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Sample Quiz: Physics Chapter 12

Rotation of a Rigid Body — Physics for Scientists and Engineers: A Strategic Approach (Randall D. Knight)

10 questions · 23 points · answer key included

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Multiple Choice

  1. A particle moves in a circular path in the xy-plane. When measuring angular momentum from the center of the circle, the angle between the position vector vecr and the velocity vector vecv is 90°. Which statement correctly describes the angular momentum vector vecL for counterclockwise motion? (2 PTS)
    1. The angular momentum vector lies in the xy-plane perpendicular to vecr
    2. The angular momentum vector points along the positive z-axis with magnitude Lz = mr2ω
    3. The angular momentum vector points along the negative z-axis with magnitude Lz = mr2ω
    4. The angular momentum vector points along the positive z-axis with magnitude Lz = mrvt
  2. A thin rod of mass M and length L rotates about an axis located one-third of the length from one end. Using the parallel-axis theorem and knowing that Icm = 1/12ML2 for a rod about its center, what is the moment of inertia about this axis? (2 PTS)
    1. I = 1/12ML2
    2. I = 1/12ML2 + M(1/3L)2 = 7/36ML2
    3. I = 1/12ML2 + M(1/6L)2 = 1/9ML2
    4. I = 1/3ML2
  3. A rotating rigid body experiences a net torque. According to the relationship between torque and angular momentum, which equation correctly describes how the net torque affects the angular momentum? (2 PTS)
    1. vecτnet = Ivecω
    2. dvecLdt = vecτnet
    3. dvecLdt = vecL × vecω
    4. vecτnet = vecL × vecω
  4. Two wheels have the same total mass M and radius R. Wheel A has its mass concentrated near the center, while Wheel B has its mass concentrated around the rim. If both are spun with the same angular velocity, which statement is correct? (2 PTS)
    1. Wheel A has a smaller moment of inertia and is easier to spin than Wheel B
    2. Wheel B has a smaller moment of inertia because the mass is farther from the axis
    3. Wheel A has a larger moment of inertia because concentrated mass increases rotational resistance
    4. Both wheels have the same moment of inertia because they have equal mass and radius
  5. A rigid body rotates about a fixed axis. The net torque about the pivot point is zero. Which of the following must be true? (2 PTS)
    1. The body must be in static equilibrium with no motion at all
    2. The angular acceleration is zero, so the angular velocity cannot change
    3. The sum of all forces acting on the body must be zero
    4. The angular velocity is constant or the body is not rotating

True or False

  1. The moment of inertia of an object is independent of the choice of rotation axis. (1 PTS)
  2. For a particle in circular motion, the angular momentum vector is perpendicular to both the position vector and the velocity vector. (1 PTS)
  3. A net torque applied to a rigid body causes its angular momentum to change at a rate equal to the magnitude of the torque. (1 PTS)

Problem Solving

  1. A jeepney wheel with moment of inertia I = 1.2 kg·m2 is spinning at an initial angular velocity of ωi = 8.0 rad/s. A brake applies a constant friction torque of τ = -2.4 N·m. How long does it take for the wheel to come to a complete stop, and what is the angular displacement (in radians) during this time? (5 PTS)
  2. A barbell consists of a 0.50 kg ball and a 2.0 kg ball connected by a massless 0.50 m rod. The barbell rotates about its center of mass at 40 rpm. Calculate (a) the location of the center of mass from the 2.0 kg ball, (b) the moment of inertia about the center of mass, and (c) the rotational kinetic energy. (5 PTS)

Answer key

Multiple Choice

  1. 1.B — For circular motion in the xy-plane with counterclockwise rotation, the right-hand rule applied to vecL = vecr × vecp gives a vector pointing in the positive z-direction. The magnitude is Lz = mrvt = mr2ω where vt is the tangential velocity.
  2. 2.C — The parallel-axis theorem states I = Icm + Md2. The distance from the center of mass (at the rod's center) to the axis one-third from one end is d = 1/6L. Thus I = 1/12ML2 + M(1/6L)2 = 1/9ML2.
  3. 3.B — The fundamental relationship between net torque and angular momentum is dvecLdt = vecτnet, which is the rotational equivalent of Newton's second law dvecpdt = vecFnet.
  4. 4.A — Moment of inertia depends on how mass is distributed: I = ∑ mi ri2. Mass concentrated near the center (smaller ri values) produces a smaller moment of inertia, making the wheel easier to spin. Mass at the rim increases the moment of inertia.
  5. 5.D — When net torque is zero, dvecLdt = 0, which means angular momentum is constant. For a rigid body, this implies constant angular velocity (which could be zero if the body is not rotating). The body need not be in static equilibrium; it could be rotating at constant angular velocity.

True or False

  1. 1.False — Moment of inertia depends critically on the axis of rotation. Different axes produce different values of I because the distances ri of mass elements from the axis change. The parallel-axis theorem explicitly shows how I varies with axis location.
  2. 2.True — Angular momentum is defined as vecL = vecr × vecp. The cross product of two vectors is perpendicular to both. Since vecp is parallel to vecv, vecL is perpendicular to both vecr and vecv.
  3. 3.False — The correct relationship is dvecLdt = vecτnet, a vector equation. The rate of change of angular momentum equals the net torque vector, not just its magnitude. The direction of the change also matters.

Problem Solving

  1. 1.Using τ = Iα: α = τ/I = (-2.4)/1.2 = -2.0 rad/s2. Using ωf = ωi + α t with ωf = 0: 0 = 8.0 + (-2.0)t, so t = 4.0 s. Using ωf2 = ωi2 + 2αΔθ: 0 = (8.0)2 + 2(-2.0)Δθ, so Δθ = 16 rad. — The angular acceleration is found from the torque equation. The stopping time uses the kinematic equation for constant angular acceleration. The angular displacement uses the kinematic equation relating final and initial angular velocities to angular acceleration and displacement.
  2. 2.(a) Using xcm = (m1 x1 + m2 x2)/(m1 + m2) with the 2.0 kg ball at origin: xcm = ((2.0)(0) + (0.50)(0.50))/(2.0 + 0.50) = 0.10 m. (b) The radii are r1 = 0.10 m and r2 = 0.40 m. Thus I = m1 r12 + m2 r22 = (2.0)(0.10)2 + (0.50)(0.40)2 = 0.020 + 0.080 = 0.10 kg·m2. (c) Converting: ω = 40 × 2π/60 = 4.19 rad/s. Then Krot = 1/2Iω2 = 1/2(0.10)(4.19)2 = 0.88 J. — The center of mass is found using the mass-weighted average position. The moment of inertia is the sum of mi ri2 for each mass about the center of mass. The rotational kinetic energy uses K = 1/2Iω2 with angular velocity converted to rad/s.