Physics 206 Examples with SolutionsCircular Motion and Gravity

Problem 1 . Consider the position vector,

r�

= − r sin(ωt) x − r cos (ω t) y

• Find the velocity vector.

v�

=dr

�

dt= − r ω cos(ωt) x + r ω sin(ωt) y

• Express the velocity vector in terms of r and/or θ .The magnitude of the velocity is

r ω

At t = 0 , r�

( 0) = − r y and v�

( 0) = − r ω x . This indicates that the particle is moving in a clockwisedirection so,

v�

= − r ω θ• Find the acceleration vector.

a�

=dv

�

dt= r ω2 sin(ω t) x + r ω2 cos(ω t) y

• Express the acceleration vector in terms of r and/or θ .

a�

= − ω2 r� − r ω2 r

Problem 2.

A) An object of mass 2 kg is moving in a circle of radius 2 m. At a given instant the object has aspeed of 3 m/s and an angular acceleration of 2 rad/s. Find the magnitude of the net force on theobject at this instant.

a�

= ar r + at θ

a�

=(− v2/r

)r + r α θ

| a� | = a =(v2/r) 2 +

(r α ) 2

√=

(3 2/ 2 ) 2 + ( 2 · 2) 2

√m/s2 = 6. 02 m/s2

F = m a = ( 2) ( 6. 02 ) N = 1 2 . 04 N

B) A car is rounding a banked curve which is part of a circle of radius 200 m. The curve is banked atan angle of 1 0 degrees. The car is not sliding and experiences no frictional force. What is the speedof the car?

r

r

y

r

~FS

y~N

~FG

θ

θ

θ

Consider the figure. There is to be no frictional force so F�

S = 0 . Summing the forces in the r -direc-tion gives,

− N sinθ = m ar = m ( − v2/r) ( 1 )

1

Summing the forces in the y-direction gives,

N cosθ − m gE = 0 ( 2 )

Solving ( 2) for N and using the result in ( 1 ) gives,( m gEcosθ

)sinθ = m v2/r

v = r gE tanθ√

= 1 8. 6 m/s ≈ 41 . 7 mph

Problem 3. Consider the position vector,

r�

= r sin(ωt) x − r cos(ω t) y

A) Find the magnitude of this position vector. Show your work.

r = ( r sin(ω t) )2

+ ( − r cos(ωt) ) 2

√= r sin2 (ωt) + cos2 (ωt)

√= r

B) Where is this particle at t = 0?

r�

( 0) = − r y

x

y

r�

( 0)

C) Express the position vector in terms of r and/or θ .

r�

= rr

D) Find the velocity vector. Show your work.

v�

=dr

�

dt= r ω cos(ωt) x + r ω sin(ωt) y

E) What is the velocity at t = 0?

v�

( 0) = r ω x

x

y

r�

( 0)

v�

( 0)

θ0 =3 π

2

F) What is the magnitude of the velocity vector? Show your work.

v = vx2 + vy

2√

= ( r ω cos(ωt) ) 2 + ( r ω sin(ω t) )2

√= r ω cos2 (ωt) + sin2 (ωt)

√= r ω

G) Express the velocity vector in terms of r and/or θ .This is uniform motion on a circle since the magnitude of the positikon vector is constant and thespeed is constant. S ince as shown in the figure above the particle is initially moving in the counter-clockwise direction. Since it has constant speed it and is on the circle it must conitnue to do so.The counter-clockwise direction is in the direction of θ so,

v�

= r ω θ

2

Another, perhaps more logically direct way to approach this problem is to note that ( see thefigure) ,

θ = θ0 + ωt =3 π

2+ ω t⇒ ωt = θ − 3 π

2

Using trig identities given on the supplements page:

cos(ωt) = cos( θ − 3 π/ 2) = cos( θ ) cos( − 3 π/2 ) − sin( θ ) sin( − 3π/2 ) = − sinθcos(ωt) = sin( θ ) cos( − 3π/ 2) + cos( θ ) sin( − 3π/ 2 ) = cosθ so

v�

= r ω ( cos(ωt) x + sin(ωt) y ) = r ω ( − sinθx + cosθ y ) = r ω θ

H) Find the acceleration vector. Show your work.

a�

=dv

�

dt= − r ω2 sin(ωt) x + r ω2 cos(ωt) y

I) What is the acceleration at t = 0?a

�

( 0) = r ω2 y

x

y

a�

( 0)

Note that the acceleration points radially inward.

