Q.A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

Showing the 12 most recent of 60 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A ring of mass 1.5 kg and radius 35 cm is rolling without slipping on a horizontal floor. If the translational kinetic energy of the ring is 100 J, then its rotational kinetic energy is (A) 150 J (B) 50 J (C) 100 J (D) 200 J

   ›Reveal solutionSolution

   A rolling ring always splits its kinetic energy 50:50 between translation and rotation, because its moment of inertia $I=MR^2$ makes the two energy expressions identical in form. Answer: (C) 100 J.

   Concept and Intuition

   For any object rolling without slipping, translational and rotational kinetic energies are linked through the no-slip condition $v=\omega R$. Whether the two energies are equal, or one dominates, depends entirely on the shape factor $k^2$ in $I=Mk^2$ (where $k$ is the radius of gyration). A ring/hoop has all its mass at radius $R$, so $I=MR^2$ exactly — this is the special case where the rotational KE expression becomes numerically identical to the translational KE expression once $v=\omega R$ is substituted. (Contrast: a solid disc has $I=\tfrac12MR^2$, so its rotational KE is only half its translational KE; a solid sphere has $I=\tfrac25MR^2$, giving an even smaller ratio.)

   Step-by-Step Solution

   1. Moment of inertia of a ring about its central axis: $I=MR^2$.
   2. Rolling without slipping condition: $v=\omega R$, i.e. $\omega=v/R$.
   3. Rotational KE:

   $KE_{rot}=\frac12I\omega^2=\frac12(MR^2)\left(\frac{v}{R}\right)^2=\frac12Mv^2.$

   4. Compare to translational KE: $KE_{trans}=\frac12Mv^2$ — identical expression!
   5. Since $KE_{rot}=KE_{trans}$ for a ring, and we're given $KE_{trans}=100\ J$, immediately $KE_{rot}=100\ J$. (Note: the given mass 1.5 kg and radius 35 cm are not even needed — this ratio is a pure shape property.)

   Common Mistakes

   • Assuming a rolling object's rotational and translational KEs are always equal for any shape — this only happens for a ring/hoop ($I=MR^2$); for other shapes the ratio is different (e.g. disc: 1:2, sphere: 2:5).
   • Getting distracted by trying to compute $\omega$ or $v$ numerically from the given mass/radius — they cancel out entirely; the answer follows purely from the ring's shape factor.

   ✓Final answer

   The correct option is (C) — 100 J.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Hollow sphere rolls on an inclined plane of height $h$ without slipping. If it starts from rest, its speed at the bottom is (A) $\sqrt{2gh}$ (B) $\sqrt{\dfrac{10gh}{7}}$ (C) $\sqrt{\dfrac{6gh}{5}}$ (D) $\sqrt{\dfrac{4gh}{3}}$

   ›Reveal solutionSolution

   Rolling without slipping converts gravitational PE into both translational and rotational KE; using the hollow sphere's moment of inertia $I=\tfrac23mr^2$ gives $v=\sqrt{6gh/5}$.

   Concept and Intuition

   When an object rolls without slipping down an incline, its total kinetic energy at the bottom is split between translational motion of the centre of mass and rotation about that centre. The fraction going into rotation depends on the object's moment-of-inertia factor ($k^2$ in $I=mk^2$) — a hollow sphere ($k^2=\tfrac23r^2$) ends up slower at the bottom than a solid sphere ($k^2=\tfrac25r^2$) because more of its mass (and hence energy) is invested in rotation.

   Step-by-Step Solution

   1. Energy conservation (starting from rest, height $h$): $mgh = \tfrac12mv^2 + \tfrac12I\omega^2$.
   2. For a hollow (thin-walled) sphere: $I = \tfrac23mr^2$.
   3. Rolling without slipping: $\omega = v/r$, so $\tfrac12I\omega^2 = \tfrac12\left(\tfrac23mr^2\right)\left(\dfrac{v^2}{r^2}\right) = \tfrac13mv^2$.
   4. Substitute: $mgh = \tfrac12mv^2+\tfrac13mv^2 = \left(\tfrac36+\tfrac26\right)mv^2 = \tfrac56mv^2$.
   5. Solve for $v$: $v^2 = \dfrac{6gh}{5} \Rightarrow v = \sqrt{\dfrac{6gh}{5}}$.

