Circular Motion

All Lessons

I. Intro to Circular Motion

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Video Instruction Part 1

a) When an object is moving in a circle, at a constant speed, it is accelerating because it is

b) The direction of the velocity is to the circle.

Ex 1) In which direction will the car go when it hits the ice?

c) The direction of the force and acceleration vector:

$$a_c = \frac{v^2}{r}$$

a/r relationship? | a/v relationship?

Ex 2) Draw the velocity, acceleration and force vector arrows for all 4 positions

F = ma =

Ex 3) A car is moving with a constant velocity in a circle with a diameter of d. It takes the car 36 seconds to make 6 trips around the circle. Find the car's acceleration.

Ex 4) Calculate the speed an earth satellite must have to maintain a circular orbit at an altitude of $2.0 \times 10^5 \text{ m}$ ($200,000 \text{ m}$) where the acceleration due to gravity is $9.2 \text{ m/s}^2$ (radius of the earth is $6,400,000 \text{ m}$).

Teacher CER Key: Fundamentals

Missing Words: (a) constant speed | accelerating | changing direction. (b) tangent. (c) toward the center. Relationships: Inverse ($1/r$) | Square ($v^2$). Formula: $F_c = \frac{mv^2}{r}$.

Ex 3 Proof: Period $T = \frac{36 \text{ s}}{6 \text{ trips}} = 6.0 \text{ s}$. Radius $r = \frac{d}{2}$. Velocity $v = \frac{2\pi r}{T} = \frac{\pi d}{6.0 \text{ s}}$. Acceleration $a_c = \frac{v^2}{r} = \frac{(\pi d / 6.0 \text{ s})^2}{d / 2} = \frac{\pi^2 d}{18 \text{ s}^2}$.

Ex 4 Proof: Total radius $r = 6,400,000 \text{ m} + 200,000 \text{ m} = 6,600,000 \text{ m}$. Gravity is the centripetal acceleration ($a_c = g = 9.2 \text{ m/s}^2$). $a_c = \frac{v^2}{r} \implies 9.2 \text{ m/s}^2 = \frac{v^2}{6,600,000 \text{ m}}$. $v = \sqrt{(9.2 \text{ m/s}^2)(6,600,000 \text{ m})} \approx 7792 \text{ m/s}$.

II. Surface & Vertical Motion

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Circular Motion Horizontal Surface w/ Friction

Ex 5) A penny with a mass $M$ is placed at a distance $R$ from the center of a moving record with a coefficient of static friction of $\mu$. The radius of the record is $R$ and it moves with a speed of $V$.

a) What is the direction of the force on the penny?

b) What is the speed of the penny in terms of the variables given?

Mass on a String - Vertical Circle

Instructional Video

Mass $M$ attached to a string moves at a constant speed of $v$ in a vertical circle of radius $r$. Find string tension $T_{top}$ and $T_{bottom}$.

a) Tension Top - Make center of circle +

b) Tension at Bottom

Ex 7) What is the minimum speed needed to maintain the object's circular motion? What is $T_{top}$ at the minimum speed?

Roller Coaster

Ex 8) Car of mass $M$, circle radius $r$, speed $v$. Find Normal Force at top.

Interactive: Circular Motion Vectors (Velocity, Acceleration, and Force)

What minimum velocity is required for the roller coaster to prevent the passengers from falling off? Minimum $v$ occurs when

1. Find the normal force on the car.
b) Above what minimum speed will the car leave the road?

Teacher CER Key: Vertical Analysis

Ex 5: (a) Directed towards the center. (b) Static friction provides the centripetal force. $\mu F_N = \frac{MV^2}{R}$. Since $F_N = Mg$, $\mu Mg = \frac{MV^2}{R} \implies V = \sqrt{\mu g R}$.

Ex 6: Using $g = 10 \text{ m/s}^2$. At the top (down is positive): $T_{top} + Mg = \frac{Mv^2}{r} \implies T_{top} = M\left(\frac{v^2}{r} - 10 \text{ m/s}^2\right)$. At the bottom (up is positive): $T_{bot} - Mg = \frac{Mv^2}{r} \implies T_{bot} = M\left(\frac{v^2}{r} + 10 \text{ m/s}^2\right)$.

