Vertical Circular Motion Practice Exam SOLUTIONS
1. A ball with a mass m is fastened to a string and is swung in a vertical
circle. When the ball is at the highest point of the circle the tension in the
string is:
(A) mg
(B) mg + ma
(C) ma -mg (D) mg/ma
Fnet = Fg + FT
Fnet = mac
[Set Fnet equations equal to each other.]
Fg + FT = mac
FT = mac – mg
2. A 4.0 kg mass is attached to one end of a rope 2 m long. If the mass is swung in a
vertical circle from the free end of the rope, what is the tension in the rope when the
mass is at its highest point if it is moving with a speed of 5 m/s?
(A) 5.4 N
(B)10.8 N
(C) 50 N
(D) 65.4 N
Fnet = Fg + FT
Fnet = mac = mv2/r
[Set Fnet equations equal to each other.]
Fg + FT = mv2/r
FT = [mv2/r] - mg = [4*(52)/2] - (4*10) = 50 - 40 = 10 [would be 10.8 if g = 9.8 was used]
3. A ball of mass m is fastened to a string. The ball swings at constant speed in a vertical
circle of radius R with the other end of the string held fixed. Neglecting air resistance,
what is the difference between the string's tension at the bottom of the circle and at the
top of the circle?
(A) mg
(B) 2mg
(C) 4mg
(D) 8mg
Top of circle: FT = mac – mg [obtained from both problems 1 and 2 above]
Bottom of circle: FT – mg = mac
FT = mg + mac
[Top of circle] – [Bottom of circle] = -2mg
4. A child whirls a ball at the end of a rope in a vertical circle. Which of the following
statements is true?
(A) The speed of the ball is constant.
(B) The velocity of the ball is constant.
Vertical Circular Motion Practice Exam SOLUTIONS
(C) The magnitude of the ball’s acceleration is constant.
(D) The acceleration of the ball is directed radially inwards, toward the center of its
path.
5. A roller coaster car is on a track that forms a circular loop of radius R in the vertical
plane. If the car is to just maintain contact with track at the top of the loop, what is the
minimum value for its velocity at this point?
(A) gR
(B) 0.5gR
(C) g/R
(D) 2gR
(E) (gR)1/2
Fnet = Fg + FN
Fnet = mac
[Set Fnet equations equal to each other.]
Fg + FN = mac
[If car just barely makes contact with the track, FN = 0 for a slight moment]
Fg + FN = mv2/R
mg = mv2/R
gR = v2
𝑣=
𝑔𝑅
Vertical Circular Motion Practice Exam SOLUTIONS
1. A ball of mass M attached to a string of length L moves in a circle in a vertical
plane as shown above. At the top of the circular path, the tension in the string is
twice the weight of the ball. At the bottom, the ball just clears the ground. Air
resistance is negligible. Express all answers in terms of M, L, and g.
a. Determine the magnitude and direction of the net force on the ball when it is
at the top.
Fnet = Fg + FT
[Fg = Mg and the problem stated that “the tension is the string is twice the
weight of the ball so FT = 2*Mg]
Fnet = Mg + 2 Mg = 3 Mg
Both the force of gravity and the force of tension are pointing directly down
when the ball is at the top of the loop.
b. Determine the speed vo of the ball at the top.
From part (a), it was found that Fnet = 3Mg. Newton’s 2nd Law of motion
serves as the bridge to connect forces with centripetal acceleration and with
the speed.
Fnet = 3 Mg [from part (a)]
Fnet = mac = m(v2/r) [from Newton’s 2nd law of motion]
[Set the right hand sides of the equations equal to each other.]
3Mg = Mv2/L
3gL = v02 [Substitute the given values and solve for v02]
3𝑔𝐿 = 𝑣!
Vertical Circular Motion Practice Exam SOLUTIONS
c. What is the centripetal acceleration of the ball at the top? ac = v2/r
ac = v02/L = 3gL/L = 3g
[This could also be found by using the two equations for Fnet in part(b).
3 Mg = Mac
3g = ac]
d. If the string snaps at the top of the path, determine the time it takes the ball to
reach the ground.
The string snaps at a height of 2L above the ground, the diameter of the loop.
This becomes a free fall problem for a height of 2L.
The relevant equation is y = yi + vit + ½ at2 and the relevant variables are equal to
the following:
y = 0 [ends on the ground]
yi =2L
vi = 0 m/s [no y-component of velocity at the top of the loop]
a = -10 m/s2
When all the values are substituted, it looks like this:
0 = 2L +0 –5 t2
-2L = -5t2
−2𝐿 = −5𝑡 ! →
2𝐿
𝟐𝑳
= 𝑡 ! → =𝒕
5
𝟓