Conservation of Angular Momentum AP Physics C Mechanics FRQ – Exam Style Questions etc.
Conservation of Angular Momentum AP Physics C Mechanics FRQ
Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
Question
3. A system consists of a small sphere of mass m and radius R at rest on a horizontal surface and a uniform rod of mass M = 2m and length attached at one end to a pivot with negligible friction, where R \(\ll \l ^{2}\). There is negligible friction between the surface and the sphere to the right of Point A and nonnegligible friction to the left of Point A. The rod is held horizontally as shown in Figure 1, then is released from rest. The total rotational inertia of the rod about the pivot is \(\frac{1}{3}M \l ^{2}\) and the rotational inertia of the sphere about its center is \(\frac{2}{5}m R ^{2}\). After the rod is released, the rod swings down and strikes the sphere head-on. As a result of this collision, the rod is
stopped, and the ball initially slides without rotating to the left across the horizontal surface.
(a) Derive an expression for the angular speed of the rod just before striking the sphere in terms of the length \(\l \) and physical constants as appropriate.
(b) Derive an expression for the linear speed \(v_{0}\) of the sphere immediately after colliding with the rod in terms of the length \(\l \) and physical constants as appropriate
After sliding a short distance, at time t = 0 the sphere encounters a region of the horizontal surface with a coefficient of kinetic friction \(\mu \), beginning at Point A as indicated in Figure 1. The sphere begins rotating while sliding and eventually begins rolling without sliding at Point B, also as indicated.
(c) In the following diagram, which represents the sphere while the sphere is traveling between Points A and B, draw and label the forces (not components) that act on the sphere. Each force must be represented by a distinct arrow starting on, and pointing away from, the point of application on the sphere.
(d) Derive an expression for each of the following as the sphere is rotating and sliding between points A and B in terms of \(v_{0}\), \(\mu \), R, t, and physical constants as appropriate.
- The linear velocity v of the center of mass of the sphere as a function of time t
- The angular velocity \(\omega \) of the sphere as a function of time t
(e)
- Derive an expression for the time it takes the sphere to travel from Point A to Point B in terms of \(v_{0}\), \(\omega \), and physical constants as appropriate.
- Derive an expression for the linear velocity of the sphere upon reaching Point B in terms of \(v_{0}\).
▶️Answer/Explanation
(a)Example Response
(a)Alternate Example Solution
(b) Example Solution
\(L_{i} = L_{f}\)
\(I\omega = mvr\)
\(\frac{2}{3} ml^{2}\sqrt{\frac{3g}{l}} =mv^{0}l\)
\(\therefore v_{0} = \sqrt{\frac{4}{3}gl}\)
Scoring Note: The last equation is not needed for scoring the item but is presented for clarity.
(c) Example Solution
Scoring note: Examples of appropriate labels for the force due to gravity include: \(F_{G}\), \(F_{g}\) , \(F_{grav}\) , W, mg, MG, “grav force”, “ F Earth on sphere” , “ F on sphere by Earth”, \(F_{Earth \\\on \\\sphere}\) , \(F_{E, sphere}\) , \(F_{sphere, E}\). The labels G or g are not appropriate labels for the force due to gravity. \(F_{n}\) , \(F_{N}\), N , “normal force”, “ground force”, or similar labels may be used for the normal force.
Scoring Note: A response that includes extraneous vectors can earn a maximum of 1 point.
Scoring Note: Horizontally displacing the \(F_{N}\) and \(F_{g}\) vectors slightly is permitted in order to show the distinct points at which those forces are exerted on the sphere.
(d)(i) Example Solution
\(\sum F = ma\)
\(-\mu mg = ma\)
\(a = -\mu g\)
\(v = v_{0}+at\)
\(\therefore v= v_{0} – \mu gt
Scoring Note: Only the final expression for velocity must have correct signs.
(d)(ii) Example Solution
(e)(i) Example Solution
Scoring Note: The last equation is not needed for scoring the item but is presented for clarity.