   Common Mistakes

   • Using the solid-sphere moment of inertia ($\tfrac25mr^2$, giving $\sqrt{10gh/7}$) instead of the hollow-sphere value — a very common mix-up between solid and hollow.
   • Forgetting the rotational KE term entirely, which would give the (wrong) free-fall answer $\sqrt{2gh}$.

   ✓Final answer

   The correct option is (C) — $\sqrt{\dfrac{6gh}{5}}$.

   ANSWER: C

3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The rotational kinetic energy of a solid sphere of mass 3 kg and radius 0.2 m rolling down an inclined plane of height 7m is (nearer to) (A) 80 J (B) 36 J (C) 40 J (D) 60 J

   ›Reveal solutionSolution

   A solid sphere rolling down an incline splits its total kinetic energy in a fixed ratio between translation and rotation; for a solid sphere the rotational share is $2/7$ of the total, giving 60 J here.

   Concept and Intuition

   For rolling without slipping, $v = \omega R$, so translational and rotational kinetic energies are always in a fixed ratio determined by the moment of inertia factor. For a solid sphere, $I = \frac{2}{5}MR^2$, and this ratio comes out to rotational KE being $2/7$ of the total KE — independent of speed, mass, or radius (radius and mass are given here mostly as distractors, though mass does scale the total energy).

   Step-by-Step Solution

   1. Total mechanical energy converted from height (energy conservation, rolling without slipping, no energy loss to friction since it's rolling not sliding): $E_{total} = Mgh = 3 \times 10 \times 7 = 210\ J$.
   2. Total KE at the bottom = translational KE + rotational KE $= \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.
   3. For a solid sphere, $I = \frac{2}{5}MR^2$ and rolling condition $v = \omega R$: $KE_{rot} = \frac{1}{2}\cdot\frac{2}{5}MR^2\cdot\frac{v^2}{R^2} = \frac{1}{5}Mv^2$ $KE_{trans} = \frac{1}{2}Mv^2$
   4. Fraction rotational $= \dfrac{\frac{1}{5}Mv^2}{\frac{1}{5}Mv^2+\frac{1}{2}Mv^2} = \dfrac{1/5}{7/10} = \dfrac{2}{7}$.
   5. $KE_{rot} = \frac{2}{7}\times210 = 60\ J$.

   Common Mistakes

   • Computing total KE but forgetting to split off just the rotational part.
   • Using the wrong moment-of-inertia fraction (e.g., using a hollow sphere's $2/3$ factor instead of a solid sphere's $2/5$).

   ✓Final answer

   The correct option is (D) — 60 J.

   ANSWER: D

4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A particle executes uniform circular motion with an angular momentum L. Its kinetic energy is doubled and the angular frequency is halved, then its angular momentum becomes (A) 2 L (B) 4 L (C) L/2 (D) L/4

   ›Reveal solutionSolution

   This tests the relation between kinetic energy, angular frequency, and angular momentum in circular motion, $KE=\tfrac12 L\omega$. The answer is (B), 4L.

   Concept and Intuition

   For a particle in circular motion, angular momentum $L=mr^2\omega$ and kinetic energy $KE=\tfrac12 mr^2\omega^2$. Notice that $KE=\tfrac12(mr^2\omega)\omega=\tfrac12 L\omega$ — this links $L$, $KE$, and $\omega$ directly without needing $m$ or $r$ individually, which is exactly what's needed here since only $KE$ and $\omega$ are said to change.

