1. Centripetal Example Problems Answers
1. r = 16.0 m v = m/s v2
( 20.0 m/s )2
ac = = r
16.0 m
400 m2/s2 =
16.0 m
ac = m/s2

2. ( ) ( ) 2. m = 2.0 kg r = 1.0 m T = 0.80 s (a) ac = ? d v2 v = t ac =
For circular path,
d = 2 π r and t = T 1 v2 = r
2 π r
( )
v = 1
2 π r 2 T = r T
( )
4 π2 r 1
4 π2 r2 =
ac = T2 r T2

3. 2. m = 2.0 kg
r = 1.0 m
T = s
(a) ac = ?
4 π2 r
ac = T2
4 π2 ( 1.0 m ) =
( 0.80 s )2
ac = m/s2

4. 2. m = 2.0 kg
r = 1.0 m
T = s
ac = m/s2
(b) Tension in string = ?

5. 2. m = 2.0 kg
r = 1.0 m
T = s
ac = m/s2
(b) Tension in string = force exerted by string
F = ?

6. 2. m = 2.0 kg
r = 1.0 m
T = s
ac = m/s2
(b) Tension in string = force exerted by string
F = m a
= ( 2.0 kg )( 61.7 m/s2 )
F = 123 N

7. v2
4 π2 r
ac =
ac = r T2
F = m a
m v2
4 π2 m r
Fc =
Fc = r T2

8. 2. m = 2.0 kg
r = 1.0 m
T = s
ac = m/s2
(b) Tension in string = force exerted by string
4 π2 m r
Fc = T2
4 π2 ( 2.0 kg )( 1.0 m ) =
( 0.80 s )2
Fc = 123 N

9. 3. m = x 102 kg T = s r = m
(a) F = ?
Car tires must exert a centripetal force on the car
in order to keep the car in circular motion
4 π2 m r
Fc = T2
4 π2 ( 6.00 x 102 kg )( 50.0 m ) =
( 10.0 s )2
Fc = N

10. 3. m = x 102 kg T = s r = m
(b) a = ?
4 π2 r
ac = T2
4 π2 ( 50.0 m ) =
( 10.0 s )2
ac = m/s2

11. 4. A force F is needed to keep an object of mass m moving
at a uniform speed v in a circular path of radius r.
(a) What force is necessary if the mass is doubled?
m v2
Use
Fc = r
A force F is needed to keep an object of mass m moving
at a uniform speed v in a circular path of radius r.
m v 2 So
F = r

12. ( ) m v 2 F = r (a) What force is necessary if the mass is doubled?
So m = 2 m
r = r v = v
m v2
Insert into
Fc =
to find new force F’ r ( )
( 2 m ) v 2 2
m v 2 =
2 F
F’ = = r r
So F’ = 2 F
New force is twice the old force;
force is doubled

13. ( ) m v 2 F = r (b) What force is necessary if the speed is doubled?
So v = 2 v
r = r m = m
m v2
m ( 2 v ) 2
Fc =
F’ = r r ( )
m 4 v 2 4
m v 2 =
4 F
F’ = = r r
New force is 4x the old force;
force is quadrupled
So F’ = 4 F

14. ( ) m v 2 F = r (c) What force is necessary if the radius is doubled?
So r = 2 r
v = v m = m ( )
m v2
m v 2 1
m v 2
Fc =
F’ = = r
2 r 2 r =
½ F
New force is half the old force;
force is halved
So F’ = ½ F

15. ( ) m v 2 F = r (d) speed and radius are both doubled? So v = 2 v
r = 2 r m = m
m v2
m ( 2 v ) 2
Fc =
F’ = r
2 r ( )
m 4 v 2 4
m v 2 =
2 F
F’ = =
2 r 2 r
New force is 2x the old force;
force is doubled
So F’ = 2 F

16. ( ) m v 2 F = r (e) mass, speed and radius are all doubled? So v = 2 v
r = 2 r m = 2 m
m v2
2 m ( 2 v ) 2
Fc =
F’ = r
2 r ( )
2 m 4 v 2 8
m v 2 =
4 F
F’ = =
2 r 2 r
New force is 4x the old force;
force is quadrupled
So F’ = 4 F

17. 5. m = kg r = 100 m laps in 240 s
(a) T = ?
T is measured in seconds per
revolution (lap)
240 s
20 laps in 240 s
T =
= T = s 20

18. 5. m = kg r = 100 m laps in 240 s
T = s
(b) ac = ? v2
4 π2 r
ac = OR
ac = r T2

19. 5. m = kg r = 100 m laps in 240 s
T = s
(b) ac = ?
4 π2 r
ac = T2
4 π2 ( 100 m ) =
( 12.0 s )2
ac = 27.4 m/s2

20. 5. m = kg r = 100 m laps in 240 s
T = s ac = m/s2
(c) Fc = ?
F = m a
Fc = m ac
Fc = ( 1000 kg )( 27.4 m/s2 )
Fc = N
Will get same answer if
values are inserted into
4 π2 m r
Fc = T2

21. 6. speed of rotation = 1600 km/hr
(a) Wt. of kg person = ?
Wt. = m g
= ( kg )( 9.8 m/s2 )
Wt. = 980 N

22. 6. speed of rotation = 1600 km/hr
m = kg Wt. = 980 N
(b) centripetal force on person = ?
m v2
Fc = r
v = speed of person = 1600 km/hr
Convert to m/s
1600 km
1000 m
1 hr
1 min
= m/s hr
1 km
60 min
60 s

23. 6. speed of rotation = 1600 km/hr = v = 444 m/s
m = kg Wt. = 980 N
(b) centripetal force on person = ?
m v2
Fc = r
r = radius of rotation
= radius of Earth
= x 106 m
( look it up! )

24. 6. speed of rotation = 1600 km/hr = v = 444 m/s
m = kg Wt. = 980 N
r = x 106 m
(b) centripetal force on person = ?
m v2
( kg )( 444 m/s )2
Fc = = r
6.37 x 106 m
Fc = N
So weight of person is much greater than the Fc
needed to keep person on Earth; gravity easily
provides enough Fc to keep person from being
thrown off into space