1 PHYS1100 Practice problem set, Chapter 7: 9, 13, 17, 21, 25, 35 ...

PHYS1100 Practice problem set, Chapter 7: 9, 13, 17, 21, 25, 35, 40, 41, 48, 53, 54
7.9. Model: The rider is assumed to be a particle.
2
Solve: Since a = v / r, we have
r
( )( )
v = a r = 98 m/s 12 m ⇒ v = 34.3 m/s
2 2
r
Assess: 34.3 m/s ≈ 70 mph is a large yet understandable speed.
7.13. Model: The vehicle is to be treated as a particle in uniform circular motion.
Visualize:
On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary centripetal
acceleration. The vertical component of the normal force balances the weight.
Solve: From the physical representation of the forces in the r-z plane, Newton’s second law can be written
2
mv
∑ Fr
= nsinθ
=
∑ Fz
= ncosθ
− mg = 0 ⇒ ncosθ
= mg
r
Dividing the two equations and making the conversion 90 km hr = 25 m/s yields:
( 25 m/s)
2
( )
2
2
v
tanθ
= = = 0.128 ⇒ θ = 7.27°
rg 9.8 m/s 500 m
Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our
experience as well.
1

7.17. Model: The satellite is considered to be a particle in uniform circular motion around the moon.
Visualize:
Solve: The radius of moon is 1.738 × 10 6 m and the satellite’s distance from the center of the moon is the same
quantity. The angular velocity of the satellite is
and the centripetal acceleration is
ar
2π
2π
rad 1min
ω = = × = ×
T 110 min 60 s
−4
9.52 10 rad/s
−
( )( ) 2
= rω = 1.738 × 10 m 9.52 × 10 rad/s = 1.58 m/s
2 6 4 2
The acceleration of a body in orbit is the local “g” experienced by that body.
7.21. Model: Model the roller coaster car as a particle undergoing uniform circular motion along a loop.
Visualize:
Notice that the r-axis points downward, toward the center of the circle.
Solve: In this problem the apparent weight is equal to the weight: wapp = n = mg.
We have
2
mv
2
∑ Fr
= n + w = = mg mg v rg ( )( )
r
+ ⇒ = 2 = 2 20 m 9.8 m/s = 19.8 m/s
2

7.25. Model: Model the particle on the crankshaft as being in nonuniform circular motion.
Visualize:
Solve: (a) The initial angular velocity is ω 0 = 2500 rpm × (1 min/60 s) × (2π rad/rev) = 261.8 rad/s. The
crankshaft slows from 261.8 rad/s to 0 in 1.5 s. Thus
ω
a
rω
(0.015 m)(261.8 rad/s)
t
0
2
1
= 0 = ω0 + t1
⇒ at
= − = − = –2.618 m/s
r
t1
1.5 s
(b) During these 1.5 s, the crankshaft turns through angle
2
a 2 2.618 m/s
2
1 0 0t
t
1
t1
θ = θ + ω + = 0 + (261.8 rad/s) (1.5 s) − (1.5 s) = 196.3 rad
2r
2(0.015 m)
In terms of revolutions, θ 1 = (196.3 rad)(1 rev/2π rad) = 31.25 rev.
3

7.35. Model: We will use the particle model for the car, which is undergoing uniform circular motion on a
banked highway, and the model of static friction.
Visualize:
Note that we need to use the coefficient of static friction µ s , which is 1.0 for rubber on concrete.
Solve: Newton’s second law for the car is
mv
r
2
∑ F cos sin
r
= fs θ + n θ =
z
θ
s
∑ F = ncos − f sinθ
− w = 0 N
Maximum speed is when the static friction force reaches its maximum value fs max
= µ
s
n.
Then
2
cos15 sin15 mv
n
s
r
Dividing these two equations and simplifying, we get
Assess:
( µ ° + ° ) = n( cos15 µ
s
sin15 )
µ 2
s
tan15 v
µ s
+ ° + tan15°
= ⇒ v = gr
1− µ tan15° gr 1− µ tan15°
s
( )( ) ( 1.0 + 0.268 )
2
( 1−
0.268)
= 9.80 m/s 70 m = 34.5 m/s
The above value of 34.5 m/s ≈ 70 mph is reasonable.
s
° − ° = mg
4

7.40. Model: Use the particle model for the ball which is in uniform circular motion.
Visualize:
Solve:
From Newton’s second law along r and z directions,
Dividing the two force equations gives
2
mv
∑ Fr
= ncosθ
= ∑ Fz
= nsinθ
− mg = 0 ⇒ nsinθ
= mg
r
From the geometry of the cone, tan θ = r y.
Thus
gr
tanθ =
2
v
r gr
= ⇒ v =
2
y v
gy
5

