Circular Motion:  1
 2 
3  4  5  6

7  8  9  10
 11  12  13  14
 15  Go up
 by Chris Murray, 2004
1. What is the centripetal acceleration of a car going 21 m/s around a corner with a radius of 45 m?
Use a_{c} = v_{t}^{2}/r: v_{t} = 21 m/s, r = 45 m, so a_{c} = (21 m/s)^{2}/(45 m)
a_{c} = 9.8 m/s/s
(Table of contents)
2. What must be the radius of curvature for a curve in a road if cars going 27 m/s never exceed a lateral acceleration of 4.2 m/s/s?
Use a_{c} = v_{t}^{2}/r: v_{t} = 27 m/s, r = ?, a_{c} = 4.2 m/s/s, so 4.2 m/s/s = (27 m/s)^{2}/r
r = 173.6 m
(Table of contents)
3. My econobox pulls .72 g's (Multiply .72 by 9.8) of lateral acceleration going around a corner with a radius of 35 m. What is its speed?
a_{c} = (.72)(9.8 m/s/s) = 7.056 m/s/s
Use a_{c} = v_{t}^{2}/r: v_{t} = ?, r = 35 m, a_{c} = 7.056 m/s/s, so 7.056 m/s/s = v^{2}/(35 m)
v_{t} = 15.7 m/s
(Table of contents)
4. A centrifuge has a tangential velocity of 34 m/s and a radius of 12 cm. What is the centripetal acceleration at its edge?
Use a_{c} = v_{t}^{2}/r: v_{t} = 34 m/s, r = .12 m, so a_{c} = (34 m/s)^{2}/(.12 m)
a_{c} = 9633 m/s/s
(Table of contents)
5. A certain car can pull .92 g's of lateral acceleration. What radius turn can it do at 40 m/s?
a_{c} = (.92)(9.8 m/s/s) = 9.016 m/s/s
Use a_{c} = v_{t}^{2}/r: v_{t} = 40 m/s, r = ?, a_{c} = 9.016 m/s/s, so 9.016 m/s/s = (40 m/s)^{2}/r
r = 177.5 m
(Table of contents)
6. Your 1963 microbus can do about 4.5 m/s/s of centripetal acceleration. How fast dare you go around a corner with a radius of 60 m?
Use a_{c} = v_{t}^{2}/r: v_{t} = ?, r = 60 m, a_{c} = 4.5 m/s/s, so 4.5 m/s/s = v^{2}/(60 m)
v_{t} = 16.4 m/s
(Table of contents)
7. What centripetal force is needed to make a 5 Kg hammer swing 3.4 m/s in a arc with radius 1.75 m?
Use : F_{c} = mv_{t}^{2}/r: F_{c} = ?, v_{t} = 3.4 m/s, r = 1.75 m, m = 5 kg, so F_{c} = (5 kg)(3.4 m/s)^{2}/(1.75 m)
F_{c} = 33 N
(Table of contents)
8. What centripetal force is needed to make a 65 Kg rider go 15 m/s on the edge of a ride with a radius of 4.5 m?
Use : F_{c} = mv_{t}^{2}/r: F_{c} = ?, v_{t} = 15 m/s, r = 4.5 m, m = 65 kg, so F_{c} = (65 kg)(15 m/s)^{2}/(4.5 m)
F_{c} = 3250 N
(Table of contents)
9. A 320 Kg space probe has jets which can exert a centripetal force of 120 N. What is the sharpest radius of a turn it can make if it is going 520 m/s?
Use : F_{c} = mv_{t}^{2}/r: F_{c} = 120 N, v_{t} = 520 m/s, r = ??, m = 320 kg, so 120 N = (320 kg)(520 m/s)^{2}/r
r = 7.2 x 10^{5} m
(Table of contents)
10. How fast can your 800 Kg car go around a corner with a radius of 13 m when the available centripetal force is 6500 N?
Use : F_{c} = mv_{t}^{2}/r: F_{c} = 6500 N, v_{t} = ??, r = 13 m, m = 800 kg, so 6500 N = (800 kg)v^{2}/(13 m)
v = 10.3 m/s
(Table of contents)
11. A car can achieve a lateral acceleration of .95g. What would be its minimum time around a track with a radius of 23 m?
a_{c} = (.95)(9.8 m/s/s) = 9.31 m/s/s
Use a_{c} = 4p^{2}r/T^{2}: a_{c} = 9.31 m/s/s, r = 23 m, T = ??, so 9.31 m/s/s = 4p^{2}(23 m)/T^{2}
T = 9.9 s
(Table of contents)
12. What is the acceleration of the edge of a record that is 15 cm in radius, and is spinning at 45 rpm? (Revolutions per minute...convert it!!  you need to find seconds per revolution)
T = (60 sec/min)/(45 rev/min) = 1.333333 sec/rev
Use a_{c} = 4p^{2}r/T^{2}: a_{c} =??, r = .15 m, T = 1.3333 s, so a_{c} = 4p^{2}(.15 m)/(1.3333 s)^{2}
a_{c} = 3.33 m/s/s
(Table of contents)
13. What is the g force experienced at the tip of a centrifuge that is 5.4 cm from the center of rotation, and is spinning at a rate of 12,000 rpm? (Find ac, divide by 9.8 to find "g" force)
T = (60 sec/min)/(12,000 rev/min) = .005 sec/rev
Use a_{c} = 4p^{2}r/T^{2}: a_{c} =??, r = .054 m, T = .005 s, so a_{c} = 4p^{2}(.054 m)/(.005 s)^{2}
a_{c} = (85273.38203 m/s/s)/(9.8 m/s/s) = 8700 "g" s
(Table of contents)
14. There is a coefficient of friction of .34 between a 1200 kg car and the pavement. What is the force of friction between the road and the tires, and what is the car's maximum speed around a 73 m radius corner?
To do this we want to set the force of friction equal to the centripetal force as the friction in this case is what is making the car go around the corner. First lets find the friction force using:
F_{f} = µ_{k}F_{n } = µmg = (.34)(1200 kg)(9.8 N/kg) = 3998.4 N
Now we're in good shape, we finally use:
Use : F_{c} = mv_{t}^{2}/r: F_{c} = 3998.4 N, v_{t} = ??, r = 73 m, m = 1200 kg, so 3998.4 N = (1200 kg)v^{2}/(73 m)
v = 15.6 m/s
As a final note it is interesting that if we set the friction equal to the centripetal force without putting in numbers you can see that the mass will cancel out:
µmg = mv_{t}^{2}/r
µg = v_{t}^{2}/r
So that larger cars both require and have greater force to go around the corner in equal proportion.
(Table of contents)
15. A 2.0 gram penny is on a turntable 13 cm from the center. As you gradually speed up the turntable, the penny flies off when it is turning one revolution per second. What is the centripetal acceleration of the penny, and what is the coefficient of friction between the penny and the turntable? (Hint  set the force of friction equal to the centripetal force)
Lets find the centripetal acceleration first:
Use a_{c} = 4p^{2}r/T^{2}: a_{c} =??, r = .13 m, T = 1 s, so a_{c} = 4p^{2}(.13 m)/(1 s)^{2}
a_{c} = 5.132194289 = 5.13 m/s/s
Next let's find the centripetal force using
F = ma = (.002 kg)(5.1322 m/s/s) = 0.01026 N.
This force must be the force of friction as that is what is keeping the penny on the turntable. Finally, I am going to use the formula for friction to find the coefficient:
F_{f} = µ_{k}F_{n } = µmg
0.01026 N = µ(.002 kg)(9.8 N/kg)
µ = .52
(Table of contents)