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JEECUP (UP Polytechnic) Systems of Particles and Rotational Motion — practice questions

Explore 67 free Systems of Particles and Rotational Motion questions in JEECUP (UP Polytechnic) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Systems of Particles and Rotational Motion questions and exercises with answers

Showing questions 1–30 of 67.

The moment of inertia of a solid sphere of mass $M$ and radius $R$ about a diameter is:Read the answer and worked solution →Two point particles of equal mass are placed at x = 2 m and x = 8 m on the x-axis. Where does the centre of mass of the system lie?Read the answer and worked solution →If $\hat{i}$, $\hat{j}$, $\hat{k}$ are right-handed orthogonal unit vectors, then $\hat{i} \times \hat{j}$ equals:Read the answer and worked solution →For a homogeneous (uniformly dense) thin circular disc, the centre of mass lies at: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 →

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. $MR^2$
Show answer and solution

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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Two point particles of equal mass are placed at x = 2 m and x = 8 m on the x-axis. Where does the centre of mass of the system lie?

  1. x = 4 m
  2. x = 5 m
  3. x = 6 m
  4. x = 10 m
Show answer and solution

Correct answer: B. x = 5 m

For two equal masses, the centre of mass lies exactly midway between them: X = (x1 + x2)/2 = (2 + 8)/2 = 5 m.

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If $\hat{i}$, $\hat{j}$, $\hat{k}$ are right-handed orthogonal unit vectors, then $\hat{i} \times \hat{j}$ equals:

  1. $\hat{i}$
  2. $\hat{j}$
  3. $\hat{k}$
  4. $-\hat{k}$
Show answer and solution

Correct answer: C. $\hat{k}$

By the right-hand rule for the cross product, $\hat{i} \times \hat{j} = \hat{k}$. The cyclic order i → j → k gives the positive sign.

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