Solution for Homework 25 Gravitation II, Torque/Angular Momentum Problems
Solution to Homework Problem 25.1()
Problem:
Explain what each term in Newton’s Law of Universal Gravitation represents:

F
g
12

=
Gm
1
m
2
r
2
12
Solution

F
g
12

is the gravitational force exerted on an object of mass
m
2
by an object of mass
m
1
when the objects are separated
by a distance
r
12
(
vector
r
12
is the displacement vector between
m
1
and
m
2
), and
G
is the gravitational constant, determined
by experiment.
Total Points for Problem: 4 Points
Solution to Homework Problem 25.2()
Problem:
What can be said about the work done by a conservative force along a closed path?
Solution
The total work is zero. Remember that work is always done by a force
outside
the system. If we instead include the
object exerting the force in the system and talk about potential energy, then the potential energy an object has at
point
A
will always be the same, no matter what route the object takes to get there. These are basically two different
ways to say the same thing; the difference is which way the system is defined.
Total Points for Problem: 3 Points
Solution to Homework Problem 25.3()
Problem:
How must the kinetic energy of an orbiting object be related to its gravitational potential energy in order for the
orbit to be “bound”?
Solution
The body’s kinetic energy must be
less
than the magnitude of its gravitational potential energy, so that the total
energy is less than zero.
Total Points for Problem: 2 Points
Solution to Homework Problem 25.4()
Problem:
A
5kg
block is placed on a smooth surface. A light string is attached to it, and placed over a solid, uniform,
500g
pulley of radius
R
= 10cm
. An identical block hangs from the end of the string. The string, being magic physics
string, does not slip on the pulley, nor does it stretch.
(a)Draw free body diagrams for the blocks, and an extended free body diagram for the pulley.
(b)Determine the magnitude of the linear acceleration of the blocks.
Solution to Part (a)
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Grading Key: Part (a) 10 Points
1 point(s) : normal force and gravity balance for mass on surface (can be left out,
as long as neither is there).
1 point(s) : force of tension along string direction on block on surface.
1 point(s) : force of tension upward on hanging block
1 point(s) : force of gravity downward on hanging block
1 point(s) : force of gravity larger than force of tension
1 point(s) : tension forces on blocks are pairs with the ones on the pulley
1 point(s) : tension forces on pulley are not the same size, to cause the angular
rotation
1 point(s) : Any forces on the pulley besides the tension forces are balanced (for
instance, gravity downward would be balanced by a normal force on the pivot).
1 point(s) : direction of acceleration indicated
1 point(s) : direction taken as positive indicated
Solution to Part (b)
“Smooth surface” in a physics problem means negligible friction so we can ignore kinetic friction, and “light string”
means we can neglect the string’s mass, so that simplifies things a little. So we know that the torque on the pulley is:
τ
=
F
⊥
R
=
Iα
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 Spring '08
 Stewart
 Physics, Angular Momentum, Force, Momentum, Potential Energy, Work, rubber wheel

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