Centripetal Force
Practice
practice problem 1
 Determine the normal and friction forces at the four points labeled in the diagram below.
 at the bottom (and rising)
 halfway to the top
 at the top
 45° past the top
 Determine the minimum coefficient of static friction needed to complete the stunt as planned.
solution
The only forces present in this problem are weight (which always points down), normal (which always points toward the center of the loop), and friction (which is always tangential to the loop). The weight of the motorcycle never changes. The normal varies from a maximum at the bottom of the loop to a minimum at the top. As the motorcycle drives up the loop, the friction force acts along the direction of motion to keep gravity from slowing it down. As the motorcycle drives down the loop, the friction force acts opposite the direction of motion to keep gravity from speeding it up. The bike isn't speeding up or slowing down, but it is changing direction. This means the net force always points toward the center of motion.
Weight points down and normal points up, so the net force is their difference. The normal force points toward the center, so it should be given the positive value. The net force is the centripetal force.
∑F = ma N_{i} − mg = mv^{2} r N_{i} = m ⎛
⎜
⎝v^{2} + g ⎞
⎟
⎠r N_{i} = 250 kg ⎛
⎜
⎝(11 m/s)^{2} + (9.8 m/s^{2}) ⎞
⎟
⎠(6 m) N_{i} = 7490 N Friction isn't necessary since the motorcycle isn't accelerating horizontally.
f_{i} = 0 N
The normal force points horizontally, toward the center of the loop. There isn't anything to balance this force at this position. This is our net force — our centripetal force.
∑F = ma N_{ii} = mv^{2} r N_{ii} = (250 kg)(11 m/s)^{2} (6 m) N_{ii} = 5040 N Weight points down and normal points inward (toward the center of the loop). They aren't opposite anymore, so they can't cancel. The motorcycle isn't accelerating vertically at this moment, so there must be something to balance the weight. That's where friction steps in. Friction and weight are equal.
f_{ii} = W = mg
f_{ii} = (250 kg)(9.8 m/s^{2})
f_{ii} = 2450 NBoth weight and normal point down, so the net force is their sum.
∑F = ma N_{iii} + mg = mv^{2} r N_{iii} = m ⎛
⎜
⎝v^{2} − g ⎞
⎟
⎠r N_{iii} = 250 kg ⎛
⎜
⎝(11 m/s)^{2} − (9.8 m/s^{2}) ⎞
⎟
⎠(6 m) N_{iii} = 2590 N Once again, friction isn't necessary since the motorcycle isn't accelerating horizontally.
f_{iii} = 0 N
The normal and friction forces are at right angles to each other. This makes them "good" vectors. Normal points toward the center and contributes to the centripetal force. Friction is tangential to the circle and contributes nothing to the centripetal force. Weight is the "bad" vector. It needs to be broken up into components. The component pointing toward the center contributes to the centripetal force.
∑F = ma N_{iv} + mg cos θ = mv^{2} r N_{iv} = m ⎛
⎜
⎝v^{2} − g cos θ ⎞
⎟
⎠r N_{iv} = 250 kg ⎛
⎜
⎝(11 m/s)^{2} − (9.8 m/s^{2}) (cos 45°) ⎞
⎟
⎠(6 m) N_{iv} = 3310 N The component of the weight tangent to the track is balanced by the friction (so that the speed stays constant).
f_{iv} = W sin θ = mg sin θ
f_{iv} = (250 kg)(9.8 m/s^{2})(sin 45°)
f_{iv} = 1730 N
As I was solving this problem, I realized there is a conceptually easier way to solve it. What stays constant in this problem? Weight and centripetal force. What changes? Direction. Compute weight and centripetal force once, then combine them as appropriate to the location.
W = mg
W = (250 kg)(9.8 m/s^{2})
W = 2450 NF_{c} = mv^{2} r F_{c} = (250 kg)(11 m/s)^{2} (6 m) F_{c} = 5040 N Weight points down and normal points up. The net force (their difference) is the centripetal force.
∑F = ma
N_{i} − W = F_{c}
N_{i} = F_{c} + W
N_{i} = 5040 N + 2450 N
N_{i} = 7490 NFriction isn't necessary since the motorcycle isn't accelerating horizontally.
f_{i} = 0 N
Normal points toward the center, which makes it the centripetal force.
∑F = ma
N_{ii} = F_{c}
N_{ii} = 5040 NFriction counteracts weight to keep the motorcycle moving with constant speed around the loop.
f_{ii} = W
f_{ii} = 2450 NBoth weight and normal point down toward the center of the loop. The net force (their sum) is the centripetal force.
