Circular Motion AP Physics 1 MCQ – Exam Style Questions etc.
Vectors and Motion in Two Dimensions AP Physics 1 MCQ
Unit: 2. Force and Translational Dynamics
Weightage : 10-15%
Exam Style Practice Questions, Circular Motion AP Physics 1 MCQ
Question
A student holds one end of a string in a fixed position. A ball of mass 0.2 kg attached to the other end of the string moves in a horizontal circle of radius 0.5 m with a constant speed of 5 m/s. How much work is done on the ball by the string during each revolution?
(A) OJ
(B) 0.5 J
(C) 1.0J
(D) \(2\pi J\)
(E) \(5\pi J\)
Answer/Explanation
Ans:A
Question
The figure above shows a particle executing uniform circular motion in a circle of radius R. Light sources (not shown) cause shadows of the particle to be projected onto two mutually perpendicular screens. The positive directions for x and y along the screens are denoted by the arrows. When the shadow on Screen 1 is at position x = -(o.5)R and moving in the +x direction, what is true about the position and velocity of the shadow on Screen 2 at that same instant?
(A) y = -(o.866)R; velocity in -y direction
(B) y = -(o.866)R; velocity in +y direction
(C) y = -(o.5)R; velocity in -y direction
(D) y = +(o.866)R; velocity in -y direction
(E) y = +(o.866)R; velocity in +y direction
Answer/Explanation
Ans: A
From the diagram below,
if R cos \(\theta\) = -( o.5)R, then \(\theta\) = 240°. Therefore, y = R sin \(\theta\) =
R sin 240° = -(o.866)R. Also, it is clear that the particle’s
subsequent motion will cause the shadow on Screen 2 to
continue moving in the -y direction.
Question
In the figure shown, a tension force \(F_T\) causes a particle of mass m to move with constant angular speed w in a horizontal circular path (in a plane perpendicular to the page) of radius R. Which of the following expressions gives the magnitude of \(F_T\)? Ignore air resistance.
(A) \(m \omega ^2 R\)
(B) \(m\sqrt{\omega ^4 R^2 – g^2}\)
(C) \(m\sqrt{\omega ^4 R^2 + g^2}\)
(D) \(m(\omega ^2 R – g)\)
(E) \(m(\omega ^2 R + g)\)
Answer/Explanation
Ans: C
The figure below shows that \(F_T sin \theta = mg\) and \(F_T cos \theta = mv^2/R = mw^2 R:\)
We can eliminate \(\theta \) from these equations by remembering that \(sin^2 \theta + cos^2 \theta\) is always to 1:
\((\frac{mg}{F_T})^2 + (\frac{m \omega ^2 R}{F_T})^2 = 1 \Rightarrow \frac{m^2g^2 + m^2 \omega ^4 R^2}{F_T}=1 \Rightarrow F_T = m \sqrt{\omega ^4 R^2 + g^2}\)
That’s enough information to get you to the correct answer,
(C), but here’s another way to think about this problem.
If you look at the image of \(F_T sin \theta \) and \(F_T cos \theta\), you might recognize that this is a right triangle, to which you can apply the Pythagorean theorem. \(F_T, x = \frac{mv^2}{R}\) and \(F_T , y = mg\), which means that \(F_T ^2 = (mg)+(m \omega ^2 R) \rightarrow F^2_T = m^2(g^2 + \omega ^4 R^2) \rightarrow m\sqrt{g^2 + \omega^2 + \omega^4 R^2}\).
Question
A ball of mass m is attached to a light string, as shown earlier. The ball is whirled counterclockwise in a vertical circle at constant speed. At which labeled position is the tension in the string largest?
(A) A
(B) B
(C) C
(D) D
(E) None of the above; the tension is constant throughout.
Answer/Explanation
Ans:
D—The net force is ma, which in circular motion is \(mv^{2}/r\). Because speed, radius, and mass don’t change throughout the motion, the net force is constant. To find the net force, add or subtract the individual forces on the ball. At the peak, both the tension in the string and the weight pull down: \(F_{net}\) = T + mg. At the bottom, though, the tension pulls up while the weight pulls down: \(F_{net}\) = T – mg. With \(F_{net}\) and mg constants, the tension at the bottom must be greater—at the bottom, tension is \(F_{net}\) + mg, whereas at the top, tension is \(F_{net}\) – mg.
Question
In a carnival ride, people of mass m are whirled in a horizontal circle by a floorless cylindrical room of radius r, as shown in the diagram above. If the coefficient of friction between the people and the tube surface is \(\mu \), what minimum speed is necessary to keep the people from sliding down the walls?
(A) \(\sqrt{\mu rg}\)
(B) \(\sqrt{\frac{rg}{\mu }}\)
(C) \(\sqrt{\frac{\mu }{rg}}\)
(D) \(\sqrt{\frac{1}{\mu rg}}\)
(E) \(\sqrt{\mu mg}\)
Answer/Explanation
Ans:
B—The free-body diagram for a person includes \(F_{N}\) toward the center of the circle, mg down, and the force of friction up:
Because the person is not sliding down, \(mg=F_{f}\). And because the motion of the person is circular, the normal force is a centripetal force, so \(F_{N}=mv^{2}/r\). The force of friction by definition is \(\mu F_{N}\). Combining these equations, we have \(mg=\mu mv^{2}/r\); solve for v to get answer choice B. Note: Without any calculation, you could recognize that only choices A and B have units of speed, so you would have had a good chance at getting the answer right just by guessing one of these two!