8.2 Curved Motion

				ri+1		UT
		−	r		AT	UN
			i	A
					AN	ri
					UT
	5						5
[ M ]					UN	[ M ]
	0						0
y						y
	−5						−5
	−10	−5		0	5	10
				x [ M ]

				221
				UT
			ri+1
	−	ri		UN
			AN	AT
			A
				ri
			UT
			UN
−10	−5	0	5	10
		x [ M ]

Fig. 8.3 Illustration of how the acceleration vector is decomposed into a tangential and normal component for a motion with increasing speed (left) and decreasing speed (right)

where the time derivative of the angle is called the angular velocity, ω. We will return to this description of rotational motion when we address rotations in Chap. 14.

8.2.1 Example: Acceleration of a Matchbox Car

Problem: A matchbox car is spun around a horizontal, semi-circular track with a radius of 20 cm. The car has an approximately constant speed v = 0.5 m/s throughout the turn. (a) Find the acceleration of the car during the motion. (b) If you make the circle half the diameter, will the acceleration of the car increase or decrease?

Solution: The car is moving with constant speed along a circular track. The motion of the car is therefore constrained to follow a circular path. Since the car is moving with constant speed, we argued above that the car only has a radial acceleration in the direction toward the center of the circle, and with a magnitude:

	v2		(0.5 m/s)2	= 1.25 m/s2 .
aN =		=			(8.20)
	R		0.2 m

The acceleration is inversely proportional to the radius of the circle. If we reduce the radius to one half, the acceleration will increase by a factor 2.

222

Fig. 8.4 Illustration of a rod of length L = 0.8 m rotating at a constant rate around an axis through one of the ends of the rod. The position of the rod is illustrated for various times given as fractions of T , the time of a complete revolution, called the period of the rotation. The point P is at a distance R from the rotation axis, and the circular path traveled by the point is illustrated by the dashed circle

8 Constrained Motion

	0.8					T
						7				T
						2			18
						.
						0		.
							0
	0.6
			T
			36
			.
				0
											T
	0.4									.09
										0
	0.2	T	45.0
M]	0										P		0.00 T
[
y										R
	−0.2		0
		T 54.
										0
	−0.4									.
											90	T
					0
				.
	−0.6		T	63
									0
						0			.
									81
						.
						7				T
						2
	−0.8					T
			−0.5			0				0.5
						x [M]

8.2.2 Example: Acceleration of a Rotating Rod

Problem: A rod of length L is rotating around an axis through one of its ends. The rod is rotating at a constant speed so that it takes a time T for the rod to complete a revolution. (a) Find the speed of a point P on the rod located a distance R from the rotation axis. (b) Find the acceleration of the point P .

Solution: The rotating rod is illustrated in Fig. 8.4. The point P is located a distance R from the rotation axis. Since the rod is rotating at a constant speed, the speed of the point P on the rod is also constant, and the average speed is equal to the instantaneous speed. We find the average speed over a complete revolution from the length the point has traveled during a complete revolution and the time a revolution takes. The rod completes a revolution in a time T , and the point P has moved a distance s = 2π R along the whole, closed circular loop. We find the average speed of P as:

v =	s	=	2π R	=	2π	R ,	(8.21)
	T		T		T

The quantity 2π/ T is often called the angular frequency, or, as we will see later, the angular velocity, and we usually use the symbol ω for this:

ω =	2π	,	(8.22)
	T
we can therefore write:
v = Rω .			(8.23)

8.2 Curved Motion

223

We find the acceleration of the point P by realizing that the point moves along a circular path of radius R and with a speed v. The acceleration is therefore directed in toward the center of the circle, which is also the rotation axis, and the magnitude of the acceleration is given by the centripetal acceleration term:

a = R =		R	= Rω2 = R	T		.	(8.24)
	v2	( Rω)2			2π	2

8.2.3 Example: Normal Acceleration in Circular Motion

Problem: A ball is spun in a string in a circular path with radius R = 1 m. (a) If the ball is spun at a constant speed of v = 2 m/s, find the tangential and normal acceleration of the ball. (b) If the ball starts at rest and increases its speed at a constant rate, d v/d t = 0.1 m/s2, find the tangential and normal accelerations of the ball as a function of time.

