Rotational Kinematics AP Physics C Mechanics MCQ – Exam Style Questions etc.
Rotational Kinematics AP Physics C Mechanics MCQ
Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
Questions (a) , (b)
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.
Question(a)
What is the angular acceleration of the cylinder between t = 0 and t = 2 seconds?
(A) 0.5 radian/s² (B) 1 radian/s² (C) 2 radian/s² (D) 4 radian/s² (E) 5 radian/s²
Answer/Explanation
Ans:A
Solution: α=∆ω/∆t
Question(b)
What is the kinetic energy of the cylinder at time t = 2 seconds?
(A) 1 J (B) 2 J (C) 3 J (D) 4 J (E) cannot be determined without knowing the radius of the cylinder
Answer/Explanation
Ans:B
Solution: Krot = ½ Iω2
Question
The radius of a collapsing spinning star (assumed to be a uniform sphere with a constant mass) decreases to \(\frac{1}{16}\) of its initial value. What is the ratio of the final rotational kinetic energy to the initial rotational kinetic energy?
(A) 4
(B) 16
(C) \(16^2\)
(D) \(16^3\)
(E) \(16^4\)
Answer/Explanation
Ans: C
Since no external torques act on the star as it collapses (just like a skater when she pulls in her arms), angular momentum, Iw, is conserved, and the star’s rotational speed
increases. The moment of inertia of a sphere of mass m and radius r is given by the equation \(I = kmr^2 \) (with \(k = \frac{2}{5}\), but the actual value is irrelevant), so we have:
\(\omega _f = \frac{I_i}{I_f} \omega _i = \frac{kmr^2_i}{kmr^2_f}\omega_i = \frac{r^2_i}{r^2_f} \omega _i = \frac{r_i^2}{(\frac{1}{16}r_i)^2} \omega _i = 16^2 \omega_i\)
Therefore,
Questions (a) , (b)
A wheel with rotational inertia I is mounted on a fixed, frictionless axle. The angular speed ω of the wheel is increased from zero to ωf in a time interval T.
Question(a)
What is the average net torque on the wheel during this time interval?
(A) \(\frac{\omega _{f}}{T}\) (B) \(\frac{\omega _{f}}{T^{2}}\) (C) \(\frac{I\omega _{f}^{2}}{T}\) (D) \(\frac{I\omega _{f}}{T^{2}}\) (E) \(\frac{I\omega _{f}}{T}\)
Answer/Explanation
Ans:E
Solution: τ = ∆L/∆t = (Iωf – 0)/T