Angular Momentum and Angular Impulse AP Physics C Mechanics MCQ – Exam Style Questions etc.
Angular Momentum and Angular Impulse AP Physics C Mechanics MCQ
Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
Question
A ball, rolling without slipping on a horizontal surface, encounters a frictionless, downward- sloping ramp, as shown above. Which of the following correctly describes the motion of the ball on the ramp?
(A) Constant translational speed with no angular speed
(B) Increasing translational speed with no angular speed
(C) Increasing translational speed with constant nonzero angular speed
(D) Increasing translational speed with decreasing angular speed
(E) Increasing translational speed with increasing angular speed
Answer/Explanation
Ans:C
Question
Two horizontal disks of mass M have the radii shown above. Disk A is attached to an axle of negligible mass spinning freely with angular velocity\( ω_0\) . Disk B, not attached to the axle and initially held at rest, is released and drops down onto disk A. When both disks spin together without slipping, the angular velocity \(ω_f\) of the disks is
(A)\(\frac{1}{3}\omega _0\)
(B)\(\frac{1}{2}\omega _0\)
(C)\(\frac{2}{3}\omega _0\)
(D)\(\frac{4}{5}\omega _0\)
(E)\(\frac{2}{\sqrt{5}}\omega _0\)
Answer/Explanation
Ans:D
Substituting into the equation for conservation of angular momentum yields
Question
The figure shows a view from above of two objects attached to the end of a rigid massless rod at rest on a frictionless table. When a force F is applied as shown, the resulting rotational acceleration
of the rod about its center of mass is kF/(mL). What is k?
(A) \(\frac{3}{8}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{5}{8}\)
(D) \(\frac{3}{4}\)
(E) \(\frac{5}{6}\)
Answer/Explanation
Ans: C
The center of mass of the system is at a distance of
\(y_{cm}=\frac{(m)(0) + (2m)(L)}{m +2m} = \frac{2}{3} L\)
below the mass m. With respect to this point, the clockwise torque produced by the force F has magnitude
\(\tau = rF = (\frac{2}{3}L – \frac{1}{4} L) F = \frac{5}{12} LF\)
Since the rotational inertia of the system about its center of mass is
\(I = \sum m_i r_i^2 = m (\frac{2}{3}L)^2 + (2m) (\frac{1}{3}L)^2 = \frac{2}{3} mL^2\)
the equation \(\tau = Ia\) becomes
\(\frac{5}{12} LF = (\frac{2}{3}mL^2) \alpha \Rightarrow \alpha = \frac{5}{8} \frac{F}{mL}\)
Question
A cylinder rotates with constant angular acceleration about a fixed axis. The cylinder’s moment of inertia about the axis is 4 kg m2 . At time t = 0 the cylinder is at rest. At time t = 2 seconds its angular velocity is 1 radian per second. What is the angular momentum of the cylinder at time t = 2 seconds?
(A) 1 kg m²/s (B) 2 kg m²/s (C) 3 kg m²/s (D) 4 kg m²/s (E) It cannot be determined without knowing the radius of the cylinder.
Answer/Explanation
Ans:D
Solution: L = Iω