Solution (a): The acceleration of the ball has a tangential and a normal component:

aT =	d v	, aN =	v2		(8.25)
				.
	d t		ρ

Because the speed is constant, we have that d v/d t = 0, and the tangential acceleration is zero. The radius of curvature, ρ for a circular path with radius R is the radius of the circle, ρ = R. The tangential acceleration is therefore:

aN =	v2	=	(2 m/s)2	= 4 m/s2 .	(8.26)
	ρ		1 m

Solution (b): The speed of the ball is v(t ). We know that the speed of the ball is v(0) = 0 m/s when t = 0 s, and that d v/d t = 0.1 m/s2. We find the speed by integration:

v(t ) − v(0) =	0	d t d t = 0	0.1 m/s d t = 0.1 m/s 0	d t = 0.1 t m/s . (8.27)
	t	d v	t	t

We use this to find the tangential and normal acceleration:

aT

d v

0.1 m/s , aN

=

v2

0.1 t m/s2 2

0.01 t 2 m/s4 .

(8.28)

= d t

=

ρ

=

=

1 m

= aN when 0.1 s2

= 0.01 t 2, that is for t = t = √

s. When

We see that aT

10

t

< t the tangential acceleration is larger than the normal acceleration, and when

t

> t the normal acceleration is larger than the tangential acceleration.

224	8 Constrained Motion

Summary

Constrained motion:

•Constrained motion is motion along a given path, for example, a bead on a wire is constrained to follow the wire.

•We describe motion along a curve by the distance, s(t ), along the path, as a function of time, t .

•Mathematically, any motion r(t ) can be decomposed into a path, r(s), and motion along the path, s(t ).

Linear motion:

•In Linear motion, the motion of an object is restricted to a straight line.

•We orient the x -axis along the line, and describe the position of the object with x (t ) = s(t ).

•The velocity and acceleration of the object are also directed along the line: v(t ) =

(d s/d t ) = (d x /d t ), a(t ) = (d x /d t ).

x’

Y’ x’

0s

Y

x

5s 4s

3s 2s

1s

) s r(4

Circular motion:

•In circular motion the object is constrained to move along a circle of radius R.

•The velocity of the object is tangential to the circle (and normal to the radius): v(t ) = v uˆ T (t )

•For motion with constant speed, the acceleration is directed in toward the center of the circle.

•For motion with varying speed, the acceleration has both tangential and normal

components: a(t ) = (d v/d t ) uˆ T (t ) + (v2(t )/ R) uˆ N (t )

8.2 Curved Motion

UT 3s

UN

U

T

2s

U
N
R	U
	N

U T

1s

225

UT

AT

A

2s

AN

UN

General motion:

•We describe a constraining curve by the r(s), parameterized by the path length, s, along the curve. The position of an object is r(s(t )), where s(t ) is the position along the curve.

•The velocity of the object is tangential to the curve: v(t ) = v(t ) uˆ T (t )

•The acceleration of the object has both tangential and normal components: a(t ) = (d v/d t ) uˆ T (t ) + (v2(t )/ρ (t )) uˆ N (t ) where ρ (s(t )) is the radius of curvature of the curve at r(s(t )).

Exercises

Discussion Questions

8.1Vertical loop. You spin a ball in a string through a vertical loop, keeping the speed constant throughout. Where is the acceleration the greatest, at the bottom or at the top of the loop. Explain your answer.

8.2On the surface of the Earth. You drive around the world with an accurate accelerometer. Sketch the direction of the acceleration you measure: at the North Pole, in London, on the equator, in New Zealand, on the south pole.

8.3In the plane. If your motion is restricted to be along a flat plane, may your acceleration be out of the plane? Explain. If your motion is restricted to be on a

surface, is your acceleration restricted to be along the surface?

226	8 Constrained Motion

8.4 Loose branch. A small monkey is climbing far out on a branch when it suddenly breaks. The branch does not snap off, but start to rotate about the point where it is broken. The monkey clings to the branch. What is the direction of the acceleration of the monkey?

Problems

8.5Skier pulled up a slope. A skier is pulled up a hill with an inclination α with the horizontal. He is pulled with a constant acceleration of a = 2 m/s2 along the hill and starts from rest at the bottom of the hill.

(a) Find the speed, v(t ), of the skier measured along the slope as a function of time, t . (b) Find the position, s(t ), of the skier measured as a distance from the starting point after a time t .

(c) Find the position, r(t ), of the skier in the x y-coordinate system, where x is the horizontal axis and y is the vertical axis.

(d) Use the vector position, r(t ), to find the speed of the skier, and compare with the results you found above.

8.6Skiing down a slope. You are skiing down a planar skislope with an inclination

αwith the horizontal. Your acceleration down along the slope is a = g sin α. You start from a height h.

