Rotational Equilibrium and Newton’s First Law in Rotational Form AP Physics C Mechanics MCQ – Exam Style Questions etc.
Rotational Equilibrium and Newton’s First Law in Rotational Form AP Physics C Mechanics MCQ
Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
Question
A wheel of 0.5 m radius rolls without slipping on a horizontal surface. The axle of the wheel advances at constant velocity, moving a distance of 20 m in 5 s. The angular speed of the wheel about its point of contact on the surface is
(A) \(2 radians.s^{-1}\)
(B) \(4radians.s^{-1}\)
(C) \(8 radians.s^{-1}\)
(D) \(16 radians.s^{-1}\)
(E) \(32 radians.s^{-1}\)
Answer/Explanation
Ans:C
Question
A disk of radius 0.1 m initially at rest undergoes an angular acceleration of 2.0 rad/ \(s^2\) . If the disk only rotates, find the total distance traveled by a point on the rim of the disk in 4.0 s.
(A) 0.4 m
(B) o.8 m
(C) 1.2 m
(D) 1.6 m
(E) 2.0 m
Answer/Explanation
Ans: D
Using the Big Five equation \(\Delta \theta = \frac{1}{2} (2 rad/s^2)(4.0s)^2 = 16 rad\)
Therefore, \(\Delta s = r\Delta \theta\) = ( 0.1 m)(16 rad) = 1.6 m. If you’re
worried about memorizing the Big Five equations, here’s
another method. The angular acceleration is 2.0 rad/\(s^2\) ,
which means it gets 2.0 rad/s faster every second. It starts
from rest. After 1 s, w = 2 rad/ s. After 2 s, w = 4 rad/ s. After 3 s, w = 6 rad/s. After 4.0 s, w = 8.o rad/s. That makes the
average angular velocity during these four seconds 4.0 rad/ s
(w increased linearly from o s to 8 s). Since the disk averages
4.0 rad/ s for 4 seconds, the angular distance traveled will be
16 rad, which brings you right back to \(\Delta s = r \Delta \theta\). One final alternative is to make a graph with w on the
vertical axis and t on the horizontal axis. The slope of an w
vs. t plot is a= 2.0 rad/\(s^2\) . The area under the graph from o
to 4 is the angular distance traveled by the point in the first
4.0 seconds.
Question
A ball with radius of 0.2 m rolls without slipping on a level surface. The center of mass of the ball moves at a constant velocity, moving a distance of 30 metres in 10 seconds. The angular speed of the ball about its point of contact on the surface is
(A) 0.6 rad/s
(B) 3 rad/s
(C) 8 rad/s
(D) 15 rad/s
(E) 60 rad/s
Answer/Explanation
Ans: D
The speed of the center of mass of the ball can be calculated using the equation, x = vt. Inserting the appropriate numbers gives
30m = v(10s)
v = 3 m/s
The angular speed of the ball about the contact point can be calculated using the equation, v = wr. The value for r is the distance from the contact point to the center of mass, which is the radius of the ball. Inserting the appropriate numbers gives
3 m/s = w (0.2 m)
w = 15 rad/s
Question
A uniform stick has length L. The moment of inertia about the center of the stick is Io. A particle of mass M is attached to one end of the stick. The moment of inertia of the combined system about the center of the stick is
(A) \(I_{0}+\frac{1}{4}ML^{2}\) (B) \(I_{0}+\frac{1}{2}ML^{2}\) (C) \(I_{0}+\frac{3}{4}ML^{2}\) (D) \(I_{0}+ML^{2}\) (E) \(I_{0}+\frac{4}{5}ML^{2}\)
Answer/Explanation
Ans:A
Solution: Itot = ΣI = I0 + IM = I0 + M(½L)2