Ex 7 & 8: Minimum speed at the top occurs when the tension or normal force drops exactly to $0 \text{ N}$. Gravity provides the entire required centripetal force. $Mg = \frac{Mv^2}{r} \implies g = \frac{v^2}{r} \implies v = \sqrt{gr} = \sqrt{(10 \text{ m/s}^2)r}$.

III. Centripetal Force Problems

Solutions 1-3 Solutions 4-6

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Find the force needed to keep a $0.50 \text{ kg}$ rat circling $15$ times every $10. \text{ seconds}$, if it is spun in a circle with a radius of $70. \text{ cm}$.

Find the force of friction needed to keep a $3000. \text{ kg}$ car traveling on an exit ramp of $100. \text{ m}$ radius, with a speed of $22 \text{ m/s}$.

A passenger cresting a hill at $15 \text{ m/s}$ ($r=30\text{m}$). Force felt from seat? ($g=10\text{m/s}^2, M=50\text{kg}$).

Loop coaster ($r=40\text{m}$). What speed at top makes $60\text{kg}$ rider feel “weightless?”

Rotor ride ($56 \text{ rpm}, d=6.0\text{m}$). Force on $90. \text{ kg}$ rider?

How many g’s does a pilot feel when executing a $100. \text{ m}$ radius turn at $100. \text{ km/hr}$?

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Teacher CER Key: Calculations

Evidence (1 & 2): (1) Rat: Period $T = \frac{10 \text{ s}}{15 \text{ rev}} = 0.667 \text{ s}$. Velocity $v = \frac{2\pi(0.70 \text{ m})}{0.667 \text{ s}} \approx 6.6 \text{ m/s}$. Force $F_c = \frac{(0.50 \text{ kg})(6.6 \text{ m/s})^2}{0.70 \text{ m}} = 31.1 \text{ N}$.
(2) Car: $F_{friction} = F_c = \frac{(3000 \text{ kg})(22 \text{ m/s})^2}{100 \text{ m}} = 14,520 \text{ N}$.

Evidence (3 & 4): (3) Seat: Using $g = 10 \text{ m/s}^2$, $Mg - F_N = \frac{Mv^2}{r} \implies F_N = 500 \text{ N} - \frac{(50 \text{ kg})(15 \text{ m/s})^2}{30 \text{ m}} = 500 \text{ N} - 375 \text{ N} = 125 \text{ N}$.
(4) Coaster: Weightless when $F_N = 0 \text{ N} \implies v = \sqrt{gr} = \sqrt{(10 \text{ m/s}^2)(40 \text{ m})} = 20 \text{ m/s}$.

Evidence (5 & 6): (5) Rotor: $56 \text{ rev/min} \implies f = \frac{56}{60} \text{ Hz} \implies T \approx 1.07 \text{ s}$. Radius $r = 3.0 \text{ m}$. Velocity $v = \frac{2\pi(3.0 \text{ m})}{1.07 \text{ s}} \approx 17.6 \text{ m/s}$. Normal Force $F_N = F_c = \frac{(90 \text{ kg})(17.6 \text{ m/s})^2}{3.0 \text{ m}} \approx 9292.8 \text{ N}$.
(6) Pilot: $v = 100 \text{ km/hr} = 27.78 \text{ m/s}$. $a_c = \frac{(27.78 \text{ m/s})^2}{100 \text{ m}} = 7.72 \text{ m/s}^2$. Dividing by $g$ ($10 \text{ m/s}^2$) yields $0.77 \text{ g}$'s.

IV. Conical Pendulum

Instructional Video

Ex) $1.00\text{-meter}$ pendulum, $m$, $r=0.50\text{ m}$.

Draw the FBD.

Find $\theta$.
Find $F_{net}$ in y and x.
Find $v$.
Find Period $T$.

