Chapter 6 · JEE Preparation Physics

System of Particles and Rotational Motion

35 practice questions · 5 marks weightage

Topics in this chapter

  • Center Of Mass Systems
  • Moment Of Inertia And Mass Distribution
  • Torque And Rotational Equilibrium
  • Angular Momentum And Collisions
  • Rolling Motion And Friction
  • Rotational Dynamics And Energy

Cheat sheet

Cheat Sheet

Practice questions

  1. Question 1Q-E9B5AD

    The center of mass of a system of two particles of masses m1m_1 and m2m_2 is at the origin. If mass m1m_1 is moved by a distance d⃗\vec{d} towards the origin, by what distance must mass m2m_2 be moved so that the center of mass remains unchanged?

    1. A.(m1/m2)d⃗(m_1/m_2)\vec{d} away from the origin
    2. B.(m1/m2)d⃗(m_1/m_2)\vec{d} towards the origin
    3. C.(m2/m1)d⃗(m_2/m_1)\vec{d} away from the origin
    4. D.zero
    See the answer and solution
  2. Question 2Q-56588F

    A thin uniform circular ring of mass MM and radius RR is rolling without slipping on a horizontal surface with a linear velocity vv. The total kinetic energy of the ring is:

    1. A.12Mv2\frac{1}{2} Mv^2
    2. B.Mv2Mv^2
    3. C.32Mv2\frac{3}{2} Mv^2
    4. D.2Mv22 Mv^2
    See the answer and solution
  3. Question 3Q-A196E0

    A solid sphere, a hollow sphere, a solid cylinder, and a hollow cylinder of the same mass and radius are allowed to roll down an inclined plane from rest at the same height simultaneously. Which of them reaches the bottom of the incline first?

    1. A.Solid sphere
    2. B.Hollow sphere
    3. C.Solid cylinder
    4. D.Hollow cylinder
    See the answer and solution
  4. Question 4Q-0AD358

    A uniform rod of length LL and mass MM is pivoted freely at one end and hangs vertically. It is displaced slightly and released. The angular velocity of the rod when it passes through the bottom-most position is:

    1. A.3gL\sqrt{\frac{3g}{L}}
    2. B.6gL\sqrt{\frac{6g}{L}}
    3. C.2g3L\sqrt{\frac{2g}{3L}}
    4. D.gL\sqrt{\frac{g}{L}}
    See the answer and solution
  5. Question 5Q-79503A

    A ballet dancer spins with angular velocity ω0\omega_0 on a frictionless turntable with her arms outstretched. If she folds her arms such that her moment of inertia decreases to 75 %75\text{ \%} of its initial value, her new angular velocity becomes:

    1. A.0.75ω00.75 \omega_0
    2. B.1.33ω01.33 \omega_0
    3. C.1.50ω01.50 \omega_0
    4. D.2.00ω02.00 \omega_0
    See the answer and solution

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