Physics 141 Problem Set 2 Corrected Solutions Nataliya Yufa (2.6) Gravel mixer We want to find ω such that at least for a moment the particles are not stuck to the cylinder. This is most likely to happen at the top, where the normal force is in the direction of gravity. For the particle to leave the cylinder, N=0. Hence the y-component of the total force on the particle is mv 2 . r
(1)
v2 = mω 2 R, r
(2)
Fg + N = Fg = Using the relation v = ωR we obtain mg = m
r
g . (3) R Thus for all values of ω less than ωc the particles will not be stuck to the walls all the time. ⇒ ωc =
(2.9) Particle in a cone The particle happens to be moving in a horizontal circle inside a cone of half-angle θ. We want to find the radius r of the circle in terms of v0 , θ and g. Consider the x- and y-components of the net force acting on the mass. Since there’s no movement in the y-direction, Fy is zero, i.e. Fy = Ny − mg = 0 ⇒ Ny = mg. From the geometry we find that tan θ = Nx =
Ny , Nx
(4)
hence
Ny mg = . tan θ tan θ
(5)
Circular motion in the horizontal direction implies that Fx = Nx = and solving for r we have r=
mg mv02 = , tan θ r
vo2 tan θ . g 1
(6)
(7)
(2.12) Pulling out the tablecloth Initially, the glass is accelerated by the force of friction due to the tablecloth for some time tmax . In this time it will reach velocity vo vo =
Ff tmax = µgtmax . m
(8)
On the other hand, the glass will be slowed down by the friction due to the table (which, because the coefficients of friction happen to be the same for the table and the tablecloth, has the same magnitude as the frictional force of the tablecloth). Therefore, during both the acceleration and the deceleration the glass will travel exactly half of the available distance. We want the glass to slow down in a given distance d/2, thus q
µgd.
v0 =
(9)
Combine the two expressions for v0 to get q
v0 = µgtmax = s
⇒ tmax =
µgd = µg
s
d = µg
s
µgd
0.5f t 1 = √ s. 2 0.5 · 32f t/s 4 2
(10) (11)
(2.25) Shortest possible period Intuitively, the shortest period of rotation will occur when the two solid spheres are as close as possible to each other. Let’s try to prove this. Because gravity is responsible for the circular motion, we have Gm2 mv 2 = (12) 4r2 r s
⇒v= The period is given by
Gm . 4r
(13)
2πr 4π 3/2 =√ r . (14) v Gm Note that the smaller the radius, the shorter the period. However, we’re limited by the radius of the spheres, as they cannot come closer than their radius R. Hence the shortest R3/2 . possible period would be T = √4π Gm T =
(2.28) Car going around a bend
2
Let’s choose the x- and y-axis to be horizontal and vertical respectively, since the car should be moving in a horizontal surface and therefore there should be a non-zero force solely in the horizontal direction. Writing the components of the total force we have Fy = 0 = N cos θ − mg ± µN sin θ
(15)
and
mv 2 . (16) R Here the different signs correspond to the maximum and minimum velocities. It follows that mg N= . (17) cos θ ± µ sin θ Fx = N sin θ ∓ µN cos θ =
Substituting this into the expression for Fx we get an equation for v 2 : v2 =
gR (sin θ ∓ µ cos θ). cos θ ± µ sin θ
(18)
Writing this in a different form gR(sin θ − µ cos θ) gR(sin θ + µ cos θ) < v2 < . cos θ + µ sin θ cos θ − µ sin θ
(19)
(2.33) Particle on a rotating rod a) The trick to this problem is to realize that there is no force in the radial direction as ˆ Newton said F = ma, thus the rod is frictionless. Hence the total force is given by F~ = F θ. h
i
˙ θˆ , F θˆ = m (¨ r − rθ˙2 )ˆ r + (rθ¨ + 2r˙ θ)
(20)
where θ˙ equals ω. It must be that the radial component of the acceleration is zero since the radial part of the force is zero: r¨ − ω 2 r = 0. (21) Now we need to find the solution to this differential equation. Let’s try the solution r = Ae−γt + Beγt . Then r¨ = γ 2 Ae−γt + γ 2 Beγt = γ 2 r.
(22)
In order to satisfy our differential equation we need to have r¨ = γ 2 r = ω 2 r,
(23)
and therefore γ = ω. b) By inspection, notice that Beωt will tend to positive or negative infinity, unless B is zero. Thus if we want the radius to be always decreasing without reaching zero, B must be zero. In all other cases where the particle doesn’t reach the origin, we must have a positive B. In that limit, we see that Aeωt will tend to zero for large t, while Beωt will go to infinity. 3
Physics 141 Problem Set 7 Eduard Antonyan
1
(7.2) Rotating flywheel on a rotating table
W
T
L0
w0
T
a
Figure 1: Side view. As it can be seen from the picture the torque is out of the picture and its magnitude is: τ = 4T αl,
(1)
where we assumed that the angle α is small. The derivative of the angular momentum That, and also its magnitude follows from: − → → − − → dL0 = Ω × L0 dt Thus τ =
dL0 dt
is also out the picture. (2)
= ΩL0 = ΩI0 ω0 , and therefore: α=
2
− → dL0 dt
I0 ω0 Ω . 4T l
(3)
(7.5) Car on a curve
→ → − − → − (a) Let us understand first why should the car tend to roll over. As one can see from the picture, N1 , f1 and f2 create − → torque (about the center of mass), that’s into the picture, and N2 creates torque out of the picture. If the car is stable, then the total torque should be zero. By Newton’s second law: f1 + f2 = M a = N1 + N2 = M g
1
M v2 r (4)
N2 d
N1
L
a f1
Mg
f2
Figure 2: View from behind. The car is turning to the left.