   Step-by-Step Solution

   1. Start from $L=mr^2\omega$ and $KE=\tfrac12 mr^2\omega^2$.
   2. Notice $KE=\tfrac12(mr^2\omega)\cdot\omega=\tfrac12 L\omega$, so $L=\dfrac{2KE}{\omega}$.
   3. Originally, $L=\dfrac{2KE}{\omega}$.
   4. After the change: $KE'=2KE$ and $\omega'=\dfrac{\omega}{2}$.
   5. New angular momentum: $L'=\dfrac{2KE'}{\omega'}=\dfrac{2(2KE)}{\omega/2}=\dfrac{4KE\times2}{\omega}=\dfrac{8KE}{\omega}$.
   6. Since $L=\dfrac{2KE}{\omega}$, we have $\dfrac{8KE}{\omega}=4\times\dfrac{2KE}{\omega}=4L$.

   Common Mistakes

   • Treating $m$ and $r$ as fixed and trying to solve for them individually — it's unnecessary and can lead to inconsistent assumptions; the relation $KE=\tfrac12 L\omega$ bypasses this.
   • Sign/inversion errors when substituting $\omega'=\omega/2$ (dividing by a half doubles instead of halves) — track this carefully.

   ✓Final answer

   The correct option is (B) — 4 L.

   ANSWER: B

5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A particle is moving in a circular path with constant angular velocity. Its initial angular momentum is L. If the radius of the circle is tripled by keeping angular velocity same, the new angular momentum is (A) 3L (B) 6L (C) 9L (D) $\dfrac{L}{3}$

   ›Reveal solutionSolution

   A point particle's angular momentum about the circle's centre is $L = mr^2\omega$ — it scales with the square of the radius, not linearly. Tripling $r$ while keeping $\omega$ fixed multiplies $L$ by $3^2=9$.

   Concept and Intuition

   It's tempting to think angular momentum scales the same way as linear momentum with radius, but for circular motion $L = I\omega$, and the moment of inertia of a point mass is $I = mr^2$ — quadratic in $r$. So even though $\omega$ is unchanged, tripling the radius has an amplified effect on $L$ because the particle is both farther from the axis (bigger lever arm) and moving faster in absolute terms (since $v=r\omega$ also increases with $r$).

   Step-by-Step Solution

   1. Angular momentum of a particle in circular motion: $L = I\omega = (mr^2)\omega$.
   2. Given: initial radius $r$, initial angular momentum $L = mr^2\omega$.
   3. New radius $r' = 3r$, same $\omega$: $L' = m(3r)^2\omega = 9mr^2\omega$.
   4. Since $mr^2\omega = L$, we get $L' = 9L$.

   Common Mistakes

   • Assuming $L$ scales linearly with $r$ (giving $3L$) — it actually scales as $r^2$ because $I \propto r^2$.
   • Confusing this with linear momentum $p=mv=mr\omega$, which would scale linearly with $r$ at fixed $\omega$.

   ✓Final answer

   The correct option is (C) — 9L.

   ANSWER: C

6. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A uniform rod of mass $m$, length $l$ falls with speed $v$ as shown in the figure. Suddenly one of its ends gets stuck to a frictionless hook. Angular velocity of rod just after its end gets stuck is: [FIGURE] (a horizontal rod of mass $m$ and length $l$ falling vertically with speed $v$; one end of the rod is approaching a fixed frictionless hook mounted on a wall to its right) (A) $\dfrac{3v}{2l}$ (B) $\dfrac{v}{l}$ (C) $\dfrac{2v}{l}$ (D) $\dfrac{3v}{4l}$

   ›Reveal solutionSolution

   The rod’s end suddenly sticks to a frictionless hook, so angular momentum about the hook is conserved during the collision. The initial angular momentum comes from the rod’s linear motion, and the final angular momentum is that of the rod rotating about the hook. Solving gives $\omega = \frac{3v}{2l}$, which corresponds to option (A).

   Concept and Intuition

   When the rod’s end snags on the frictionless hook, the hook exerts an impulsive force on that end. Because the hook is frictionless, this force passes through the hook point — it has no torque about the hook itself. Therefore, angular momentum of the rod about the hook is conserved during the very brief collision. The rod initially has no rotation, only downward translation. After the catch, it rotates about the hook as a rigid body. We equate the angular momentum just before and just after the catch.