J) What is the magnitude of the acceleration vector. Show your work.

a�

= − ω2 ( rsin(ωt) x + rcos(ωt) y ) = − ω2 r�

a =∣∣ a�∣∣ =

∣∣ − ω2 r� ∣∣ = ω2

∣∣ r�∣∣ = ω2r

K) Express the acceleration vector in terms of r and/or θ .a

�

= − ω2 ( rsin(ωt) x + rcos(ωt) y ) = − ω2 r�

= − rω2 r

Problem 4. On earth, when supported by a stationary surface, we experience a normal force that is aresponse to the gravitational pull of the earth. This normal force is equal to our weight and in some sense,our feeling of having weight is really due to this normal force. In free fall, for example, we still haveweight but we might feel weightless. In addition, our bodies have evolved to work properly when experi-encing this normal force. Astronauts in orbit are not subject to this normal force, they are falling aroundthe earth together with the the satellite they are in and do not experience a contact force between them-selves and the satellite. ( Both the satellite and the astronaut still have a weight in orbit, they are just infree fall around the earth together. ) Astronauts in orbit for a prolonged period experience a variety ofdeleterious effects due to the lack of a normal force. It has been proposed that using a giant rotating ringship of th1 e form shown below could be used to simulate gravity ( really the ship would provide the samenormal force we feel on earth. ) On earth the normal force is a response to the gravitational pull of theearth, in the ship the normal force provides the centripetal force needed to keep the astronaut moving in acircular path. Suppose a ship of outer radius 1 km was built.

A) What would be the needed angular speed in order that the astronauts experience the same normalforce that they experience on earth?On earth N = m gE . The normal force supplies the centripeatl acceleration for the people on therotating ship, so,

Fnet , r = m ar

− N = − m(r ω2

)

m gE = m r ω2

ω = gE/r√

= 9 . 8/ 1 000√

rad/ s = . 099 rad/s

3

B) How long would it take the ship to rotate once?

θ − θ0 = ωt⇒ t =θ − θ0

ω=

2 π

0. 099s = 63. 5 s

ω

r

Problem 5. A road goes over a hill. The top of the hill forms the top segment of a circle of radius 30 m.What is the greatest speed that a car could have without losing contact with the ground as it drives overthe hill. Ignore the effects of air resistance.Solution : As the car goes over the top of the hill the radial direction is upwards. The forces acting on thecar are the normal force upward and the gravitational force downward. These forces together mustprovide the needed radial acceleration if the car is to remain on the road. The car leaves the road if thenormal force is zero:

Fnet , r = m ar = − m v2/r

N − m gE = − m v2/r

Max speed whenN = 0

vmax = r gE√

= ( 30 ) ( 9. 8 )√

m/svmax = 1 7 m/s ≈ 38 mph

Problem 6. Consider the position vector and corresponding velocity vectors given as equations 9 . 1 and9. 4 in the text.

A) Show that r�

and v�

are perpendicular. ( Hint consider the dot product. )Solution

r� · v� = rx vx + ry vy

r� · v� = ( rcos(ωt) ) ( − rωsin(wt) ) + ( rsin(ωt) ) ( rωcos(ωt) ) = 0

r� · v� = 0 = r v cos( θrv )

θrv = 90◦

B) Show that v�

points in the direction of the counter-clockwise tangent to the circular path. ( Con-sider the cross product r

� × v� . )Solution : By the right hand rule if v

�

points in the counter-clockwise tangential direction then r� ×

v�

is proportional to z . S imilarly, if v�

points in the clockwise tangential direction then r� × v� is pro-

portional to - z . So,

r� × v� = ( r cos(wt) x + r sin(ωt) y ) × ( − r ω sin(ωt) x + r ω cos(ωt) y )

r� × v� = r2 ω cos2

(ωt) ( x × y ) − r2 ω sin2

(ωt) ( y × x )

r� × v� = r2 ω

(cos2 (ω t) + sin2 (ωt)

)z

r� × v� = r2 ω z

4

So v�

points in the counter-clockwise tangential direction.

Problem 7. A block of mass 1 0 kg is attached to the end of the rod. The rod is attached to a post thatturns so as to turn the rod and move the block in a circle of radius 4 m. The block traces out a circle ona frictionless surface. See the figure. Assume that the block starts from rest. The bolt connecting the rodto the block is known to break if the rod exerts a force equal to 2000 N on the block.