Teacher CER Key: Conical

Evidence: Given $L = 1.0 \text{ m}$ and $r = 0.50 \text{ m}$. Angle $\theta$ from vertical: $\sin\theta = \frac{r}{L} = 0.50 \implies \theta = 30^\circ$.
y-direction (equilibrium): $T \cos(30^\circ) = mg \implies T = \frac{mg}{\cos(30^\circ)}$.
x-direction (centripetal): $T \sin(30^\circ) = \frac{mv^2}{r}$.
Substitute $T$: $mg \tan(30^\circ) = \frac{mv^2}{r} \implies v = \sqrt{rg \tan(30^\circ)}$. Using $g = 10 \text{ m/s}^2$, $v = \sqrt{(0.50 \text{ m})(10 \text{ m/s}^2)(0.577)} \approx 1.7 \text{ m/s}$.

V. Concept Questions (MCQ)

Instructional Video I Instructional Video II

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1. A ball is fastened to a string and is swung in a vertical circle. When the ball is at the highest point of the circle its velocity and acceleration directions are:

(A)

(B)

(C)

(D)

(E)
2. A ball with a mass $m$ is fastened to a string and is swung in a vertical circle. When the ball is at the highest point of the circle the tension in the string is:

(A) $mg$

(B) $mg + ma$

(C) $ma - mg$

(D) $mg/ma$

(E) $ma/mg$
3. An object moves with uniform circular motion. Which arrow best depicts the net force acting on the object at the instant shown?

(A) A

(B) B

(C) C

(D) D

(E) E
4. A motorcyclist moves at a constant speed down one hill and up another hill along the smooth curved surface. When the motorcyclist reaches the lowest point of the curve its velocity and acceleration directions are:

(A)

(B)

(C)

(D)

(E)
5. A car moves along the curved track. What is the apparent weight of the driver when the car reaches the lowest point of the curve?

(A) $mg$

(B) $mg + ma$

(C) $ma - mg$

(D) $mg/ma$

(E) $ma/mg$
6. What is the direction of $F_N$ of the driver at the lowest point?

(A) Upward

(B) Downward

(C) Forward

(D) Backward

(E) No acceleration
7. A car is traveling on a road in hilly terrain. Assume the car has speed $v$ and the tops/bottoms have radius $R$. The driver is most likely to feel weightless:

A. at the top of a hill when $v = \sqrt{gR}$

B. at the bottom of a hill when $v > \sqrt{gR}$

C. going down a hill when $v = \sqrt{gR}$

D. at the top of a hill when $v < \sqrt{gR}$
8. What is the centripetal acceleration of a $0.2 \text{ kg}$ ball rotating at $3 \text{ m/s}$ on a $1.2 \text{ m}$ string?

(A) $1.2 \text{ m/s}^2$

(B) $3.0 \text{ m/s}^2$

(C) $7.5 \text{ m/s}^2$

(D) $3.2 \text{ m/s}^2$

(E) $2.4 \text{ m/s}^2$
9. What is the centripetal force exerted on the object in #8?

(A) $1.0 \text{ N}$

(B) $1.2 \text{ N}$

(C) $0.2 \text{ N}$

(D) $1.5 \text{ N}$

(E) $3.0 \text{ N}$
10. When a student stands on a rotating table, the frictional force exerted on the student by the table is:

(A) greater than table-on-student

(B) less than table-on-student

(C) equal to table-on-student

(D) directed away from center

(E) zero at constant speed
11. A child whirls a ball in uniform circular motion. Which is NOT true?

(A) Speed is constant

(B) Velocity is constant

(C) Radius is constant

(D) Acceleration magnitude is constant
12. Turntable overhead view, coin moves counterclockwise. Which vector represents frictional force?

(A)

(B)

(C)

(D)

(E)
13. Force $F$, speed $v$, radius $r$. If $F$ is constant but $r$ is halved, the speed becomes:

(A) $2v$

(B) $v/2$

(C) No change

(D) $v\sqrt{2}$

(E) $v/\sqrt{2}$
14. If radius is quadrupled and speed is doubled, centripetal force:

(A) Doubles

(B) Halves

(C) No change

(D) Quadruples

(E) Quartered
15. Snapshot of three cars on elliptical track. Smallest displacement?

(A) A

(B) B

(C) C

(D) None
16. Which car in #15 must have non-zero acceleration?

A

B

C

All three
17. Which car in #15 must have centripetal force directed to the center of curvature?