   Watch out

   A common mistake is to try to use conservation of linear momentum or energy. The hook exerts an external force, so linear momentum is not conserved. Energy is also not conserved because the collision is inelastic (the end sticks). Only angular momentum about the hook is conserved.

   ---

   Step-by-step solution

   1. Set up the coordinate system and initial conditions

      The rod is horizontal, length $l$, mass $m$, falling straight down with speed $v$. Its center of mass (CM) is at the midpoint. Just before the catch, the right end is about to touch the hook. The rod has no rotation — it’s purely translating downward.

   2. Angular momentum about the hook just before the catch

      The hook is at the right end of the rod. For a point mass $dm$ at a horizontal distance $x$ from the hook (measured leftward along the rod), its velocity is $v$ downward. The angular momentum of that mass element about the hook is

   $dL = (x)(v)\,dm$

   because the perpendicular distance from the hook to the velocity vector is $x$, and the direction (into the page) is the same for all elements.

   Integrate over the rod: $dm = \frac{m}{l} dx$, with $x$ from $0$ (at the hook) to $l$ (at the left end).

   $L_{\text{initial}} = \int_0^l x v \frac{m}{l} dx = \frac{m v}{l} \int_0^l x\,dx = \frac{m v}{l} \cdot \frac{l^2}{2} = \frac{1}{2} m v l.$

   Tip

   This is the same as the angular momentum of a point mass $m$ at the rod’s center of mass, because the CM is at distance $l/2$ from the hook: $L = (l/2)(m v) = \frac{1}{2} m v l$. That’s a quick check.

   3. Angular momentum about the hook just after the catch After the end sticks, the rod rotates rigidly about the hook with angular velocity $\omega$ (clockwise, say). The moment of inertia of a uniform rod about one end is

   $I_{\text{end}} = \frac{1}{3} m l^2.$

   Hence the final angular momentum is

   $L_{\text{final}} = I_{\text{end}} \, \omega = \frac{1}{3} m l^2 \, \omega.$

   4. Conservation of angular momentum No external torque about the hook acts during the collision, so

   $L_{\text{initial}} = L_{\text{final}} \quad \Rightarrow \quad \frac{1}{2} m v l = \frac{1}{3} m l^2 \, \omega.$

   Cancel $m$ and $l$ (both nonzero):

   $\frac{1}{2} v = \frac{1}{3} l \, \omega \quad \Rightarrow \quad \omega = \frac{3v}{2l}.$

   5. Match with options The result $\frac{3v}{2l}$ corresponds to option (A).

   For a rod of length $l$ falling with speed $v$ and catching at one end, the angular speed after the catch is $\omega = \frac{3v}{2l}$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

7. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A uniform rod of mass 20 kg and length 1.6 m is pivoted at its one end and can swing freely in the vertical plane. The angular acceleration of the rod just after the rod is released from rest in the horizontal position is (g – acceleration due to gravity) (A) $\dfrac{15}{16}g$ (B) $\dfrac{17}{16}g$ (C) $\dfrac{16}{15}g$ (D) $\dfrac{9}{16}g$

   ›Reveal solutionSolution

   Use $\tau=I\alpha$ for a rod pivoted at one end and released from horizontal; $\alpha=\dfrac{3g}{2L}=\dfrac{15}{16}g$ for $L=1.6\,\mathrm m$.

   Concept and Intuition

   When a uniform rod pivoted at one end is released from horizontal, gravity acting at its centre of mass produces a torque about the pivot. Newton's second law for rotation, $\tau=I\alpha$, with $I$ the moment of inertia of the rod about the pivoted end, gives the initial angular acceleration.

