Download 13.11. Visualize: Solve: Torque by a force is defined as τ = Frsinφ

13.11. Visualize:
r
Solve: Torque by a force is defined as τ = Frsinφ where φ is measured counterclockwise from the r vector to the
r
F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the
20 N force:
(30 N)r1 sinφ1 + (20 N)r2 sinφ2 = (30 N)(0.02 m) sin ( − 90°) + (20 N)(0.02 m) sin (90°)
= ( − 0.60 N m) + (0.40 N m) = −0.20 N m
Assess: A negative torque causes a clockwise motion of the pulley.
13.14. Model: The disk is a rotating rigid body.
Visualize:
The radius of the disk is 10 cm and the disk rotates on an axle through its center.
Solve: The net torque on the axle is
τ = FArA sin φA + FBrB sin φB + FCrC sin φC + FDrD sin φD
= (30 N)(0.10 m) sin (−90°) + (20 N)(0.05 m) sin 90° + (30 N)(0.05 m) sin 135° + (20 N)(0.10 m) sin 0°
= −3 N m + 1 N m + 1.0607 N m = −0.939 N m
Assess: A negative torque means a clockwise rotation of the disk.
13.17. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the
rotation axis passes through mass A and is perpendicular to the page.
Visualize: Please refer to Figure Ex13.17.
m x m x + mB xB + mC xC + mD xD
Solve: (a) xcm = ∑ i i = A A
mA + mB + mC + mD
∑ mi
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
= 0.0571 m
100 g + 200 g + 200 g + 200 g
m y + mB yB + mC yC + mD yD
= A A
mA + mB + mC + mD
=
ycm
=
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g)(0 m)
= 0.0571 m
700 g
(b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page is
I A = ∑ miri2 = mArA2 + mBrB2 + mCrC2 + mDrD2
i
= (0.100 kg)(0 m)2 + (0.200 kg)(0.10 m)2 + (0.200 kg)(0.1414 m)2 + (0.200 kg)(0.10 m)2 = 0.0080 kg m2
13.26. Model:
Visualize:
Solve:
The massless rod is a rigid body.
r
To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium (τnet = 0 Nm). We have
( Fnet )y = (40 N) − (100 N) + (60 N) = 0
N, so the object is in translational
equilibrium. Measuring τnet about the left end,
τnet = (60 N)(3.0 m) sin (+90°) + (100 N)(2.0 m) sin (−90°) = −20 N m
The object is not in equilibrium.
13.29. Model: The triangle is a rigid body rotating about an axis through the center.
Visualize: Please refer to Figure Ex13.28. Each 200 g mass is a distance r away from the axis of rotation, where r
is given by
0.20 m
0.20 m
= cos30° ⇒ r =
= 0.2309 m
r
cos30°
Solve: The moment of inertia of the triangle is I = 3 × mr2 = 3(0.200 kg)(0.2309 m)2 = 0.0320 kg m2. The
frequency of rotation is given as 5.0 revolution per s or 10π rad/s. The rotational kinetic energy is
Krot. =
1 2 1
Iω = (0.0320 kg m 2 )(10.0π rad/s)2 = 15.8 J
2
2
13.33. Model: The can is a rigid body rolling across the floor. Assume that the can has uniform mass distribution.
Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass.
The moment of inertia of the can about the center of mass is 12 MR 2, where R is the radius of the can. Also vcm =
Rω, where ω is the angular velocity of the can. The total kinetic energy of the can is
K = K cm + K rot
=
1
1
1
2
2
= Mvcm
+ I cmω 2 = Mvcm
+
2
2
2
3
3
2
Mvcm
= (0.50 kg)(1.0 m/s)2 = 0.375 J
4
4
1 1
 v 
MR 2   cm 
2  2
 R 
2
(iˆ × ˆj ) × iˆ = kˆ × iˆ = ˆj
iˆ × ( ˆj × iˆ) = iˆ × ( − kˆ ) = −iˆ × kˆ = −(− ˆj ) = ˆj
13.37. Solve: (a)
(b)
13.46. Model: The bar is a rotating rigid body. Assume that the bar is thin.
Visualize: Please refer to Figure Ex13.46.
Solve: The angular velocity ω = 120 rpm = (120)(2π)/60 rad/s = 4π rad/s. From Table 13.3, the moment of
