Physics 101
Classical Physics
Spring 2010
Homework 4 Solutions
DUE IN CLASS ON FEBRUARY 11
1. A string is tied to a rock and the rock is swung in a circle of radius 2 m at a
constant linear speed of 3 m/s. The tension in the rope is mainted at 5 N. What
is the radial (centripetal) acceleration of the rock? (a) 1.67 m/s
2
,
(b) 4.5 m/s
2
,
(c) 5 m/s
2
, (d) 1.33 m/
s
2
, (e) cannot be determined without knowing the mass
of the rock.
a
c
=
v
2
r
=
(3 m/s)
2
2 m
= 4
.
5 m/s
2
2. A stone at the end of a string is twirled in a vertical circle.
If its speed is
kept constant, which of the following statements is true about the tension in the
string?
(a) It must be largest at the bottom, to support the stone against gravity and
also provide the centripetal acceleration of the stone.
(b) It must be largest at the top, because it is the sum of the centripetal accel
eration and gravity. (c) It must be the same at every point in the circle, or else
the string would break. (d) It is equal to the force of gravity at the bottom.
(e) It is equal to the inward centripetal force at the bottom.
At the top, the sum of the force due to gravity and the tension provides the centripetal
acceleration, but at the bottom the tension minus the force due to gravity provides the
centripetal acceleration.
Since the speed is kept constant, the centripetal acceleration is
constant, therefore the tension must be greater at the bottom.
Use the following information for the next 3 problems.
A toy ‘matchbox car’ is on a track that includes a looptheloop (like a roller
coaster).
3. If the car
just barely
makes it through the looptheloop, what forces act on it at
the top?
(a) Gravity and friction
(b) Gravity and the centripetal force
(c) Gravity and the centrifugal force
(d) Just gravity
(e) Just the normal force
Since the car just makes it, there is no normal force at the top, so gravity is the only force
acting on the car. Also, note that the normal force is equal to zero, so there is no frictional
force acting on the car.
4. If the mass of the car is 0.1 kg, the radius of the loop is 0.1 m, and it is traveling
at 2 m/s, what is the normal force exerted at the bottom?
(a) 4N, (b) 0N, (c) 3 N, (d) 1 N,
(e) 5 N
1
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F
N
=
mg
+
m
v
2
r
= 5 N
5. If the car is halfway up the loop, and is still traveling at 2 m/s, what is the
magnitude of the normal force?
(a) 4N
, (b) 0N, (c) 3 N, (d) 1 N, (e) 5 N
F
N
=
m
v
2
r
= 4 N
6. A stone of mass
2
kg is on a string and traveling in a horizontal circle.
The
tension in the string is
T
= 10
N.
(a) What is the centripetal acceleration of the ball?
Tension is the only force acting on the ball, and it is in the direction of the centripetal
acceleration. Therefore,
T
=
ma
c
⇒
a
c
=
T
m
=
10 N
2 kg
= 5 m/s
2
This is the magnitude of the centripetal acceleration, so the centripetal acceleration is
5 m/s
2
towards the center of the circle, or equivalently
~a
c
=

(5 m/s
2
)ˆ
r
.
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