∑F = ma
N_{iii} + W = F_{c}
N_{iii} = F_{c} − W
N_{iii} = 5040 N − 2450 N
N_{iii} = 2590 NAgain, friction isn't necessary since the motorcycle isn't accelerating horizontally.
f_{iii} = 0 N
The position for this part of the problem was rigged so the two components would have the same magnitude and there'd be fewer things to calculate.
W_{∥} = W_{⊥} = mg sin θ = mg cos θ
W_{∥} = W_{⊥} = (250 kg)(9.8 m/s^{2})/√2
W_{∥} = W_{⊥} = 1730 NThe normal force and a component of the weight point toward the center of the loop (the component perpendicular to the loop), so they get to be the centripetal force.
N_{iv} + W_{⊥} = F_{c}
N_{iv} = F_{c} − W_{⊥}
N_{iv} = 5040 N − 1730 N
N_{iv} = 3310 NThe other component of the weight (the component parallel to the loop) is balanced by friction so the net force tangential to the loop is zero and the speed does not change.
f_{iv} = W_{∥}
f_{iv} = 1730 N
Static friction needs to be greatest when the motorcycle is vertical. The tires need to grip the loop to balance out the weight and keep the bike from accelerating tangential to the loop. At these two points (one where the bike is going up and the other where the bike is going down) friction equals weight and normal provides the centripetal force. I'll solve this both ways — once using the fancy, formal method…
μ_{min} = f = mg = gr N mv^{2}/r v^{2} μ_{min} = (9.8 m/s^{2})(6 m) (11 m/s)^{2} μ_{min} = 0.486 and again using the conceptually easier method…
μ_{min} = f = W N F_{c} μ_{min} = 2450 N 5040 N μ_{min} = 0.486
practice problem 2
 meters per second
 Earth days per rotation
 rotations per Earth year
solution
Use the centripetal acceleration equation and solve for speed. Substitute values for the acceleration due to gravity on Earth and the radius of the Earth's orbit (also known as an astronomical unit).
v = √a_{c}r ⇐ a_{c} = v^{2} r v = √ [(9.8 m/s^{2})(1.50 × 10^{11} m)] v = 1.21 × 10^{6} m/s Sounds fast, but things in space tend to move fast anyway. How does this compare to the yearly motion of the Earth around the Sun? That's the goal of the next parts of the question.
We'll solve this practice problem two ways. First we'll use the definition of speed and substitute the value calculated above and the distance traveled in one rotation (the circumference).
T = 2πr ⇐ v = ∆s = 2πr v ∆t T T = 2π(1.50 × 10^{11} m) 1.21 × 10^{6} m/s T = 780,000 s ≈ 9 days The fancier way is to start from the centripetal acceleration equation, replace speed with circumference over period, and simplify. This gives us an equation that some people like so much they memorize it.
a_{c} = v^{2} = (2πr/T)^{2} = 4π^{2}r r r T^{2} Solve for period and substitute.
T = 2π√ r a_{c} T = 2π√ 1.50 × 10^{11} m 9.8 m/s^{2} T = 780,000 s ≈ 9 days That's a quick "year" — if we use the astronomical definition of the year as the period of one trip around the Sun. How does it compare to a regular calendar year?
This problem is best solved by dimensional analysis.
n = 365 days/year 9 days/rotation n = 40.5 rotations/year
We will return to this problem in the section on power.
practice problem 3
The curve radii of modern highspeed systems result in dependence on the speed and the maximum possible superelevation of the guideway to compensate for the centrifugal forces occurring. The Transrapid's guideway can have a maximum superelevation of 12 degree (up to 16 degree in special cases) which allows smaller radii at higher speeds than in the case of conventional wheelonrail systems.
 Minimal radius: 350 m
 200 km/h:0705 m
 400 km/h: 2825 m
 500 km/h: 4415 m
Determine…
 the maximum centripetal acceleration (in m/s^{2} and g) implied by these specifications
 the speed limit (in m/s and km/h) on a curved section of track with the minimal radius
solution
Once you get past reading the awkward translation from German to English, this is a conceptually easy question. Set up a table like the one below and complete the missing parts. Use a graphing calculator as it eliminates the drudgery of repeated calculation.
maximum speed  radius  centripetal acceleration  