inertial of a rod about its center is I = 121 ML2 . The angular momentum is
 1
L = Iω =   (0.50 kg)(2.0 m)2 (4π rad/s) = 2.09 kg m 2 /s
 12 
If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the
page. Consequently,
r
L = (2.09 kg m 2 /s, out of the page)
13.47. Model: The disk is a rotating rigid body.
Visualize: Please refer to Figure Ex.13.47.
Solve: From Table 13.3, the moment of inertial of the disk about its center is
I=
1
1
MR 2 = (2.0 kg)(0.020 m)2 = 4.0 × 10 −4 kg m 2
2
2
The angular velocity ω is 600 rpm = 600 × 2π/60 rad/s = 20π rad/s. Thus, L = Iω = (4.0 × 10−4 kg m2)(20π rad/s) =
0.0251 kg m2/s. If we wrap our right fingers in the direction of the disk’s rotation, our thumb will point in the –x
direction. Consequently,
r
L = −0.0251 iˆ kg m 2 /s = (0.0251 kg m 2 /s, into page)
13.54. Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is thin.
Visualize: Please refer to P13.54.
Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1, and mass m2.
Using Table 13.3 for the moment of inertia of the rod, we get
2
I rod = I rod about center + I m1 + I m2 =
=
1
L
L
ML2 + m1   + m2  
12
2
4
1
1
1
L2  M
m 
ML2 + m1L2 + m2 L2 =  + m1 + 2 
12
4
16
4 3
4 
Assess: With m1 = m2 = 0 kg, I rod = 121 ML2 , as expected.
2
13.56. Model: The beam is a rigid body of length 3.0 m and the student is a particle.
Visualize:
Solve:
r
r
To stay in place, the beam must be in both translational equilibrium ( Fnet = 0 N) and rotational equilibrium
(τ net = 0 Nm) . The first condition is
∑F
y
= − wbeam − wstudent + F1 + F2 = 0 N
⇒ F1 + F2 = wbeam + wstudent = (100 kg + 80 kg)(9.80 m/s2) = 1764 N
Taking the torques about the left end of the beam, the second condition is
−wbeam (1.5 m) – wstudent (2.0 m) + F2 (3.0 m) = 0 N m
− (100 kg)(9.8 m/s2)(1.5 m) – (80 kg)(9.8 m/s2)(2.0 m) + F2 (3.0 m) = 0 N m
⇒ F2 = 1013 N
From F1 + F2 = 1764 N, we get F1 = 1764 N – 1013 N = 751 N.
Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any
point on the body of interest.
13.58. Model: Model
the forearm as a rigid body. For the forearm parallel to the floor, we require both transr
lational equilibrium ( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m) .
Visualize:
Solve:
τnet = Fbicep(0.025 m) – wF(0.16 m) – wB(0.32 m) = 0 N m
Fbicep (0.025 m) – (1.2 kg)(9.8 m/s2)(0.16 m) – (0.500 kg)(9.8 m/s2)(0.32 m) = 0 N m
⇒ Fbicep = 138 N
Assess: 138 N is much larger than the 4.9 N weight of the ball because the moment arm is so much shorter.
13.68. Model: Assume the string does not slip on the pulley.
Visualize:
The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward
force on the block m2, but a downward force on the outer edge of the pulley. Similarly the string exerts a force on
block m1 to the right, but a leftward force on the outer edge of the pulley.
Solve: (a) Newton’s second law for m1 and m2 is T = m1a1 and T – m2g = m2a2. Using the constraint –a2 = +a1 = a,
we have T = m1a and −T + m2g = m2a. Adding these equations, we get m2g = (m1 + m2)a, or
a=
m2 g
mmg
⇒ T = m1a = 1 2
m1 + m2
m1 + m2
(b) When the pulley has mass m, the tensions (T1 and T2) in the upper and lower portions of the string are different.
Newton’s second law for m1 and the pulley are:
T1 = m1a and T1R – T2R = − Iα
We are using the minus sign with α because the pulley accelerates clockwise. Also, a = Rα. Thus, T1 = m1a and
T2 − T1 =
I a aI
=
R R R2
Adding these two equations gives
I 

