Download Winch & Satellite Mechanics: Energy Potential, Kinetic, & Conservation and more Exams Physics in PDF only on Docsity!
PHY6938 Proficiency Exam Spring 2003
March 28, 2003
Mechanics
- The drum of a winch has a mass M and a radius R. A cable wound around the drum suspends a load of mass m. The entire cable has a length L and a density (mass per unit length) λ, with a total mass mc = Lλ. The load starts from rest, at the bottom of the winch, and begins to fall toward the ground, unwinding the cable as it moves.
mass=M
R
m
d
(a) Write an expression for the change in potential energy of the load and the cable that has been paid out from the drum, between the initial situation and when the load has fallen a distance d.
To find the potential, it is useful to choose a reference level or point for which the potential is zero. This selection of the reference point is arbitrary. But it always helps to choose a point which make the solution easier. Below that point the potential energy has negative value and above that point the potential energy is positive. In this problem the center of the drum is chosen as the point with potential energy V = 0. With this choice the potential energy of the drum and the part of the cable wrapped around the drum will be zero. When the
V=
m
M
d=
R
load is at the highest level, its potential energy is given by
Vl−initial = −mgR. (1)
The initial potential energy of the cable is
Vc−initial = 0. (2)
When the load has fallen a distance d the final potential energy of the load is Vl−final = −mg(R + d). (3)
The final potential energy of the cable can be obtained as following. Since the cable is homogeneous the center of the mass of the part paid out from the drum is located at the center of this part.
Vc−final = −dλ
d 2
d^2 2 L
mc. (4)
The change in the potential energy for each part will be ∆Vl = Vl−final − Vl−initial = −mgd (5)
∆Vc = Vc−final − Vc−initial = −
d^2 2 L
= mc, (6)
and the total change in the potential is
∆Vtotal = ∆Vl + ∆Vc = −(m +
d^2 2 L
mc)g. (7)
(b) Write an expression for the kinetic energy of the cable and the load. Assume that the load has speed v and there is no slipping between the cable and the drum. Since the cable is connected to the load its speed will also be v. The kinetic energy of the load and the cable are given by
Tl =
mv^2 (8)
Tc =
dλv^2
d 2 L
mcv^2 , (9)
and the total kinetic of the drum and the cable is
Tl+c =
(m +
d L
mc)v^2. (10)
(c) Express the rotational kinetic energy of the winch in terms of I and ω and convert this expression in terms of M and v. Since there is no slipping between the drum and the cable, the points of the drum which are in contact with the cable will rotate with the same speed v. The rotational energy of the drum is
Td =
Iω^2. (11)
The moment of inertia I of the drum and its angular velocity are given by
I =
MR^2 (12)
ω =
v R
Substitute Eqs.12 and 13 in Eq.11 it yields
Td =
Mv^2 (14)
From Eqs.3, 4, and 5 1 2
mv 22 −
GMearthm v 1 v 2 rmax
mv 12 −
GMearthm rmax
v 22 − v^21 =
2 GMearth rmax
v 2 − v 1 v 1
From Eq.7 solve for rmax, it yields
rmax =
2 GM
v 1 (v 2 + v 1 )
rmin =
2 GM
v 2 (v 2 + v 1 )
- A wedge of mass M = 4. 5 kg sits on a horizontal surface. Another mass m = 2. 3 kg sits on the sloping side of the wedge. The incline is at an angle of 31. 7 ◦^ with respect to the horizontal. All surfaces are frictionless. The mass m is released from rest on mass M, which is also initially at rest. What are the accelerations of M and m once the mass is released?
m
M
θ
g
This problem can be solved in different ways. One of them is drawing free body diagram for each body, introducing a coordinate system, writing the Newton equations plus the constraints. But this can be complicated. Instead, we use Lagrange formalism to solve this problem. Introduce the generalized coordinates as in the figure below.
m
M
z
x
y
The Lagrangian for this system will be given by
L = Twedge + Tm − Vwedge − Vm. (1)
and
Twedge =
M x˙^22 (2)
Vwedge =
Mgy 2 = constant, since the wedge does not move in the y direction. (3)
Tm =
m( ˙x^21 + ˙y 12 ) (4)
Vm = mgy 1. (5)
But x 1 and y 1 can be written as following
x 1 = x 2 + z 1 cos(θ) (6) y 1 = z 1 sin(θ) (7)
Substitute Eqs.6 and 7 in Eqs.4 and 5, then replace T ’s and V ’s in Eq.1., and rename x 2 = x, z 1 = z for simplicity. The result is given by
L =
m[ ˙x^2 + ˙z^2 + 2 ˙x zcos˙ (θ)]
M x˙^2 − mgzsin(θ) (8)
The equations of motions are obtained by
d dt
∂L
∂ q˙
∂L
∂q
= 0, where q = x, z. (9)
We get
mx¨ + mzcos¨ (θ) + M x¨ = 0 (10) m¨z + mxcos¨ (θ) + mgsin(θ) = 0. (11)
From Eq.11 solve for ¨z, then substitute in Eq.
x¨ =
mgsin(θ)cos(θ) m + M − mcos^2 (θ)
Substituting the numerical values for different parameters, we get
¨x = 1. 9624
m s^2
¨z = − 6. 82
m s^2
To find the acceleration of the mass m, derivate Eqs.6 and 7 twice and use the numerical values from Eq.13 and 14
x¨ 1 = − 3. 85
m s^2
y¨ 1 = − 3. 58
m s^2
