Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Chapter 7 P74 Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #)
978-049511374-3
THE PROBLEM STATEMENT
Ch 7 P74. A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal
plane, with the cord making a 30° angle with the vertical. (See Fig. P7.74.)
(a) Determine the ball's speed.
(b) If, instead, the ball is revolved so that its speed is 4.0 m/s, what angle
does the cord make with the vertical?
(c) If the cord can withstand a maximum tension of 9.8 N, what is the highest speed at which the
ball can move?
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Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
L
Ch 7 P74. A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal
plane, with the cord making a 30° angle with the vertical. (See Fig. P7.74.)
(a) Determine the ball's speed.
(b) If, instead, the ball is revolved so that its speed is 4.0 m/s, what angle does
the cord make with the vertical?
(c) If the cord can withstand a maximum tension of 9.8 N, what is the highest
R
speed at which the ball can move?
BRAINSTORMING-Definitions, concepts , principles and
Discussion
Given: m = 0.50 kg, L = 1.5 m,  = 30o.
Find: v, then  if v = 4.0 m/s, then v if Tmax = 9.8N
THE GEOMETRY OF THE SITUATION
The picture at the top right is the picture in the text with the
v
length, L, of the string and radius, R, of the horizontal circle drawn
over it.
O
R
The picture on the left shows;
ac
Horizontal Plane
View From The Top
at the top (a)
the Horizontal Plane, the view looking down from the above.
at the bottom (b)
the Vertical Plane,
the side view showing
the length L of the
string and the Tension
T along the string.
(a)
L

T
R
ac
Vertical Plane
Side View
(b)
(d)
(c)
string.
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The middle pictures
show
(c) the omponents of
the Tension along the
string and
(d) the components of
the Length of the
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 7 P74. A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal
plane, with the cord making a 30° angle with the vertical. (See Fig. P7.74.)
(a) Determine the ball's speed.
(b) If, instead, the ball is revolved so that its speed is 4.0 m/s, what angle
does the cord make with the vertical?
(c) If the cord can withstand a maximum tension of 9.8 N, what is the highest speed at which the
ball can move?
Basic Solution (including BRAINSTORMING-Definitions, concepts , principles and Discussion
Given: m = 0.50 kg, L = 1.5 m,  = 30o.
Find: v, then  if v = 4.0 m/s, then v if Tmax = 9.8N
v
O
R ac
Horizontal Plane
View From The Top
(e)
T

Tcos

Lcos
L
R
ac
Lsin Tsin
Vertical Plane showing
components of L and T
(f)
Some geometry:
1. R is needed for ac = mv2/R, R = Lsin 

2. Components of T Th = Tsin , Tv = Tcos

(2)
Newton’s 2nd Law
Horizontal circle (Horizontal direction)
(direction of ac +<-) Fh = ma = mac = mv2/R .
Th = mv2/R = mv2/ Lsin[see eq(1)]
So,
Tsin = mv2/ Lsin
(3)
And dividing both sides by singives,
T= mv2/ Lsin2
(4)
Vertical direction – in equilibrium.
(Up+) Fv = 0;
Tv – mg =0; from Eq. (2)
Tcos = mg
(5)
SOLUTION TO (a)
Eliminate T by dividing (4) by (5) ,
Tsin/Tcos = (mv2/ Lsin)/mg gives
(v2/Lsin)/g = sin/cosso
v2 = gLsin2/cos
(6)
gives
(7)
v = sqrt(gLsin2/cos)
v = sqrt(9.8m/s2*1.5m*sin230o/cos30o = sqrt(16.97*0.25/0.866) m/s = sqrt(4.99) =
v =2.2m/s.
SOLUTION TO (b)
Now we need to find  . Since we have v use Eq(6) sin2/cos= v2/gL and the trig identity
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Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
2
2
2
sin cos = 1, to get sin /cos= (1-cos2)/cos. The makes Eq 6 become
(1-cos2)/cos= v2/gL
,
(8)
After multiplying through by cosyields the quadratic equation
1 - v2/gL coscos2 =0 ,
This gives
cosθ = -
v2
v2 2
± (
) +1
2Lg
2Lg
(9)
Notes: Because or the +- before the sqrt, the Quadratic Equations has two mathematical solutions.
There are three possibilities, (1) both are possible for the situation under study, (2) only one is, (3)
neither. Here it turns out that only one is possible.
Solution
v2
v2 2
θ = cos ( ± (
) +1 )
2Lg
2Lg
-1
(10)
Some intermediate calculations.
The first term, v2/(2Lg) = (4.0 m/s)2/(2*1.5m*9.8m/s2) = (16 m2/s2 )/(29.4m2/s2)= 0.544.
The stuff inside the sqrt
(v2/(2Lg))2+1 = 0.5442 +1 = 1.296
Sqrt((v2/(2Lg))2+1) = Sqrt(1.296) = 1.138
Using the “+” in Eq. 10 gives
cos-1(-0.544+ sqrt(1.296)) = cos-1(-0.544+1.138) = cos-1(0.594);  = 53.6o = 54o .OK!
Using the “-” in Eq. 10 gives
cos-1(-0.544- sqrt(1.296)) = cos-1(-0.544-1.138) = cos-1(1.67) ; This gives cos=1.67. But -1<=cos <=+1,
so we must reject the solution to the Quadratic Equation.
SOLUTION TO (c)
Here we need a relationship between  and T without v.
Eq(5), Tcos = mg , does it! It gives
cos = mg/T
So,
 = cos-1(mg/T)
=cos-1(0.50kg*9.8m/s2/9.8N)
 = cos-1(0.50) =60o .
(11)
(12)
(13)
Substitute this value of  into Eq(7), v = sqrt(gLsin2/cos) , with (9) gives
v = sqrt(gLsin2/(mg/T)) = sqrt(TLsin2/m)
= sqrt(9.8N*1.5m sin2o/0.50kg)
= sqrt(9.8kg m/s2*1.5m*0.8662/0.50kg)
= sqrt((9.8*1.5*0.8662/0.50)(m/s)2) =sqrt(22.09)m/s =
v= 4.7 m/s.
(14)
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