(a) Find your speed, v(t ), measured along the slope as a function of time, t . (b) Find your position, s(t ), along the slope as a function of time, t .

(c) Find your position, r(t ), relative to the point you started at.

(d) How long time does it take until you reach the ground at y = 0?

8.7Bead on a line. A bead is inserted onto a thin line with an inclination α with the vertical. When the bead is released, its acceleration along the line is a = g cos α. (a) Find the speed, v(t ), of the bead as a function of time.

(b) Find the position s(t ) of the bead along the line as a function of time, t . (c) Find the height, h(t ) of the bead as a function of time, t .

8.8Acceleration of 200 m sprinter. During a 200 m race, a sprinter is running with a speed of 10 m/s through the first curve. The length of the curve is 100 m.

(a) Find the radius, R, of the curve (it is a perfect half-circle).

(b) Find the magnitude and direction of the acceleration a of the sprinter.

8.9Velocity of point on helicopter rotor blade. The rotor blade of an helicopter has a radius of 5 m and is rotating 200 times a minute.

(a) What is the velocity of the outer tips of the rotor blade?

(b) What is the acceleration of the outer parts of the rotor blade?

8.10Turning a high-speed train. A high speed train holds a constant speed of 200 km/h. Your job is to design a 90◦ turn. Let us assume that you design the turn as a part of a circle.

8.2 Curved Motion

227

(a) Find an expression for the acceleration of the train while turning.

(b) How large must the radius of the circle be for the acceleration to be smaller than 0.1 g, where g = 9.8 m/s2?

(c) How long time does the turn take?

8.11Acceleration on the equator. You are standing on the equator of the Earth. The radius of the Earth is R = 6378 km.

(a) What is your velocity in space due to the rotation of the Earth?

(b) What is your acceleration due to the rotation of the Earth? How large is this compared to g, the acceleration of gravity.

8.12Artificial gravity in space travel. Your spaceship has been designed with a large rotating wheel to give an impression of gravity. The radius of the wheel is R = 50 m.

(a) How many rotation per minutes must the wheel execute for the acceleration at

the outer end of the wheel to correspond to the acceleration of gravity at the Earth, g = 9.8 m/s2?

(b) What is the difference in acceleration of your feet and your head if you are standing with your feet at the outer end of the rotating wheel? You can assume that you are approximately 2 m high.

8.13 Probe in tornado. A probe caught in a tornado is moving in a circular path in the horizontal plane with approximately constant speed. You have three observations of the position of the probe:

t	r
0.0 s	35.7 m i + 35.6 m j
1.0 s	12.2 m i + 49.3 m j
2.0 s	−14.6 m i + 44.9 m j

(a) Find the average acceleration of the probe.

(b) Find the center of the circle and the radius of the circle. You can use approximations as you see fit.

(c) Find an expression for the position of the probe as a function of time.

8.14 Bead on ring. A bead is attached to a ring of radius R. The ring is rotating n times per minute around an axis that intersects the ring at two positions and that also passes through the center of the ring, as illustrated in Fig. 8.5. The bead is attached at a position given by the angle θ .

(a) Find the velocity of the bead as a function of θ .

(b) Find the acceleration of the bead as a function of θ . Indicate the direction of the acceleration in the figure.

8.15 Acceleration of Jupiter (Open). Find the magnitude of the acceleration of Jupiter in its motion around the Sun.

228	8 Constrained Motion
Fig. 8.5 A bead attached to		z
a rotating ring
			R
		θ	x

8.16Car in a wire. High speed model cars are often run in circular paths by attaching

them to a wire. Here we address a car attached to a steel wire of length 8 m. The car starts from rest and accelerates with a tangential acceleration at = 0.5 m/s2.

(a) Find the speed of the car as a function of time.

(b) Find the radial acceleration of the car as function of time.

(c) At what speed is the radial acceleration 100 times larger than the tangential acceleration?

8.17Driven pendulum. A rigid pendulum consists of a 1 m long rod with a weight attached at the end. The motion of the pendulum is fixed by a motor driving the rod. The position of the weight along its circular path is s(t ):

s(t ) = A sin	2π t	,	(8.29)
	T

(a) Show that the position of the weight can be written as:

r(t ) = R	cos R ß + sin		R æ .	(8.30)
		s	s

What position on the rod does the origin correspond to? (b) What is the velocity of the weight?

(c) What is the speed of the weight? Where is the speed maximum, and where it is minimum?

(d) What is the acceleration of the weight?

(e) Decompose the acceleration into a tangential and a normal acceleration. Compare the two accelerations throughout the motion and comment on the results.