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PUCET (Punjab) System of Particles and Rotational Motion — practice questions

Explore 40 free System of Particles and Rotational Motion questions in PUCET (Punjab) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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System of Particles and Rotational Motion questions and exercises with answers

Showing questions 1–30 of 40.

The moment of inertia of a solid sphere of mass $M$ and radius $R$ about a diameter is:Read the answer and worked solution →The **moment of inertia** of a thin uniform rod of mass $M$ and length $L$ about a perpendicular axis through its **centre** is:Read the answer and worked solution →The **angular momentum** of a particle of momentum $\vec p$ at position $\vec r$ (from origin) is:Read the answer and worked solution →For a body **rolling without slipping** on a flat surface with centre-of-mass velocity $v$, the **velocity of the topmost point** is:Read the answer and worked solution →A solid sphere ($I = \tfrac{2}{5}MR^2$) rolls without slipping down a plane inclined at $\theta$. Its **linear acceleration** is:Read the answer and worked solution →By **parallel-axis theorem**, the moment of inertia of a thin ring of mass $M$, radius $R$, about a tangent in its plane is:Read the answer and worked solution →A uniform rod of length $L$ is suspended from one end. Its **time period** for small oscillations is:Read the answer and worked solution →A **yo-yo** (solid uniform cylinder) of mass $M$, radius $R$ unwinds under gravity. Its downward acceleration is:Read the answer and worked solution →Angular displacement theta in a circular motion is related to arc length s and radius r as:Read the answer and worked solution →Angular velocity omega in terms of time period T:Read the answer and worked solution →Linear speed v and angular speed omega relation for circular motion:Read the answer and worked solution →Torque tau acting on a body produces:Read the answer and worked solution →Angular momentum L of a body rotating with angular velocity omega:Read the answer and worked solution →A wheel starting from rest reaches 20 rad/s in 4 s. Angular acceleration alpha:Read the answer and worked solution →Moment of inertia of a uniform solid sphere of mass M and radius R about an axis through center:Read the answer and worked solution →Moment of inertia of a thin uniform disc (mass M, radius R) about axis perpendicular to disc through center:Read the answer and worked solution →Parallel axis theorem states I_parallel = I_cm + Md². Here d is:Read the answer and worked solution →For a thin rod of mass M and length L, I about axis through center perpendicular to rod:Read the answer and worked solution →Kinetic energy of a rotating body with moment of inertia I and angular velocity omega:Read the answer and worked solution →For a body rolling without slipping with velocity v of center of mass:Read the answer and worked solution →A solid cylinder and a hollow cylinder of same mass and radius roll down an incline. Which reaches bottom first?Read the answer and worked solution →Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicular to rod:Read the answer and worked solution →A figure skater pulls in her arms during a spin. Her moment of inertia decreases. By conservation of L (I × omega), her angular velocity:Read the answer and worked solution →For a hoop, disc, and solid sphere of same mass and radius rolling down an incline. Order of acceleration (fastest first):Read the answer and worked solution →A solid sphere of mass M, radius R rolls without slipping down an incline of angle theta. Acceleration:Read the answer and worked solution →Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicular to rod:Read the answer and worked solution →For a thin rod (mass M, length L) about axis through ONE END perpendicular to rod, I equals:Read the answer and worked solution →Torque on a body produces angular acceleration. For tau = 20 N m and I = 5 kg m², alpha equals:Read the answer and worked solution →A 50 kg flywheel of radius 0.5 m rotates at 60 rev/min. Angular momentum (I = MR²/2):Read the answer and worked solution →A solid sphere rolls down without slipping. Fraction of KE that is rotational:Read the answer and worked solution →

Worked practice questions

The moment of inertia of a solid sphere of mass $M$ and radius $R$ about a diameter is:

  1. $\dfrac{1}{2}MR^2$
  2. $\dfrac{2}{3}MR^2$
  3. $\dfrac{2}{5}MR^2$
  4. $MR^2$
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Correct answer: C. $\dfrac{2}{5}MR^2$

Standard result for a solid sphere about a central axis: $I = \dfrac{2}{5}MR^2$. Compare with a hollow (thin-shell) sphere where $I = \dfrac{2}{3}MR^2$ — that's option D and a common trap.

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The **moment of inertia** of a thin uniform rod of mass $M$ and length $L$ about a perpendicular axis through its **centre** is:

  1. $ML^2/3$
  2. $ML^2/12$
  3. $ML^2/2$
  4. $ML^2$
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Correct answer: B. $ML^2/12$

**Standard result:** $I_{\text{cm}} = ML^2/12$. **About one end (parallel axis):** $I = ML^2/12 + M(L/2)^2 = ML^2/3$.

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The **angular momentum** of a particle of momentum $\vec p$ at position $\vec r$ (from origin) is:

  1. $\vec p \cdot \vec r$
  2. $\vec r \times \vec p$
  3. $\vec p \times \vec r$
  4. $|\vec r||\vec p|$ always
Show answer and solution

Correct answer: B. $\vec r \times \vec p$

$\vec L = \vec r \times \vec p$ (right-hand rule). Note $\vec r \times \vec p = -\vec p \times \vec r$, order matters.

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