T2 = a  m1 + 2 
R


Newton’s second law for m2 is T2 – m2g = m2a2 = − m2a. Using the above expression for T2,
I 
m2 g

a  m1 + 2  + m2 a = m2 g ⇒ a =
R 
m1 + m2 + I / R2

Since I = 12 mp R2 for a disk about its center,
a=
m2g
m1 + m2 + 21 mp
With this value for a we can now find T1 and T2:
T1 = m1a =
Assess:
m1m2 g
m1 + m2 + 12 mp
T2 = a(m1 + I / R2 ) =
(
(
)
For m = 0 kg, the equations for a, T1 and T2 of part (b) simplify to
a=
m2 g
m1 + m2
These agree with the results of part (a).
and T1 =
)
1
1  m2 m1 + 2 mp g

m1 + mp  =

2  m1 + m2 + 12 mp
m1 + m2 + 12 mp 
m2 g
m1m2 g
m1 + m2
and T2 =
m1m2 g
m1 + m2
13.69. Model: The disk is a rigid spinning body.
Visualize: Please refer to Figure P13.69. The initial angular velocity is 300 rpm or (300)(2π)/60 = 10π rad/s.
After 3.0 s the disk stops.
Solve: Using the kinematic equation for angular velocity,
ω1 = ω 0 + α (t1 − t0 ) ⇒ α =
ω1 − ω 0
t1 − t0
=
(0 rad/s − 10π rad/s) −10π
=
rad/s2
(3.0 s − 0 s)
3
Thus, the torque due to the force of friction that brings the disk to rest is
τ = Iα = − fR ⇒ f = −
Iα
=−
R
(
1
2
)
mR 2 α
R
π
1
1


= − ( mR)α = − (2.0 kg)(0.15 m)  −10 rad/s2  = 1.57 N
2
2
3


The minus sign with τ = − fR indicates that the torque due to friction acts clockwise.
13.73. Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque
on the hoop causes it to rotate, transforming the gravitational potential energy of the hoop’s center of mass into
rotational kinetic energy.
Visualize:
We placed the origin of the coordinate system at the hoop’s edge on the axle. In the initial position, the center of
mass is a distance R above the origin, but it is a distance R below the origin in the final position.
Solve: (a) Applying the parallel-axis theorem, Iedge = Icm + mR2 = mR2 + mR2 = 2mR2. Using this expression in the
energy conservation equation Kf + Ugf = Ki + Ugi yields:
1
1
I edgeω12 + mgy1 = I edgeω 02 + mgy0
2
2
1
2g
(2mR 2 )ω12 − mgR = 0 J + mgR ⇒ ω1 =
2
R
(b) The speed of the lowest point on the hoop is
2g
(2 R) = 8gR
R
Assess: Note that the speed of the lowest point on the loop involves a distance of 2R instead of R.
v = (ω1 )(2 R) =
13.75. Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical
and the sphere solid.
Visualize: Please refer to Figure P13.75. The sphere rotates because the string wrapped around the rod exerts a torque τ.
Solve: From the parallel-axis theorem and Newton’s second law,
2
2
MR 2 13
R
I off center = I cm + M   = MR2 +
=
MR 2
5
4
20
2
α=
τ
I
=
τ
2
(13MR /20)
=
20τ
13MR 2
13.84. Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop.
The mechanical energy of the marble is conserved.
Visualize:
Solve: The ball’s center of mass moves in a circle of radius R – r. The free-body diagram on the marble at its highest
position shows that Newton’s second law for the marble is
mv12
R−r
The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when
the normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration
needed for the circular motion. For n → 0 N,
mg + n =
mg =
mv2
⇒ v12 = g( R − r )
(R − r)
Since rolling motion requires v12 = r 2ω12 , we have
ω12r 2 = g( R − r ) ⇒ ω12 =
g( R − r )
r2
The conservation of energy equation is
( K f + Ugf )top of loop = (K i + Ugi )initial ⇒
1 2 1 2
mv1 + Iω1 + mgy1 = mgy0 = mgh
2
2
Using the above expressions and I = 25 mr 2 the energy equation simplifies to
1
1 2
 g( R − r ) 
mg( R − r ) +   mr 2 
 + mg2( R − r ) = mgh ⇒ h = 2.7 ( R − r )
2
2
2 5
 r