A) What maximum initial acceleration could be given to the block without breaking the bolt?Solution : The acceleration of the block can be written as,

a�

= ar r + at θ = − r ω2 r + r α θ

Since the block starts from rest the initial angular velocity ω( t = 0) is zero. Therefore, initially theacceleration is just

a�

= r α θ

The force on the block is due to the rod and cannot exceed 2000 N. The maximum tangentialacceleration this force could supply is then

| F�

max | = Fmax = m | a� max | = m amax = m r αmax

αmax =Fmax

m r=

2000 N( 1 0 kg) ( 4 m)

= 50 rad/ s2

B) Assume that the block is accelerated from rest with a constant angular acceleration of α = 0. 5 rad/s. What will be the rotation rate of the block when the bolt breaks?Solution : In this case, from the last part,

a = ar2 + at

2√

= r2 ω4 + r2 α2√

| F�

max | = Fmax = m | a� max | = m amax = m r2 ωmax4 + r2 α2

√

Fmax2 = m2

(r2 ωmax

4 + r2 α2)

ωmax =

{Fmax

2 − m2 ( r2 α2)

m2 r2

} 1 /4

ωmax =

{20002 − 1 02 ( 42 0. 52

)

1 02 42

} 1 /4

rad/ s

ωmax = 7. 07 rad/s

Side View

Top View

(Different Scale)

5

Problem 8. ( 20 points) A curve of radius 1 50 m is banked at an angle of 1 0 degrees. An 800 kg cardrives around the curve with a speed of 85 km/hr without skidding.

A) Find the normal force exerted by the pavement on the car’ s tires.

B) Find the frictional force exerted by the pavement on the tires.

C) Find the minimum coefficient of static friction between the pavement and the tires.

Solution : . Use a the coordinate axes r pointing radially outward from the circle of motion, and y beingthe vertical direction. The forces acting on the car are the gravitational pull of the earth F

�

G , the normalforce due to contact with the incline N

�

and the static frictional force that keeps the car from sliding up ordown the incline.

r

r

y

r

~FS

y~N

~FG

θ

θ

θ

In the chosen coordinate system these forces can be written as

F�

G = − m gE y

N�

= − N sinθ r + N cosθ yF

�

S = FS cosθ r + FS sinθ y(FS could be + or − )

There is no acceleration in the vertical direction so applying Newton’ s second law in this direction gives,

Fnet , y = m ay = 0

− m gE + N cosθ + FS sinθ = 0 ( 3)

Applying the second law in the r-direction gives,

Fnet , r = m ar

− N sinθ + FS cosθ = − m v2/r ( 4)

A) Equations 1 and 2 represent two equations with two unknowns (N andFS ) . Solving ( 1 ) for N gives,

N = (m gE − FS sinθ ) /cosθUsing this in ( 2) gives,

− m gE tanθ + FS sinθ tanθ + FS cosθ = − m v2/r

FS = mgE tanθ − v2/r

sinθ tanθ + cosθ

Using the given information ( 85 km/hr = 23. 6 m/s) gives,

FS = − 1 560 N

Since FS is negative, using my definition, the static frictional force is directed down the incline.

6

B) From above,N = (m gE − FS sinθ ) /cosθ = 8240 N

C)

µS =| FS |N

= 0. 1 9

Problem 9. In a carnival ride, the passengers sit on a seat in a compartment that rotates with constantspeed in a vertical circle of radius r = 5 m. The heads of the seated passengers always point toward thecenter of the circle. Find the slowest rotation rate for which the seat belt need exert no force on the pas-senger at the top of the ride.If the seat belt does not help then the radial force must be supplied by gravity alone. In this case,

Fnet , r = m ar

− m gE = − m v2/r

v = gE r√

= 9 . 8 · 5√

m/sv = 7 m/s

ω = v/r = 1 . 4 rad/ s

Problem 10. Starting from Newton’ s second law of motion and the relation ar = − v2/r show that theperiod ( time for one orbit) of a satellite in circular orbit around the earth at an altitude of h above theearth’ s surface is

T =2 π

(RE + h ) 3/ 2

GME

√

where ME and RE are, respectively, the mass and radius of the earth. Show all of the steps you need todo to arrive at this result.

Solution : In orbit the only force is the inward gravitational pull. Summing the forces along the radialdirection gives,

Fnet , rad ial = m ar

− GMEm

r2= m

(− v2/r

)

GME

r= v2

v =GME

r

√=

2 π r

T

T=2 π r3/ 2

GME

√

T=2 π (RE + h )

3/ 2

GME

√

Problem 1 1 . Find the speed of a satellite in circular orbit around the moon if the radius of the orbit istwice the radius of the moon. See the useful information.

Solution : Using a relation developed in the solution to the previous problem:

v =GMmoon

r

√=

GMmoon

2 Rmoon

√= 1 1 90 m/s

Problem 1 2 .An object weighs 1 00 N on the surface of the earth.

A) What is the gravitational force that the earth exerts on this object when the object is at a positionthat is two earth radii from the center of the earth. See the figure.

F�

G =GMEm

( 2 RE ) 2 =1

4

GMEm

RE2 =

1

4WE = 25 N

7

B) What would be the weight of the object on the surface of a planet that had the same radius as theearth but was twice as massive as the earth.

F�

G =G ( 2ME ) m

RE2 = 2

GMEm

RE2 = 2 WE = 200 N

RE

2RE

RE

ME2ME

A) B)

ob ject

ob ject

Problem 13.

