Mastering Rotational Dynamics
This quiz explores the fundamental principles governing objects in rotation. You will be tested on your understanding of torque, the conservation of angular momentum, and the complex mechanics of rolling motion without slipping.
Expect questions that require you to:
- Apply Newton's Second Law in its rotational form.
- Analyze how changes in moment of inertia affect angular velocity.
- Calculate the distribution of kinetic energy in rolling objects.
- Determine the direction and magnitude of rotational vectors using the right-hand rule.
Good luck applying these multi-step reasoning skills to solve problems in rotational mechanics!
Q1. A solid cylinder and a hollow cylinder of the same mass and radius roll down an incline without slipping. Which reaches the bottom first?
The solid cylinder has a smaller moment of inertia (##1/2 MR^2##) compared to the hollow cylinder (##MR^2##). This means a smaller fraction of potential energy is converted to rotational kinetic energy, leaving more for translational speed.
Q2. An ice skater spins with her arms outstretched. When she pulls her arms in, her angular velocity increases. Which principle best explains this?
When the skater pulls her arms in, her moment of inertia decreases. Since no external torque acts on her, angular momentum (##L = I\omega##) is conserved, causing her angular velocity to increase.
Q3. A force is applied perpendicular to a door at a distance ##r## from the hinges. If the same force is applied at a distance ##r/2##, how does the torque change?
Torque is calculated as ##\tau = rF \sin(\theta)##. If the distance ##r## is halved while the force and angle remain constant, the torque is also halved.
Q4. For an object rolling without slipping, what is the velocity of the point of the object currently in contact with the ground?
In pure rolling, the point of contact is instantaneously at rest relative to the surface to satisfy the condition of no slipping.
Q5. A solid sphere rolls without slipping. What is the ratio of its rotational kinetic energy to its translational kinetic energy?
Rotational KE is ##1/2 I\omega^2 = 1/2(2/5 MR^2)(v/R)^2 = 1/5 Mv^2##. Translational KE is ##1/2 Mv^2##. The ratio is ##(1/5)/(1/2) = 2/5##.
Q6. What is the direction of the angular momentum vector for a wheel rotating clockwise in the plane of this screen?
Using the right-hand rule, curl your fingers in the direction of rotation (clockwise). Your thumb points into the screen, representing the direction of the angular momentum vector.
Q7. If the net torque acting on a system is zero, which of the following must be true?
According to the rotational version of Newton's Second Law, ##\tau_{net} = dL/dt##. If the net torque is zero, the rate of change of angular momentum is zero, meaning ##L## is conserved.
Q8. A disk has a moment of inertia ##I## and rotates with angular velocity ##\omega##. If a second identical disk (initially at rest) is dropped onto it and they stick together, what is the new angular velocity?
By conservation of angular momentum, ##I\omega = (I + I)\omega_{new}##. Therefore, ##I\omega = 2I\omega_{new}##, which simplifies to ##\omega_{new} = \omega / 2##.
Q9. A constant torque of 10 Nm is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration?
Using ##\tau = I\alpha##, we get ##\alpha = \tau / I = 10 / 2 = 5## rad/s².
Q10. Which of the following objects of mass ##M## and radius ##R## has the largest moment of inertia about its central axis?
A thin-walled hollow cylinder has ##I = MR^2##, which is larger than a solid cylinder (##1/2 MR^2##), a hollow sphere (##2/3 MR^2##), or a solid sphere (##2/5 MR^2##).
Q11. In pure rolling motion, the work done by static friction is:
In pure rolling, the point of contact has zero instantaneous velocity. Since work is ##F \cdot d## (or power is ##F \cdot v##), and ##v = 0## at the point of contact, static friction does no work.
Q12. A particle moves in a straight line with constant momentum. Does it have angular momentum relative to a fixed point not on the line of motion?
Angular momentum ##L = r \times p##. For a particle in straight-line motion, ##L = mvd##, where ##d## is the perpendicular distance (impact parameter) from the point to the line of motion. Since ##m##, ##v##, and ##d## are constant, ##L## is constant.