   Step-by-Step Solution

   1. Moment of inertia of a uniform rod about one end: $I=\dfrac{mL^2}{3}$.
   2. Weight $mg$ acts at the centre, a distance $L/2$ from the pivot; when horizontal, this is the full perpendicular lever arm, so torque $\tau=mg\cdot\dfrac{L}{2}$.
   3. Newton's second law for rotation: $\tau=I\alpha \Rightarrow mg\dfrac{L}{2}=\dfrac{mL^2}{3}\alpha$.
   4. Solve: $\alpha=\dfrac{3g}{2L}$.
   5. Substitute $L=1.6\,\mathrm m$: $\alpha=\dfrac{3g}{3.2}=0.9375\,g=\dfrac{15}{16}g$.

   Common Mistakes

   • Using $I=mL^2/12$ (about the centre) instead of $mL^2/3$ (about the end).
   • Forgetting mass cancels out of the equation, leading to unnecessary numeric mass substitution errors.

   ✓Final answer

   The correct option is (A) — $\dfrac{15}{16}g$.

   ANSWER: A

8. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.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) $\dfrac{F}{2mR}$ (B) $\dfrac{3F}{2mR}$ (C) $\dfrac{2F}{3mR}$ (D) $\dfrac{F}{3mR}$

   ›Reveal solutionSolution

   Combining Newton's second law, the torque equation about the centre, and the rolling constraint for a cylinder pulled by a horizontal force at its centre gives $\alpha = 2F/3mR$. Answer: (C).

   Concept and Intuition

   A force applied exactly through the centre of mass produces zero torque about that centre by itself — so on its own it cannot make the cylinder spin. It is friction at the contact point that supplies the torque needed to keep the cylinder rolling without slipping as it's dragged forward. The trick is to write both the translational equation (net force = $ma$) and the rotational equation (net torque about the centre = $I\alpha$) and tie them together with the no-slip condition $a=\alpha R$.

   Step-by-Step Solution

   1. Translational: $F - f = ma$ (friction $f$ opposes the forward pull as it acts backward at the contact point to prevent slipping).
   2. Rotational about the centre: $fR = I\alpha = \tfrac12 mR^2\alpha$ (moment of inertia of a solid cylinder about its axis is $\tfrac12 mR^2$).
   3. Rolling constraint: $a = \alpha R \Rightarrow \alpha = a/R$.
   4. From step 2: $f = \tfrac12 mR\alpha = \tfrac12 m a$.
   5. Substitute into step 1: $F - \tfrac12 ma = ma \Rightarrow F = \tfrac32 ma \Rightarrow a = \dfrac{2F}{3m}$.
   6. $\alpha = \dfrac{a}{R} = \dfrac{2F}{3mR}$.

   Common Mistakes

   • Assuming the pulling force itself produces the torque (it can't — it passes through the centre, moment arm zero); it's friction that does.
   • Forgetting the rolling constraint $a=\alpha R$, which is what links the translational and rotational equations into one solvable system.

   ✓Final answer

   The correct option is (C) — $\dfrac{2F}{3mR}$.

   ANSWER: C

9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.A solid sphere of mass 4 kg and radius 28 cm is on an inclined plane. If the acceleration of the sphere when it rolls down without sliding is $3.5\ \mathrm{m\,s^{-2}}$, then the acceleration of the sphere when it slides down without rolling is (A) $2.5\ \mathrm{m\,s^{-2}}$ (B) $3.5\ \mathrm{m\,s^{-2}}$ (C) $1.7\ \mathrm{m\,s^{-2}}$ (D) $4.9\ \mathrm{m\,s^{-2}}$

   ›Reveal solutionSolution

   Rolling acceleration is reduced from the sliding value by the sphere's moment-of-inertia factor; back-solving the given rolling acceleration recovers $g\sin\theta$, which is exactly the sliding (frictionless) acceleration — 4.9 m/s².