(km/h)  (m/s)  (m)  (m/s^{2})  (g) 
0350  
200  0705  
400  2825  
500  4415  
mean centripetal acceleration → 
Start by converting the maximum speeds from km/h to m/s.
⎡ ⎢ ⎣ 
m  =  km  1000 m  1 h  ⎤ ⎥ ⎦ 

s  h  1 km  3600 s 
Then apply the equation for centripetal acceleration.
a_{c} =  v^{2} 
r 
The resulting values will be in m/s^{2}. To convert to g, divide by the standard value for the acceleration due to gravity.
a_{c}[g] = a_{c}[m/s^{2}] ÷ 9.80665 m/s^{2}
This procedure gives a set of numbers that are reasonably close to one another: three values in m/s^{2} and three in g. Find the mean of both of these triplets. A partially completed table is provided below.
maximum speed  radius  centripetal acceleration  

(km/h)  (m/s)  (m)  (m/s^{2})  (g) 
0350  
200  056  0705  4.38  0.446 
400  111  2825  4.37  0.446 
500  139  4415  4.37  0.446 
mean centripetal acceleration →  4.37  0.446 
For the second part of this question, follow the logic of the first part in reverse order. Assume that the maximum permissible centripetal acceleration is the same for all curves, regardless of size. Use the mean value we just calculated to determine the speed limit on a curve with a 350 m radius.


v = √(4.372 m/s^{2})(350 m)  
v = 39.1 m/s  
Then convert from m/s to km/h.
39.1 m  1 km  3600 s  = 140 km/h  
1 s  1000 m  1 h 
Add these results to the table.
maximum speed  radius  centripetal acceleration  

(km/h)  (m/s)  (m)  (m/s^{2})  (g) 
140  039  0350  4.37  0.446 
200  056  0705  4.38  0.446 
400  111  0825  4.37  0.446 
500  139  4415  4.36  0.446 
mean centripetal acceleration →  4.37  0.446 
A more advanced technique is to solve the problem graphically.
Starting from the equation for centripetal acceleration.
a_{c} =  v^{2} 
r 
Makev^{2} the subject and compare to the equation for a straight line.
v^{2} = a_{c}r  ⇔  y = mx + b 
The equation is telling us we should put speed squared on they axis and radius on thex axis.
y =  v^{2} 
x =  r 
This makes the centripetal acceleration equal the slope of the line of best fit.
m =  slope = a_{c} 
a_{c} =  4.37 m/s^{2} 
a_{c} =  0.445 g 
b =  intercept 
b =  7.28 m^{2}/s^{2} 
b ≈  0 
The intercept value of 7.28 m^{2}/s^{2} is effectively equal to zero. Since the speed squared values are all on the order of several thousand, an intercept that's less than ten is "small" in comparison.
Use the coefficients from the line of best fit to find the speed limit on the minimum radius curve. Finish by converting the speed back to km/h from m/s.
y =  mx + b 
v^{2} =  a_{c}r + 0 
v^{2} =  (4.37 m/s^{2})(350 m) 
v =  39 m/s = 140 km/h 
practice problem 4
 Complete the first two columns using astronomical data from a reliable source. Be sure to specify the units used for each entry.
 Complete the last two columns using a calculator. Be sure to state your answers in SI units.
radius  period  speed  acceleration  

Moon  
Mercury  
Earth  
Pluto  
Sun 
solution
The data for the moon and planets came from a page in this book. The data for the Sun came from a page in The Physics Factbook. The distance from the Earth to the Sun is called an astronomical unit (au). I plan on discussing the au in the section of this book on miscellaneous units (1 au = 149.6 × 10^{6} km). I know you know the period of Earth's orbit. You may just have to pause and think for a second. The first two columns of the table are done.
radius  period  speed  acceleration  

Moon  384,400 km  27.32 days  
Mercury  57.91 × 10^{6} km  87.97 days  
Earth  1 au  1 year  
Pluto  5,906 × 10^{6} km  90,465 days  
Sun  30,000 light years  225 million years 
Speed is the rate of change of distance with time, which in circular motion means circumference divided by period.
v =  2πr 
T 
We are going to use this equation over and over again. Please watch those units. Convert to meters and seconds as appropriate.
Moon  


Mercury  

Earth  


Pluto  

Sun  

Now that we have all the speeds, we can compute the centripetal accelerations.
a_{c} =  v^{2} 
r 
Again we'll do it over and over again…
Moon  


Mercury  

Earth  


Pluto  

Sun  

The table is now ready for completion.
radius  period  speed  acceleration  

Moon  384,400 km  27.3217 days  001,023 m/s  0.00272 m/s^{2} 
Mercury  57.91 × 10^{6} km  87.969 days  047,900 m/s  0.03958 m/s^{2} 
Earth  1 au  1 year  029,800 m/s  0.00593 m/s^{2} 
Pluto  5906.38 × 10^{6} km  90,465 days  004,750 m/s  3.82 × 10^{−6} m/s^{2} 
Sun  30,000 light years  225 million years  251,000 m/s  2.22 × 10^{−10} m/s^{2} 