AP Physics C Mechanics: 5.1 Torque and Rotational Statics – Exam Style questions with Answer- MCQ

Question

 The arm is held in the horizontal position and the hand is bent at the wrist so the fingers point up, as shown in the figure above. The torque exerted by the weight of the hand with respect to the shoulder is most nearly
(A) 6 N .m
(B) 10 N .m
(C) 30 N. m
(D) 60 N. m
(E) 70 N .m

Answer/Explanation

 

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,

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
\(M = \int dm = \int_{0}^{L} kx dx = kx^2 |_0^L = \frac{1}{2}kL^2\)

Question

 A homogeneous bar is lying on a flat table. Besides the gravitational and normal forces (which cancel), the bar is acted upon by exactly two other external forces, \(F_1\) and \(F_2\) , which are
parallel to the surface of the table. If the net force on the rod is zero, which one of the following is also true?
(A) The net torque on the bar must also be zero.
(B) The bar cannot accelerate translationally or rotationally.
(C) The bar can accelerate translationally if \(F_1\) and \(F_2\) are not
applied at the same point.
(D) The net torque will be zero if \(F_1\) and \(F_2\) are applied at the
same point.
(E) None of the above

Answer/Explanation’

Ans: D

Since \(F_{net}=F_1 + F_2 = 0\), the bar cannot accelerate translationally, so (C) is false. The net torque does not need to be zero, as the following diagram shows, eliminating (A) and (B).

However, since \(F_2 = -F_1\), (D) is true; one possible illustration of this given below:

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,
\(L.F_1 = \frac{1}{3}L. \frac{1}{2} Mg + \frac{1}{2}Mg + \frac{1}{2} L.Mg \Rightarrow F_1 = \frac{5}{6} Mg\)

Scroll to Top