   Concept and Intuition

   For an object rolling without slipping down an incline, part of the gravitational potential energy goes into rotational kinetic energy, so its linear (translational) acceleration is less than $g\sin\theta$: $a_{\text{roll}}=\dfrac{g\sin\theta}{1+I/(mr^2)}$. For a solid sphere, $I=\dfrac{2}{5}mr^2$, so $a_{\text{roll}}=\dfrac{g\sin\theta}{1+2/5}=\dfrac{5}{7}g\sin\theta$. If instead it slides down without rolling (frictionless, no rotation), all the energy goes into translational KE, giving simply $a_{\text{slide}}=g\sin\theta$ — independent of mass, radius, or moment of inertia.

   Step-by-Step Solution

   1. Rolling acceleration: $a_{\text{roll}}=\dfrac{5}{7}g\sin\theta = 3.5\ \mathrm{m/s^2}$.
   2. Solve for $g\sin\theta$: $g\sin\theta = \dfrac{7}{5}\times3.5 = 4.9\ \mathrm{m/s^2}$.
   3. Sliding (frictionless) acceleration is exactly this value: $a_{\text{slide}} = g\sin\theta = 4.9\ \mathrm{m/s^2}$.

   Common Mistakes

   • Trying to use the sphere's actual mass (4 kg) or radius (28 cm) — they're irrelevant distractors since $a_{\text{roll}}$ and $a_{\text{slide}}$ depend only on $g\sin\theta$ and the moment-of-inertia factor.
   • Mixing up the factor for a solid sphere ($2/5$) with that of a disc ($1/2$) or shell ($2/3$).

   ✓Final answer

   The correct option is (D) — $4.9\ \mathrm{m\,s^{-2}}$.

   ANSWER: D

10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.A thin uniform circular disc rolls with a constant velocity without slipping on a horizontal surface. Its total kinetic energy is (A) three times its rotational kinetic energy (B) three times its translational kinetic energy (C) one and half times its rotational kinetic energy (D) twice its translational kinetic energy

    ›Reveal solutionSolution

    For a rolling disc, total KE splits as $\frac34Mv^2$ (total), $\frac14Mv^2$ (rotational), $\frac12Mv^2$ (translational) — total is 3× the rotational part.

    Concept and Intuition

    Rolling without slipping links the disc's rotational and translational motion via $v=\omega R$. A disc's moment of inertia ($I=\frac12MR^2$) is smaller than, say, a ring's ($MR^2$), which means for a disc the translational KE dominates over the rotational KE — the exact ratio is what this question checks.

    Step-by-Step Solution

    1. Moment of inertia of a disc about its central axis: $I=\dfrac12MR^2$.
    2. Rolling without slipping: $v=\omega R \Rightarrow \omega=\dfrac{v}{R}$.
    3. Rotational KE: $KE_{rot}=\frac12I\omega^2=\frac12\cdot\frac12MR^2\cdot\frac{v^2}{R^2}=\frac14Mv^2.$
    4. Translational KE: $KE_{trans}=\frac12Mv^2.$
    5. Total KE: $KE_{total}=KE_{trans}+KE_{rot}=\frac12Mv^2+\frac14Mv^2=\frac34Mv^2.$
    6. Ratio total to rotational: $\dfrac{KE_{total}}{KE_{rot}}=\dfrac{3/4}{1/4}=3$. So total KE is three times the rotational KE. (Check against translational: $\dfrac{KE_{total}}{KE_{trans}}=\dfrac{3/4}{1/2}=1.5$, i.e. total is 1.5× translational — not matching options (B)/(C), confirming (A) is the only correct statement.)

    Common Mistakes

    • Using a ring's or sphere's moment of inertia instead of a disc's $\frac12MR^2$.
    • Comparing total KE to translational KE and picking option (B) or (C), which describe the wrong ratio.

    ✓Final answer

    The correct option is (A) — three times its rotational kinetic energy.

    ANSWER: A

11. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.A thin circular ring and a circular disc of equal mass are rolling without sliding. If their linear velocities are equal and the total kinetic energy of the disc is 6 J, then the total kinetic energy of the ring is (A) 6 J (B) 3 J (C) 8 J (D) 4 J

    ›Reveal solutionSolution

    This tests total kinetic energy (translational + rotational) for rolling bodies with different moments of inertia but equal mass and linear speed. Answer: (C).

