Download Chap6. Circular Motion

Chap6. Circular Motion
Level : AP Physics
Teacher : Kim
Chap. 6 Newton’s 2nd Law Applied to Uniform Circular Motion
- According to Newton’s 1st law, an object in motion will move in a straight line at a constant speed
unless an unbalance force acts upon it
- If there is an unbalance force, continuously acting
toward a common center, then the motion of the
object is called circular motion
A ball tied to a string. The ball wants to move in a straight
line, but it is pulled towards a common center, resulting in
circular motion
- An acceleration of this type of motion is called
centripetal acceleration (or radial acceleration)
ar=
ar :radial acceleration
𝑣2
𝑟
string
where r is the radius of the circle
- Uniform Circular Motion; when the object is
moving with a constant speed while the direction of
the motion is continuously changing
ar exists not because there is a change in speed ‘v’ but
because the ball is changing direction continuously
- If we apply Newton’s 2nd law along the radial direction, we find
that the value of the net force causing the centripetal
acceleration(=radial acceleration) can be
ΣFr = mar =
*~velocity vector is always tangent to
the path
𝑚𝑣 2
𝑟
- The force that causes an object to have centripetal acceleration is
called the centripetal force
vf
Δv
- Centripetal force is not a new type of force. It is simply a force that
causes circular motion
Warm-up) Identify the centripetal force over the following examples
i) A ball tied to a string and spun over the head
ii) Satellite circling around a planet
iii) A car making a curve on a flat road
iv) A ball placed in a bucket and spun over your head
v) A car making a curve on slippery banked highway
vi
vi
vf
ΣFr = mar =
𝑚𝑣 2
𝑟
where ar=
𝑣2
𝑟
1. A 4kg mass on the end of a string rotates in a circular
motion on a horizontal frictionless table. The mass has a
constant speed of 2m/s and the radius of the circle is 0.8m.
i) Identify the centripetal force
ii) What is the magnitude of the centripetal force?
a) 28N
b) 25N
c) 20N
d) 16N
2. A car travels around an unbanked highway curve(radius 150m) at
a constant speed of 25m/s without slipping. The mass of the car is
2100kg.
i) What is the centripetal force?
Top view
ii) What is the magnitude of the centripetal force acting on the car?
a) 3400N
b) 5000N
c) 6820N
d) 8750N
iii) If the driver was 80kg, what is the magnitude of the resultant
centripetal force on the driver?
a) 333N
b) 240N
c) 184N
d) 121N
v
Fr
Rear view of the car
FN
Fr
Fg
iv) What are the possible centripetal forces acting on the driver?
3. A small ball is fastened to a string 0.24m long and suspended from a fixed
point P to make a conical pendulum. The ball describes a horizontal circle
about a center and the string makes an angle of 15° with the vertical.
i) Identify the centripetal force. (*~Remember that the centripetal always act
toward the center of the' plane' of motion)
Find the speed of the ball.
a) 1.02m/s
b) 0.912m/s
c) 0.740m/s
d) 0.404m/s
P
15°
ΣFr = mar =
𝑚𝑣 2
𝑟
where ar=
𝑣2
𝑟
4. The figure below shows a rear view of the car on the banked curved roadway. The designated speed for
the car to round the curve without slipping is v=13.4m/s and the radius of the curve is 35.0m. If the car
does not rely on friction, at what angle should the curve be banked?
θ
Non-uniform Circular Motion
5. i) A rock is attached to a string swings in a vertical circle. At the highest point
a) Two force act on the rock, and their resultant is not zero
b) Only one force acts on the rock
c) Two force act on the rock, and their resultant is zero
d) No forces act on the rock
ii) If you spin the rock fast enough, there could be a point where the string snaps. If so, then
a) the string would likely snap when the rock is at the bottom
b) the string would likely snap when the rock is at the top
v
v
6. A 0.5kg mass attached to the end of a string swings in a vertical circle(r=2m). When the mass is at the
lowest point on the circle, the speed of the mass is 12m/s
i) What is the resultant force?
a) 12N
b) 24N
c) 36N
d) 42N
ii) What is the magnitude of the force of the string(=tension force) on the mass at the lowest position?
a) 13N
b) 25N
c) 37N
d) 41N
ΣFr = mar =
𝑚𝑣 2
𝑟
where ar=
𝑣2
𝑟
7. A 0.2kg object attached to the end of a string swings in a vertical circle(r = 80cm). At the top of the
circle the speed of the object is 4.5m/s. What is the magnitude of the tension in the string at this position?
a) 3.1N
b) 2.5N
c) 3.7N
d) 4.1N
Fictitious Force in Linear Motion see p.161
Review ex) 6.7