A

B

C

All three
18. Roller coaster loop radius $R$. To just maintain contact at top, minimum velocity is:

(A) $gR$

(B) $0.5gR$

(C) $g/R$

(D) $2gR$

(E) $(gR)^{1/2}$
19. Coin distance $r$ on turntable speed $v$. Minimum coefficient of static friction:

(A) $v^2rg$

(B) $v^2/rg$

(C) $rg/v^2$

(D) $v^2/r$

(E) $v^2g/r$
20. Car curve radius $r$ speed $v$ friction $\mu$. Maximum velocity to prevent skidding:

(A) $\mu rg$

(B) $\mu r/g$

(C) $(\mu rg)^{1/2}$

(D) $(\mu rg)^2$

(E) $g/\mu r$
21. Object $m$ moves on frictionless table connected to $M$ through hole. Value of $M$ if stays in equilibrium:

(A) $mv^2/rg$

(B) $v^2/rmg$

(C) $rg/mv^2$

(D) $mv^2r/g$

(E) $mg/rv^2$
22. Rough turntable $2\text{m}$ from center, $2\text{kg}$ object, $5\text{s}$ period. $\mu_k=0.5, \mu_s=0.8$. Force of friction?

(A) $19.6\text{N}$

(B) $16.0\text{N}$

(C) $9.8\text{N}$

(D) $6.3\text{N}$

(E) $0\text{N}$

MCQ Answer Key

1:B | 2:C | 3:B | 4:C | 5:B | 6:A | 7:A | 8:C | 9:D | 12:D | 13:E | 14:C | 15:A | 17:All | 18:E | 19:B | 20:C | 21:A | 22:D

(Click the options above to check your answers!)

VI. Free Response

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Determine minimum angle at which a frictionless road should be banked for $20.0 \text{ m/s}$ at radius $200.0 \text{ m}$. (Use $g=10 \text{ m/s}^2$).

Determine the velocity that a car should have while traveling around a frictionless curve of radius $100 \text{ m}$ and that is banked $20^\circ$.

If the road was not frictionless, which way would friction point if the car exceeded this velocity? Slower?

Radius $50 \text{ m}$, banked $25^\circ$, $\mu_s = 0.3$.
- Speed without friction.
- Max speed before sliding up.
- Min speed before sliding down.

Ball $M$, length $L$ in vertical plane. At top $T=2Mg$.
- Net force at top.
- Speed $v_o$ at top.
- Time to hit ground if breaks at top.

Ball $M$, $R$, clockwise vertical circle.
- Draw forces at P and Q.
- Derive $v_{min}$ at Z.
- Derive $v_{max}$ at Q without breaking ($T_{max}$).

$0.10\text{-kg}$ rubber ball, $0.80\text{-m}$ thread. $v_{top}=6.0 \text{ m/s}$.
- Mechanical energy.
- Speed at P.
- Tension at top and bottom.

Hand-held device swinger. $0.200 \text{ kg}$, radius $0.500 \text{ m}$.
- Explain speed determination.
- Work in one revolution?
- Tension at $3.7\text{m/s}$.
- Angle if horizontal is impossible.

String breaks at $v_o$, radius $R$, height $h$.
- Time hit ground.
- Horizontal distance.
- Draw forces acting on object.

Teacher CER Key: FRQ

Banked Curve: $\tan\theta = \frac{v^2}{rg} = \frac{(20.0 \text{ m/s})^2}{(200.0 \text{ m})(10 \text{ m/s}^2)} = \frac{400}{2000} = 0.20 \implies \theta = 11.3^\circ$.

Curve Velocity: $v = \sqrt{rg \tan\theta} = \sqrt{(100 \text{ m})(10 \text{ m/s}^2)\tan(20^\circ)} \approx 19.1 \text{ m/s}$.

Ball 5 Analysis: At top, $F_{net} = T_{top} + Mg$. Since $T_{top} = 2Mg$, $F_{net} = 3Mg$ downwards.
$3Mg = \frac{Mv^2}{L} \implies v = \sqrt{3gL}$.
If string breaks, it acts as a horizontal projectile from height $h = 2L$.
$\Delta y = \frac{1}{2}gt^2 \implies 2L = \frac{1}{2}gt^2 \implies t = 2\sqrt{\frac{L}{g}}$.