1. More Centripetal Force Examples Answers
1. A car of mass 1800 kg is traveling at 15 m/s around a
curve of radius 35 m. The coefficient of friction between
the tires and road is 1.0. Determine if the car
successfully rounds the curve.
m = kg v = 15 m/s r = 35 m μ = 1.0
For the car to round the curve, a centripetal force
is required
; friction provides the force necessary for cars to change direction
Is there enough friction in this case?

2. m = kg v = 15 m/s r = 35 m μ = 1.0
Centripetal force needed:
m v2
( 1800 kg )( 15 m/s )2
Fc = = r
35 m
Fc = N
Friction force present:
Ff = μ FN
= μ m g
= ( 1.0 )( 1800 kg )( 9.8 m/s2 )
Ff = N

3. Ff = N
Fc = N
Ff > Fc
Friction force is greater than the centripetal force
There is more than enough friction force available for
centripetal force, so car successfully rounds the curve

4. 2. A motorcycle and rider have a combined mass of 400 kg
and wish to traverse a curve of radius 25 m. If the
coefficient of friction between the tires and road is 0.80,
what is the maximum speed the motorcycle can go and still
successfully negotiate the curve?
Centripetal force provided by friction
Friction present:
Ff = μ FN
= μ m g
= ( 0.80 )( 400 kg )( 9.8 m/s2 )
Ff = N
So 3136 N available for centripetal force
Fc = N

5. Fc = N
m = 400 kg r = 25 m
m v2 r r
Fc = r
m v2 = r Fc m m
r Fc
v2 = m
r Fc
( 25 m )( 3136 N )
v = = m
400 kg
v = 14 m/s

6. 3. Show that the maximum speed to get around a curve is
independent of mass—that is, a dump truck has the same
maximum speed as a motorcycle, assuming the coefficient
of friction is the same for both vehicles.
Set centripetal force = friction force
Fc = Ff
m v2
Mass term
cancels out
= μ m g r r v2
= μ g r r
v2 = μ g r
Max speed to negotiate curve
of radius r
Independent
of mass
v = μ g r

7. Apply to Example 2 :
m = 400 kg r = 25 m μ = 0.80
v = μ g r
= ( 0.80 )( 9.8 m/s2 )( 25 m )
v = 14 m/s
Same result
as earlier

8. 4. A box of mass 12 kg is on the bed of a flatbed truck. It is
unsecured. The coefficient of friction between the box and
truck bed is The truck turns a corner of radius 40 m
at a speed of 25 mph. Does the box stay on the truck, or
does the box fly off?
m = 12 kg r = 40 m μ = v = 25 mph
v = μ g r
= ( 0.40 )( 9.8 m/s2 )( 40 m )
v = m/s
This is the max speed the truck can have and still have the box stay on the truck
Is 12.5 m/s greater than or less than 25 mph?
Need to convert 25 mph to m/s

9. Conversion factor: 1 mi/hr = 0.447 m/s
look it up!
25 mi/hr
0.447 m/s
= m/s
1 mi/hr
So truck (and box) are moving at 11.2 m/s
; max speed for the box to stay on truck is 12.5 m/s
Box stays on truck