Question: (12 points, suggested time 25 minutes)
The left end of a rod of length d and rotational inertia I is attached to a frictionless horizontal surface by a frictionless pivot, as shown above. Point C marks the center (midpoint) of the rod. The rod is initially motionless but is free to rotate around the pivot. A student will slide a disk of mass mdisk toward the rod with velocity v0 perpendicular to the rod, and the disk will stick to the rod a distance x from the pivot. The student wants the rod-disk system to end up with as much angular speed as possible.
(a) Suppose the rod is much more massive than the disk. To give the rod as much angular speed as possible, should the student make the disk hit the rod to the left of point C, at point C, or to the right of point C ?
____ To the left of C ____ At C ____ To the right of C
Briefly explain your reasoning without manipulating equations.
(b) On the Internet, a student finds the following equation for the postcollision angular speed w of the rod in this situation: ω = \(\frac{m_{disk}xv_{0}}{I}\) . Regardless of whether this equation for angular speed is correct, does it agree with your qualitative reasoning in part (a) ? In other words, does this equation for ω have the expected dependence as reasoned in part (a) ?
____ Yes ____ No
Briefly explain your reasoning without deriving an equation for ω.
(c) Another student deriving an equation for the postcollision angular speed w of the rod makes a mistake and comes up with w = \(\frac{Ixv_{0}}{m_{disk}d^{4}}\) . Without deriving the correct equation, how can you tell that this equation is not plausible—in other words, that it does not make physical sense? Briefly explain your reasoning.
For parts (d) and (e), do NOT assume that the rod is much more massive than the disk.
(d) Immediately before colliding with the rod, the disk’s rotational inertia about the pivot is mdisk x2 and its angular momentum with respect to the pivot is mdisk v0x . Derive an equation for the postcollision angular speed ω of the rod. Express your answer in terms of d, mdisk , I, x, v0 , and physical constants, as appropriate.
(e) Consider the collision for which your equation in part (d) was derived, except now suppose the disk bounces backward off the rod instead of sticking to the rod. Is the postcollision angular speed of the rod when the disk bounces off it greater than, less than, or equal to the postcollision angular speed of the rod when the disk sticks to it?
____ Greater than ____ Less than ____ Equal to
Briefly explain your reasoning.
▶️Answer/Explanation
Ans:
(a)
__√__ To the right of C
Due to the rod being much more Massine than the disk, the collision can be approximated as an impulse defined to the rod. To maximize angular speed and thus angular momentum, the largest possible angular impulse should be delivered, so the lever ann should be maximized which happens when the collision point is to the right of C.
(b)
√ Yes
In this equation, increasing x, the lever arm, means increasing ω. This agrees with part (a) because the lever arm is longer when the disk hits to the right of C.
(c)
In this equation, increasing mdisk would decrease ω, all the remaining constant.
This is impossible because a longer disk delivers a longer impulse to the rod due to it having more momentum to begun with, thus measuring ω.
(d)
by conservation at angular momentum:
mdisk v0x = (I + mdisk x2) ω
ω = \(\frac{m_{disk}v_{0}x}{I + m_{disk}x^{2}}\)
(e)
√ Greater than
By conservation at angular momentum, total angular momentum is constant before and after collision. When the disk sticks, the disk’s final angular momentum is positive while it’s negative if it faunas back. Thus for the sum of the disk and rod’s angular momentum to be equal, the angular momentum and thus the angular speed of the rod must be greater if the disk faunas back.
Question
A billiard ball has mass M, radius R, and moment of inertia about the center of mass Ic = 2 MR²/5 The ball is struck by a cue stick along a horizontal line through the ball’s center of mass so that the ball initially
slides with a velocity vo as shown above. As the ball moves across the rough billiard table (coefficient of sliding friction μk), its motion gradually changes from pure translation through rolling with slipping to rolling without slipping.
a. Develop an expression for the linear velocity v of the center of the ball as a function of time while it is rolling with slipping.
b. Develop an expression for the angular velocity ω of the ball as a function of time while it is rolling with slipping.
c. Determine the time at which the ball begins to roll without slipping.
d. When the ball is struck it acquires an angular momentum about the fixed point P on the surface of the table.
During the subsequent motion the angular momentum about point P remains constant despite the frictional force. Explain why this is so.
▶️Answer/Explanation
Ans:
a. \(\sum F\)= ma; Ff = μFN;
– μMg = Ma
a = –μg
v = v0 + at
v = v0 – μgt
b. τ = Iα where the torque is provided by friction Ff = μMg
μMgR = (2MR2/5)α
α = (5μg/2R)
ω = ω0 + αt = (5μg/2R)t
c. Slipping stops when the tangential velocity si equal to the velocity of the center of mass, or the condition for
pure rolling has been met: v(t) = ω(t)R
v0 – μgt = R(5g/2R)t, which gives T = (2/7)(v0/μg)
d. Since the line of action of the frictional force passes through P, the net torque about point P is zero. Thus, the time rate of change of the angular momentum is zero and the angular momentum is constant.
Question
A bicycle wheel of mass M (assumed to be concentrated at its rim) and radius R is mounted horizontally so it may turn without friction on a vertical axle. A dart of mass mo is thrown with velocity vo as shown above and sticks in the tire.
a. If the wheel is initially at rest, find its angular velocity ω after the dart strikes.
b. In terms of the given quantities, determine the ratio:
\(\frac{final kinetic energy of the system}{initial kinetic energy of the system}\)
▶️Answer/Explanation
Ans:
a. Li = Lf
m0v0R sin θ = Iω
ω = m0v0R sin θ/I; I = (M + m0)R2
ω = m0v0 sin θ/(M + m0)R
b. Ki = ½ m0v02
Kf = ½ Iω2 = ½ (M + m0)R2(m0v0 sin θ/(M + m0)R)2
Kf/Ki = m0 sin2θ/(M + m0)