7.41. Model: Consider the passenger to be a particle and use the model of static friction.
Visualize:
Solve: The passengers stick to the wall if the static friction force is sufficient to support their weight: f s = w. The
minimum angular velocity occurs when static friction reaches its maximum possible value f s max = µ s n. Although
clothing has a range of coefficients of friction, it is the clothing with the smallest coefficient (µ s = 0.6) that will slip
first, so this is the case we need to examine. Assuming that the person is stuck to the wall, Newton’s second law is
∑
The minimum frequency occurs when
Using this expression for f s in the z-equation gives
s
∑
2
Fr
= n = mω
r Fz
= fs
− w = 0 ⇒ fs
= mg
f = f = µ n = µ mrω
2
s s max s s min
f = µ mrω
= mg
2
s s min
2
g 9.80 m/s 1 rev 60 s
⇒ ωmin
= = = 2.56 rad/s = 2.56 rad/s × × = 24.4 rpm
µ r 0.60(2.5 m) 2 π rad 1 min
Assess: Note the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is
not necessary.
6

7.48. Model: Model the chair and the rider as a particle in uniform circular motion.
Visualize:
Solve:
Since
Newton’s second law along the r-axis is
∑
r = Lsinθ
, this equation becomes
F = T + w = ma
r r r r
2
⇒ T sinθ
+ 0 N = mrω
Thus, the 3000 N chain is not strong enough for the ride.
2 ⎡ 2π
rad ⎤
T = mLω
= ( 150 kg)( 9.0 m)
⎢ = 3330 N
4.0 s
⎥
⎣ ⎦
2
7

7.53. Model: Model the car as a particle on a circular track.
Visualize:
Solve: (a) Newton’s second law along the t-axis is
F F ma
2
∑ t
=
t
=
t
⇒ 1000 N = ( 1500 kg) at
⇒ at
= 2 3 m/s
With this tangential acceleration, the car’s tangential velocity after 10 s will be
The radial acceleration at this instant is
2
( ) ( )( )
v1 t
= v0t + at
t1 − t0 = 0 m s + 2 3 m/s 10 s − 0 s = 20 3 m/s
( ) 2
2
v 20 3 m/s
1t
16 m/s
2
ar
= = =
r 25 m 9
The car’s acceleration at 10 s has magnitude
2 2
2 2 2 2 2
−1 at
−1⎛ 2 3 ⎞
a1 = at
+ ar
= ( 2 3 m/s ) + ( 16 9 m/s ) = 1.90 m/s θ = tan = tan ⎜ ⎟ = 20.6°
ar
⎝ 16 9 ⎠
where the angle is measured from the r-axis.
(b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value
f = µ n . That is,
s max
s
2
mv2
t
2
∑ Fr
= fs max
= µ
sn = µ
smg
= v
t
rg ( )( )
r
⇒
2
= = 25 m 9.8 m/s = 15.7 m/s
In the above equation, n = mg follows from Newton’s second law along the r-axis. The time when the car begins to
slide can now be obtained as follows:
= + ( − ) ⇒ 15.7 m/s 0 m s ( 2 3 m/s 2
)( t 0) 2
v v a t t
2t 0t t 2 0
= + − ⇒ t2 = 23.5 s
2
8

7.54. Model: Model the steel block as a particle and use the model of kinetic friction.
Visualize:
Solve:
(a) The components of thrust ( F r ) along the r-, t-, and z-directions are
Fr
Newton’s second law is
= F sin20° = ( 3.5 N)
sin20° = 1.20 N F F ( )
2
( F ) T F mrω
net
r
= + = ( )
r
( )
net
t
= cos20° = 3.5 N cos20° = 3.29 N F z = 0 N
F = F − f = ma
net
F = n − mg = 0 N
The z-component equation means n = mg. The force of friction is
z
2
( )( )( )
fk = µ
kn = µ
kmg
= 0.60 0.5 kg 9.8 m/s = 2.94 N
Substituting into the t-component of Newton’s second law
2
( 3.29 N) − ( 2.94 N) = ( 0.5 kg) a ⇒ a = 0.70 m/s
Having found a t , we can now find the tangential velocity after 10 revolutions = 20π rad as follows:
1 ⎛ a ⎞ 2rθ
θ = t ⇒ t = = 18.95 s
t 2 1
1 ⎜ ⎟ 1 1
2 ⎝ r ⎠
at
⎛ at
⎞
ω1 = ω0 + ⎜ ⎟ t
1
= 6.63 rad/s
⎝ r ⎠
(b) Substituting ω 1 into the r-component of Newton’s second law yields:
T F mrω ⇒ T + 1.20 N = 0.50 kg 2.0 m 6.63 rad/s ⇒ T = 42.8 N
+ = ( ) ( )( )( ) 2
2
1 r 1
1 1
t
t
t
t
k
t
9