So the faster the car is moving the larger are f1 and f2 , and thus the torque into the page. To have equilibrium, N1 will have to decrease and N2 will have increase. For high enough velocities N1 will become zero and the car will start rolling over. Now the reason putting a spinning flywheel can help is because we can put it in such way that the torque will be used to change the angular momentum of the flywheel and not of the car! → − Now let’s consider the case where the car is turning to the left with angular velocity Ω which will point vertically → − upward. Let’s put the flywheel so that its angular momentum L is pointing radially outward (to the right on the → − → − − → picture). To make that angular momentum rotate with the car, one needs torque equal to ddtL = Ω × L , i.e. pointing forward with respect to the car, which is exactly what we’ve got. Note, however, that if the car is turning to the right, we should not reverse the direction of the angular momentum, i.e. if the car is turning to the right the angular momentum should point radially inward. This is because in that case → − Ω is pointing vertically downward, and the torque is pointing out of the page (backward w.r.t. the car). Thus the → − → − − → right choice for L would be pointing again to the right on the picture, so that Ω × L is again in the direction of the → − torque. (More mathematically the reason the direction of L doesn’t change is because torque, angular velocity and angular momentum are axial vectors, not real vectors). (b) If the loading on the wheels is the same, then N1 = N2 ≡ N , thus f1 = f2 ≡ f . Since we’ve already discussed directions of the derivative of the angular momentum and the torque, let’s just write down the scalar version of torque equation: dL τ= . dt Now τ = f1 d sin α + N1 d cos α + f2 d sin α − N2 dcosα = 2f d sin α = 2f L = M vΩL,
the the (5) (6)
where in the last equality we used Newton’s second law (4). Finally for a disk-shaped flywheel using mR2 dL = ΩL = ΩIω = Ω ω, dt 2 we have: ω=
3
2M vL . mR2
(7)
(8)
(7.8) Deflecting hoop
(a) The torque induced by the force acting on the top of the hoop will be in the direction shown in the picture. Since → − → − τ = ddtL that means that the angular momentum will start rotating in the horizontal plane and if we look from the → top it will be rotating counterclockwise (− ω pointing vertically upward). So in the gyroscope approximation the force will not incline the plane of the hoop, but will instead deflect the line of rolling. 2
I Mv′
f
Mv
L t I (a)
(b)
Figure 3: (a) “Normal” view. (b) View from above.
By the conservation of momentum we have: → − → → M− v + I = M− v 0,
(9)
− where → v 0 is the velocity of the hoop after the tap. Thus the line of the rolling of the hoop will change by an angle: φ ≈ tan φ =
I . Mv
(10)
(b) The gyroscope approximation is valid if dL dt Lω, i.e. if the change in the angular momentum is small compared to the initial angular momentum. In our case this would mean that τmax = F b Lω = M b2 ω 2 = M v 2 .
(11)
Thus for the gyroscope approximation to hold, we need the peak applied force to satisfy: M v2 . b
(12)
ω1 2π 2 Hz = ≈ 0.21 s−1 . Q 60
(13)
F
4
(10.2) Oscillating spring
For the damping constant we have: γ= Then from
r ω0 =
we have: k=
5
mω02
=m
ω12
γ2 + 4
k , m
(0.21 s−1 )2 2 = 0.3kg (2π 2 Hz) + ≈ 47.377 N/kg. 4
(14)
(15)
(10.6) Falling masses and critical damping
(a) At the rest position we have equilibrium of forces, thus: M g = kx0 ,
(16)
where m is the mass of the platform and x0 is the distance from the initial position of the platform to its final position. Thus Mg k= = 980 N/kg. (17) x0 3
(b) The velocity of the mass just before it hits the platform v is determined from energy conservation: M gh =
M v2 , 2
(18)
where h is the height of the mass above the platform. Then we have an inelastic collision of the mass with the platform, so by momentum conservation: M v = (M + m)v 0 , (19) where v 0 is their velocity just after they stick together. q k ≈ 18.07 s−1 . Since we want critical damping we need γ = 2ω0 = 2 M +m Now the general equation of motion for critical damping is given by: γ
γ
x(t) = Ae− 2 t + Bte− 2 t .
(20)
Here x(t) is the height of the platform w.r.t. the final position. Since we want x(0) = x0 and coefficients A and B:
dx dt |0
= v 0 , we get for the
A = x0 γ M p 2gh + B=v + = 2 M +m 0
r
g M . x0 M + m
(21)
So finally the equation of motion is: q − xg t
x(t) = x0 e
0
+
M p 2gh + M +m
4
r
g M x0 M + m
!
q − xg t
te
0
.
(22)
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.012
Fall Term 2005 PROBLEM SET 8
Due: Friday, November 18 at 4:00pm. Reading: Finish Kleppner & Kolenkow, Chapter 6 0. Collaboration and discussion. Please give a brief statement at the top of your homework telling us the names of all the students with whom you discussed the homework problems. 1. Kleppner & Kolenkow, Problem 6.9 2. Kleppner & Kolenkow, Problem 6.11 3. Kleppner & Kolenkow, Problem 6.13 4. Kleppner & Kolenkow, Problem 6.18 5. Kleppner & Kolenkow, Problem 6.19 6. Kleppner & Kolenkow, Problem 6.23 7. Kleppner & Kolenkow, Problem 6.29 8. Kleppner & Kolenkow, Problem 6.30
9. Kleppner & Kolenkow, Problem 6.33
10. Kleppner & Kolenkow, Problem 6.35