Question
A large circular disk of mass m and radius R is initially stationary on a horizontal icy surface. A person of mass m/2 stands on the edge of the disk. Without slipping on the disk, the person throws a large stone of mass m/20 horizontally at initial speed v0 from a height h above the ice in a radial direction, as shown in the figures above. The coefficient of friction between the disk and the ice is m . All velocities are measured relative to the ground. The time it takes to throw the stone is negligible. Express all algebraic answers in terms of m, R, v0 , h, m , and
fundamental constants, as appropriate.
(a) Derive an expression for the length of time it will take the stone to strike the ice.
(b) Assuming that the disk is free to slide on the ice, derive an expression for the speed of the disk and person immediately after the stone is thrown.
(c) Derive an expression for the time it will take the disk to stop sliding.
The person now stands on a similar disk of mass m and radius R that has a fixed pole through its center so that it can only rotate on the ice. The person throws the same stone horizontally in a tangential direction at initial speed v0 , as shown in the figure above. The rotational inertia of the disk is mR2/2 .
(d) Derive an expression for the angular speed ω of the disk immediately after the stone is thrown.
(e) The person now stands on the disk at rest R/2 from the center of the disk. The person now throws the stone horizontally with a speed v0 in the same direction as in part (d). Is the angular speed of the disk immediately after throwing the stone from this new position greater than, less than, or equal to the angular speed found in part (d) ?
____ Greater than ____ Less than ____ Equal to
Justify your answer.
Answer/Explanation
Ans:
(a)
\(\Delta X = V_{0}t + \frac{1}{2}at^{2}\)
\(h = 0 + \frac{1}{2}gt^{2}\)
\(\frac{2h}{g}= t^{2}\)
\(t = \sqrt{\frac{2h}{g}}\)
(b)
Pi = Pf (conservation A momentum)
\(0 = \frac{M}{20}(v_{0})+\left ( m+\frac{m}{2} \right )s\)
\(0 = \frac{M}{20}(v_{0})+\frac{3m}{2}(s)\)
\(\frac{1}{20}(v_{0})=\frac{3}{2}(s)\)
\(\frac{1}{30}(v_{0})=s\) Speed of disk and person = \(\frac{1}{30}v_{0}\)
(c)
Fnet = Ff
\(\left ( m + \frac{m}{2} \right )a = \mu \left ( m + \frac{m}{2} \right )g\)
a = μg
vf = vi – at
\(0 = \frac{1}{30}v_{0}- \mu gt\)
\(\mu gt = \frac{1}{30}v_{0}\)
\(t = \frac{\mu _{0}}{30\mu g}\)
(d)
Li = Lf v = wR \(w = \frac{v_{0}}{R}\)
\(0 = \left ( \frac{m}{20} \right )R^{2}w_{1}+\frac{1}{2}mR^{2}w_{2}+\left ( \frac{m}{2} \right )R^{2}w_{2}\)
\(0 = \left ( \frac{m}{20} \right )R^{2}\left ( \frac{v_{0}}{R} \right )+mR^{2}w_{2}\)
\(0 = \frac{mv_{0}R}{20}+mR^{2}w_{2}\)
\(\frac{v_{0}R}{20}= R^{2}w\)
\(w = \frac{v_{0}}{20R}\)
(e)
√ Less than
\(0 = \left ( \frac{m}{20} \right )\left ( \frac{R}{2} \right )^{2}\left ( \frac{v_{0}}{R} \right )+ \frac{1}{2}mR^{2}w_{2}+\left ( \frac{m}{2} \right )\left ( \frac{R}{2} \right )^{2}w_{2}\)
\(0 = \left ( \frac{m}{20} \right )\left ( \frac{R^{2}}{4} \right )\left ( \frac{v_{0}}{R} \right )+ \frac{1}{2}mR^{2}w_{2}+\frac{1}{4}mR^{2}w_{2}\)
\(0 = \frac{mRv_{0}}{80}+\frac{3}{4}mR^{2}w_{2}\)
\(\frac{3}{4}R^{2}w_{2}=\frac{Rv_{0}}{80}\)
\(Rw_{2}=\frac{Rv_{0}}{60}\)
\(w=\frac{Rv_{0}}{60R}\)