A) Sketch the path of an object in circular orbit with constant speed around the earth. At some pointin the circular orbit indicate the direction of the velocity and acceleration of the object.

v�

a�

B) Why doesn’ t a satellite in circular orbit around the earth fall and crash into the earth’ s surface dueto the gravitational pull of the earth? The satellite accelerates towards the earth but since it hasjust the proper tangential velocity it is pulled by the earth into a circular path. That is it is pulledtowards the earth but follows a “projectile motion” path that matches the curvature of the earth.

C) What is the range of launch speeds that can lead to orbits that come near the surface of the earth?Objects launched with speeds between 7900 m/s and 1 1 , 200 m/s could orbit the earth. ( Neglectingair resistance. )

Problem 14. Starting from an expression of conservation of energy show all the steps leading to theresult that the escape speed from the earth is given by the relation

vesc = 2 gE RE√

Solution : I wish to find the minimum launch speed that will allow the object to barely reach a pointwhich is “infinitely far” ( r � RE ) from the earth. Any speed larger that this will allow the object to reacha point “infinitely far” from the earth with some speed left over. Take configuration 1 to be when theobject is on the earth with the minimum escape speed and configuration 2 to be the situation in whichthe object has reached a point very far away from the earth with no speed left. ( In the limit that r→ ∞ ,the gravitational force tends to zero so that the object will remain at rest far from the earth if it reachesconfiguration 2 . ) Conservation of energy gives:

E1 = E2

K1 + UG 1 = K2 + UG2

1

2m vesc

2 − GMEm

RE=

1

2m v2

2 − GMEm

∞ = 0

vesc = 2GME

RE

√= 2 gE RE√

8

In the lass step I used the definition

gE ≡ GME

RE 2

Problem 1 5. Find the escape speed from the moon.The logic displayed in the previous solution leads to the conclusion,

vesc = 2GMM

RM

√= 2

6. 67 × 1 0− 1 1 × 7. 349 × 1 02 2

( 1 . 74 × 1 06 )

√= 2370 m/s

Problem 16.

A) Find the speed of a satellite in circular orbit a distance of 6 · 1 06 m above the surface of the earth.

Fr = m ar

− GMEm

(RE + h ) 2= m

(− v2

RE + h

)

v =GME

RE + h

√=

6. 67 × 1 0− 1 1 ×(5. 98 × 1 02 4

)

( 6. 37 + 6) × 1 06

√= 5680 m/s

B) What should be the radius of a circular orbit around the earth for a satellite to circle the earthonce every 1 0 days?The period of the orbit would be

T= 1 0 · 24 · 3600 s = 8 . 64 × 1 05 s

The angular velocity of the satellite would be,

ω =2 π

T= 7. 27 × 1 0− 6 rad/s

The second law gives,

Fr = m ar

− GMEm

r2= m

(− ω2r

)

r =

(GME

ω2

) 1 / 3

=

(6. 67 × 1 0− 1 1 × 5. 98 × 1 02 4

( 7. 27 × 1 0− 6 ) 2

) 1 / 3

= 1 . 96 × 1 08 m

Problem 17.

A) An object is launched radially away from the earth’ s surface with a speed of 9000 m/s from thesurface of the earth. What will be the speed of the object when it has reached a height of 1 07 mabove the earth’ s surface? (Neglect the effects of air resistance. )

RE

RE + h

1 2

In situation 1 :

v1 = 9000 m/s , r1 = RE

In situation 2 :

v2 = ? , r2 = RE + 1 07 m

9

Then,

E1 = E2

K1 + U1 = K2 + U2

1

2m v1

2 − GMEm

RE=

1

2m v2

2 − GMEm

r2

v22 = v1

2 − 2 GME

(1

RE− 1

r2

)

I’ ll let maxima do the arithmetic:

( %i 9) RE: 6 . 37e6$

( %i 1 1 ) ME: 5 . 98e24$

( %i 1 2) v1 : 9000$

( %i 1 3) r2 : RE+1 e7$

( %i 1 4) G: 6 . 67e- 1 1 $

( %i 1 5) v2 : sqrt( v1 ^ 2- 2* G*ME* ( 1 /RE- 1 /r2) ) ;

( %o1 6) 21 21 . 01 031 3941 964

The speed is v2 = 21 20 m/s.

B) An object is launched radially away from the earth’ s surface with a speed of 1 2000 m/s from thesurface of the earth. How high does the satellite go before it starts to fall back to earth? Showyour work starting from conservation of energy considerations. ( Neglect the effects of air resis-tance. )This speed exceeds the escape speed from the earth and so moves further and further from theearth never stopping. It does not reach a maximum height. The work to show this would be thework leading to the escape speed shown in the text.

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