK^2$ (radius of gyration $K$ ), the friction force can be eliminated between the two equations, giving

a=\frac{g\sin\theta}{1+\dfrac{K^2}{R^2}}.

Speed at the bottom. Using $v^2=u^2+2as$ with $u=0$ and incline length $s=h/\sin\theta$ (for vertical drop $h$ ):

v=\sqrt{\dfrac{2gh}{1+K^2/R^2}}.

Time to reach the bottom. Using $v=u+at$ with $u=0$ :

t=\sqrt{\dfrac{2h\!\left(1+K^2/R^2\right)}{g\sin^2\theta}}.

The racing order. Both $a$ and $v$ decrease, and $t$ increases, as $K^2/R^2$ increases — and crucially, both the object's mass $m$ and the incline angle $\theta$ cancel out of these expressions entirely, so the outcome depends only on shape (via $K^2/R^2$ ), never on mass or incline steepness. Using the standard $K^2/R^2$ values, a solid sphere ( $K^2/R^2=2/5$ ) always reaches the bottom first, then a disc ( $1/2$ ), then a hollow sphere ( $2/3$ ), and a ring ( $1$ ) last — regardless of their individual masses or radii, as long as they start together from rest at the same height.

Kinetic energy split. The total kinetic energy in pure rolling always splits between translational and rotational parts in the fixed ratio $KE_{TRANS}:KE_{ROT}=1:(K^2/R^2)$ — for a solid sphere this is $1:\tfrac25$ , i.e. $5:2$ , so of the total kinetic energy, $\tfrac57$ is translational and $\tfrac27$ is rotational.

Step 1. List the given data in SI units. Mass $M=100\ \text{g}=0.1\ \text{kg}$ ; diameter $=10\ \text{cm}$ , so radius $R=5\ \text{cm}=0.05\ \text{m}$ ; velocity of the center of mass (along the table) $v=20\ \text{cm s}^{-1}=0.2\ \text{m s}^{-1}$ .

Step 2. Identify the correct energy formula. Since the disc rolls (does not slide) along the table, its total kinetic energy is the sum of a translational part and a rotational part:

KE=\frac12Mv^2\left(1+\frac{K^2}{R^2}\right).

For a uniform disc about its central axis, $\dfrac{K^2}{R^2}=\dfrac12$ (from the standard moment-of-inertia table).

Step 3. Substitute.

KE=\frac12(0.1)(0.2)^2\left(1+\frac12\right)=\frac12(0.1)(0.04)(1.5).

Step 4. Compute. $\frac12\times0.1=0.05$ ; $0.05\times0.04=0.002$ ; $0.002\times1.5=0.003\ \text{J}$ .

Step 5. State the result. Note the radius $R$ itself never actually appears in the final answer, because the ratio $K^2/R^2$ (which is all that matters for a disc) is a fixed number ( $1/2$ ) independent of the disc's actual size — only the mass and the speed of the center of mass matter here.

✓Final answer

=3\times10^{-3}\ \text{J}=3\ \text{mJ}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set ANNUAL5 marks
$Q. (a) Discuss rolling on inclined plane and arrive at the expression for the acceleration. OR$
(b) Explain how overtones are produced in a closed organ pipe.
›Reveal solutionSolution

A body rolling down an incline both slides down (translates) AND spins (rotates); part of the gravitational potential energy/force goes into each motion, so the linear acceleration down the slope is less than gsinθ (the value for pure sliding without friction) — the exact reduction depends on the body's radius of gyration.

Setup: A rigid body (sphere, cylinder, etc.) of mass m, radius R, and moment of inertia I = mk² about its own axis through the centre of mass (k = radius of gyration) rolls WITHOUT SLIPPING down an incline of angle θ.

Forces along the incline: component of gravity mg sinθ (down the slope, driving the motion), and a static friction force f (up the slope, at the point of contact) which provides the torque needed to make the body rotate as it rolls.

Translational equation of motion (Newton's second law along the incline):

mg sinθ − f = ma ... (1)

Rotational equation of motion (torque about the centre of mass, produced only by friction f, since gravity and the normal force pass through/near the axis):

f R = I α

Since rolling without slipping means a = Rα, i.e., α = a/R:

f R = I (a/R)

f = I a / R² = (mk²) a / R² ... (2)

Substitute (2) into (1):

mg sinθ − (mk²a/R²) = ma

mg sinθ = ma + (mk²a/R²)

mg sinθ = ma [1 + k²/R²]

Solving for acceleration:

a = g sinθ / [1 + k²/R²]

This shows the rolling body's acceleration down the incline is always LESS than g sinθ (the acceleration of a body sliding down a frictionless incline without rotating), because some of the driving effect of gravity goes into producing the body's rotation rather than purely accelerating its centre of mass forward. The exact factor 1/(1 + k²/R²) depends on the body's shape — for example, a solid sphere (k² = 2R²/5) accelerates faster than a hollow cylinder (k² = R²), because more of a hollow cylinder's mass is farther from the axis, giving it a larger moment of inertia to overcome.

✓Final answer
$A body rolling without slipping down an incline of angle θ has acceleration a = g sinθ / (1 + k²/R²), where k is its radius of gyration — always less than g sinθ, because part of gravity's effect goes into producing rotation rather than pure translation.$

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Q.A uniform disc of mass 100 g has a diameter of 10 cm. Calculate the total kinetic energy of the disc when rolling along a horizontal table with a velocity of 20 cm s−120\ \text{cm s}^{-1}20 cm s−1. (Take the surface of the table as reference.)

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Q.A uniform disc of mass 100 g has a diameter of 10 cm. Calculate the total kinetic energy of the disc when rolling along a horizontal table with a velocity of $20\ \text{cm s}^{-1}$ . (Take the surface of the table as reference.)

Unlock everything free for 14 days