Torque AP Physics C Mechanics MCQ – Exam Style Questions etc.
Torque AP Physics C Mechanics MCQ
Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
Question
The rod shown above can pivot about the point x = o and rotates in a plane perpendicular to the page. Its linear density,\(\lambda\) , increases with x such that \(\lambda \).(x) = kx, where k is a positive constant. Determine the rod’s moment of inertia in terms of its length, L, and its total mass, M.
(A) \(\frac{1}{6} ML^2\)
(B) \(\frac{1}{4}ML^2\)
(C) \(\frac{1}{3} ML^2\)
(D) \(\frac{1}{2} ML^2\)
(E) \(2ML^2\)
Answer/Explanation
Ans: D
Consider an infinitesimal slice of width dx at position x; its
mass is dm = \(\lambda\) dx = kx dx. Then, by definition of rotational
inertia,
\(I = \int r^2 dm = \int_{0}^{L}x^2 (kx dx)= \frac{1}{4}kx^4 |_0^L = \frac{1}{4}kL^4\)
Because the total mass of the rod is
\(M = \int dm = \int_{0}^{L} kx dx = kx^2 |_0^L = \frac{1}{2}kL^2\)
we see that \(I = \frac{1}{2}ML^2\).
Question
The figure above shows a uniform bar of mass \(\frac{1}{2} M\) resting on two supports. A block of mass M is placed on the bar twice as far from Support 2 as from Support 1. If \(F_1\) and \(F_2\) denote the downward forces on Support 2 as Support 1 and Support 2, respectively, what is the value of \(F_2/F_1\)?
(A) \(\frac{1}{2}\)
(B) \(\frac{2}{3}\)
(C) \(\frac{3}{4}\)
(D) \(\frac{4}{5}\)
(E) \(\frac{5}{6}\)
Answer/Explanation
Ans: D
First note that if Lis the total length of the bar, then the
distance of the block from Support 1 is \(\frac{1}{3} L\) and its distance from Support 2 is \(\frac{2}{3} L\). Let \(P_1\) denote the point at which Support 1 touches the bar. With respect to \(P_1\), the upward force exerted by Support 1, \(F_1\), produces no torque, but the
upward force exerted by Support 2, \(F_2\) , does. Since the net
torque must be zero if the system is in static equilibrium, the
counterclockwise torque of \(F_2\) must balance the total
clockwise torque produced by the weight of the block and of
the bar (which acts at the bar’s midpoint).
\(L.F_2 = \frac{1}{3} L. \frac{1}{3} L . \frac{1}{2}Mg + \frac{1}{2} L . Mg \Rightarrow F_2 = \frac{2}{3} Mg\)
Now, with respect \(P_2\), the point at which Support 2 touches the bar,
\(L.F_1 = \frac{1}{3}L. \frac{1}{2} Mg + \frac{1}{2}Mg + \frac{1}{2} L.Mg \Rightarrow F_1 = \frac{5}{6} Mg\)
These results give \(F_2/F_1 = (\frac{2}{3}Mg) / (\frac{5}{6}Mg) = \frac{4}{5}\)/
Question
A rod of negligible mass is pivoted at a point that is off-center, so that length l 1 is different from length l2. The figures above show two cases in which masses are suspended from the ends of the rod. In each case the unknown mass m is balanced by a known mass, M1 or M2, so that the rod remains horizontal. What is the value of m in terms of the known masses?
(A) M1 + M2 (B) 1/2 (M1 + M2) (C) M1 M2 (D) 1/2 M1 M2 (E) \(\sqrt{M_{1}M_{2}}\)
Answer/Explanation
Ans:
E
Solution: Applying rotational equilibrium to each diagram gives
Question
The figure above shows a square metal plate of side length 40 cm and uniform density, lying flat on a table. A force F of magnitude 10 N is applied at one of the corners, as shown. Determine the
torque produced by F relative to the center of rotation.
(A) o N•m
(B) 1.0 N•m
(C) 1.4N•m
(D) 2.0 N•m
(E) 4.0 N•m
Answer/Explanation
Ans: D
The center of rotation is the center of mass of the plate, which is at the geometric center of the square because the plate is homogeneous. Since the line of action of the force coincides with one of the sides of the square, the lever arm of
the force, l, is simply equal to \(\frac{1}{2}s\). Therefore,
\(t = lF = \frac{1}{2} sF = \frac{1}{2} (0.40m)(10 N) = 2.0 N.m\)