Q13. What happens to the rotational kinetic energy of a system if the moment of inertia is halved while the angular momentum remains constant?
Kinetic energy can be expressed as ##K = L^2 / (2I)##. If ##L## is constant and ##I## is halved, ##K## becomes ##L^2 / (2 \cdot I/2) = L^2 / I##, which is double the original energy.
Q14. A uniform rod of length ##L## and mass ##M## is pivoted at one end. What is its moment of inertia?
The moment of inertia of a rod about its center is ##1/12 ML^2##. Using the parallel axis theorem (##I = I_{cm} + Md^2##) with ##d = L/2##, we get ##1/12 ML^2 + M(L/2)^2 = 1/3 ML^2##.
Q15. A wheel of radius 0.5 m rolls without slipping at a linear speed of 2 m/s. What is its angular velocity?
For rolling without slipping, ##v = \omega R##. Therefore, ##\omega = v / R = 2 / 0.5 = 4## rad/s.
Q16. When a torque is applied to an object, the resulting change in angular momentum is proportional to:
The angular impulse-momentum theorem states that ##\Delta L = \tau \Delta t##. Thus, the change in angular momentum is proportional to the time interval.
Q17. A sphere is rolling without slipping on a horizontal surface. If the total kinetic energy is 70 J, what is the translational kinetic energy?
Total KE ##K = 1/2 Mv^2 + 1/2 I\omega^2##. For a sphere, ##K = 0.5 Mv^2 + 0.2 Mv^2 = 0.7 Mv^2##. If ##0.7 Mv^2 = 70##, then ##0.5 Mv^2 = 50## J.
Q18. Which force provides the torque necessary for a ball to roll down an incline without slipping?
Gravity acts through the center of mass (zero torque about CM), and the normal force also acts through the center (zero torque). Static friction acts at the surface, providing the torque needed for rotation.
Q19. If a planet moves in an elliptical orbit around the sun, which quantity remains constant?
Since the gravitational force acts along the line connecting the planet and the sun, the torque is zero. Therefore, the angular momentum of the planet relative to the sun is conserved.
Q20. A hoop of mass ##M## and radius ##R## rolls without slipping. What is its total kinetic energy if its center of mass speed is ##v##?
Total KE = ##1/2 Mv^2 + 1/2 I\omega^2##. For a hoop, ##I = MR^2##. So, ##KE = 1/2 Mv^2 + 1/2 (MR^2)(v/R)^2 = 1/2 Mv^2 + 1/2 Mv^2 = Mv^2##.
Q21. Two forces of equal magnitude act on a beam. Force A acts at the end perpendicular to the beam. Force B acts at the same point but at a 30-degree angle to the beam. Which creates more torque?
Torque is maximized when the force is perpendicular (##\sin(90) = 1##). Force B has a torque proportional to ##\sin(30) = 0.5##, which is half of the torque from Force A.
Q22. What is the angular momentum of a 2 kg point mass moving at 3 m/s in a circle of radius 4 m?
For a point mass in circular motion, ##L = mvr = 2 \cdot 3 \cdot 4 = 24## kg·m²/s.
Q23. A cylinder is released from rest at the top of an incline. If the incline is frictionless, the cylinder will:
Without friction, there is no torque acting on the cylinder about its center of mass. Therefore, it cannot start rotating and will simply slide down the incline.
Q24. The work done by a constant torque ##\tau## acting through an angular displacement ##\theta## is given by:
The rotational work formula is analogous to translational work (##W = Fd##). In rotation, ##W = \int \tau d\theta##, which for a constant torque is ##W = \tau \theta##.
Q25. How does the acceleration of a solid sphere rolling down an incline compare to its acceleration if it were sliding down a frictionless incline?
When rolling, some potential energy is converted to rotational kinetic energy, and the static friction force acts against the direction of motion. This results in a lower translational acceleration compared to sliding without friction.
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