Horizontal Circular Motion Problems
1) A 2.0 kg box is placed at the edge of a merry-go-round of radius of 6.0 m. The
coefficient of friction between the box and the merry-go-round is 0.30.
(a) Draw a free-body diagram of the box.
FN is the normal force, FW is the weight, and FCP is the centripetal force provided
by friction.
(b) Determine the speed at which the box slides off the edge.
m = 2.0 kg
r = 6.0 m
μ = 0.30
μ = Ff/FN = FCP/FN = mv2/r/mg
v = (μrg)1/2 = (0.30*6.0 m*9.80 m/s2)1/2 = 4.2 m/s
What is the significance of the mass canceling out?
The mass and the weight of the box can be any value, however large or small,
and the maximum velocity will always be 4.2 m/s. The velocity is only dependent
on μ, the radius of the circle, and the acceleration due to gravity.
(c) How would your answer to (b) change if the mass of the box was doubled?
Support your answer.
FCP = mv2/r = m(2v)2/r = 4mv2/r
Doubling the velocity will increase the centripetal force by a factor of 4.
2) The Rotor-Ride, a ride found at many amusement parks, consists of a hollow
cylindrical 2.5 m in radius. After the room rotates to a certain speed, the floor drops
away. A rider is left stationary because they are supported by static friction (μs).
Assume the rider has a mass of 75 kg and the coefficient of friction is 0.35.
(a) Calculate the speed required to prevent the rider from slipping down the wall
FN is the normal force which is the wall pushing on the rider towards the center of
the circle (centripetal force), Ff is the frictional force which is equal to Fw, which
keeps the rider from sliding down.
m = 75 kg
r = 2.5 m
μ = 0.35
FCP = FN
mv2/r = Ff/μ = Fw/μ = mg/μ
v = (gr/μ)1/2 = (9.80 m/s2*2.5 m/0.35)1/2 = 8.4 m/s
(b) Calculate the frequency of rotation corresponding to this speed
ω = v/r = 8.4 m/s/2.5 m = 3.4 rad/s
3) A car with a mass of 850. kg traveling at 18 m/s approaches a curve with a radius of
75.0 m.
(a) Determine the banking angle such that no friction is required between the car's
tires and the banked curve.
m = 850. kg
r = 75.0 m
v = 18 m/s
cosθ = FN/FN where θ is the angle between FN and the vertical dashed line.
FN = FNcosθ
FW = FNcosθ
(1)
sinθ = FCP/FN where θ is the angle between FN and the vertical dashed line.
FCP = FNsinθ
(2)
At this point, there are two unknowns, cos θ and sin θ. To eliminate one of the
unknowns, divide eq(2) by eq(1) giving:
FCP/FW = FNsinθ/FNcosθ
mv2/r/mg = tan θ
θ = tan-1(v2/rg) = tan-1((18 m/s)2/75.0 m/9.80 m/s2) = 24°
(b) Does this angle apply for all cars? Justify your answer.
The obvious answer is yes because any car must be able to negotiate the curve.
The more astute answer is that the mass cancels in the angle calculation meaning
that it is not a factor.
(c) Does the banking angle increase or decrease with an increase of speed? Justify
your answer.
The banking angle would have to increase with an increase in speed. Increasing the
speed increases the centripetal force, FCP, as evident in the mv2/r term. Using the
diagram, the only way FCP increases is by θ increasing.
(d) Does the banking angle increase or decrease with a decrease of radius? Justify your
answer.
Similar logic as in (c).
4) A 1.54 kg mass is attached to a 0.60 m rope and is swung in a horizontal circle. The
string makes an angle of 12° with the vertical and the speed remains constant.
(a) Determine the tension in the string.
m = 1.54 kg
r = 0.60 m
θ = 12.0°
sinθ' = FW/T = mg/T = 1.54 kg*9.80 m/s2/sin 78° = 15.4 N
(b) Determine the speed of the mass.
FCP = mv2/r
sinθ = FCP/F
v = (FCP*r/m)1/2 = (Fsinθ*r/m)1/2 = (15.4 N*sin 12°*0.80 m/1.54 kg)1/2 = 1.29 m/s
Note there are two inward forces, FCP and T, but only FCP is directed toward the
center of the circle.