    Concept and Intuition

    For any body rolling without slipping, total KE $=\tfrac12mv^2\left(1+\dfrac{I}{mr^2}\right)$, where $\dfrac{I}{mr^2}$ is a shape-dependent factor (a pure number): $\tfrac12$ for a disc, $1$ for a ring. So bodies with the same mass and linear speed will still have different total KE purely because of this shape factor.

    Step-by-Step Solution

    1. Total KE of a rolling body: $KE = \tfrac12mv^2 + \tfrac12 I\omega^2$, and for rolling without slipping $\omega = v/r$.
    2. For the disc, $I_{disc}=\tfrac12mr^2$:

    $KE_{disc} = \frac12mv^2 + \frac12\left(\frac12mr^2\right)\left(\frac{v}{r}\right)^2 = \frac12mv^2+\frac14mv^2 = \frac34mv^2.$

    Given $KE_{disc}=6\ \text{J}$, so $\tfrac34mv^2=6 \implies mv^2 = 8$.

    3. For the ring, $I_{ring}=mr^2$:

    $KE_{ring} = \frac12mv^2 + \frac12(mr^2)\left(\frac{v}{r}\right)^2 = \frac12mv^2+\frac12mv^2 = mv^2.$

    4. So $KE_{ring} = mv^2 = 8\ \text{J}$ (using the value found in step 2).

    Common Mistakes

    • Assuming equal KE for equal mass and speed — the rotational contribution depends on the mass distribution (moment of inertia), so a ring (all mass at the rim) always has more rotational KE than a disc of the same mass, radius and linear speed.
    • Forgetting the rotational KE term entirely and treating this as pure translational motion.

    ✓Final answer

    The correct option is (C) — 8 J.

    ANSWER: C

12. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
    Q.If a solid sphere of mass 50 g and diameter 20 cm rolls without slipping with a velocity 5 cm s$^{-1}$ on a surface, then its total kinetic energy is (A) $6.25 \times 10^{-7}$ J (B) $2.50 \times 10^{-7}$ J (C) $8.75 \times 10^{-5}$ J (D) $3.75 \times 10^{-5}$ J

    ›Reveal solutionSolution

    Tests total (translational + rotational) kinetic energy of a rolling solid sphere; the diameter given is a distractor (not needed, since $v$ is given directly).

    Concept and Intuition

    For any body rolling without slipping, the point of contact is instantaneously at rest, so the body's motion is a pure translation of the center of mass plus a pure rotation about that center — both energies add. Writing $\omega = v/R$, the rotational KE becomes $\tfrac12 I\omega^2 = \tfrac12\left(\dfrac{k^2}{R^2}\right)mv^2$, so total KE is $\tfrac12mv^2\left(1+\dfrac{k^2}{R^2}\right)$. For a solid sphere, $I=\dfrac25MR^2$, i.e. $k^2/R^2=2/5$.

    Step-by-Step Solution

    1. $m = 50\text{ g} = 0.05\text{ kg}$, $v = 5\text{ cm/s} = 0.05\text{ m/s}$.
    2. For a solid sphere, $\dfrac{k^2}{R^2}=\dfrac25$, so total KE factor $=1+\dfrac25=\dfrac75$.
    3. Total KE $=\tfrac12mv^2\times\dfrac75 = \tfrac12(0.05)(0.0025)\times1.4$.
    4. $\tfrac12(0.05)(0.0025) = 6.25\times10^{-5}$; multiplying by $1.4$ gives $8.75\times10^{-5}$ J.

    Common Mistakes

    • Using only translational KE ($\tfrac12mv^2$), ignoring the rolling constraint's rotational contribution.
    • Using the wrong $k^2/R^2$ (e.g. $2/3$ for a hollow sphere instead of $2/5$ for solid).
    • Getting distracted by the diameter value — it is not needed since $v$ is already given.

    ✓Final answer

    The correct option is (C) — $8.75\times10^{-5}$ J.

    ANSWER: C