Homework 8. problems: 10.40, 10.73, 11.55, 12.43

Homework 8

problems: 10.40, 10.73, 11.55, 12.43

Problem 10.40

A block of mass m

1

=2 kg and a block of mass m

2

=6 kg are connected by a massless strint over a pulley in the shape of a solid disk having radius R=0.25 m and mass M=10 kg. These blocks are allowed to move on a fixed wedge of angle θ=30° as shown in the figure. The coefficient of kinetic friction is 0.36 for both blocks. Draw free-body diagrams of both blocks and the pulley. Determine (a) the acceleration of the two blocks and (b) the tension in the string on both sides of the pulley.

It is convenient to consider each block in a different reference frame. In the reference frame in the figure, a number of scalar components of the considered vector quantities assume a zero value. Using Newton’s second law of motion, we can relate the interaction of each object with its motion.

Only horizontal component (x

₁

) of the acceleration of block’s m

₁

center of mass has a non-zero value a. Therefore

0 g m N 0 0

a m 0 0 N T

1 1

1 1

k 1

=

− + +

= + + μ

−

m1

m2

θ x1

y1

m

1

g N

1

T

1

f

k,1

m

2

g

N

2

-T

2

f

k,2

Mg -T

₁

N

T

2

x2

y2

Similarly, only the component along the ramp of the acceleration of the center of mass on the second block has a nonzero value of a (same as the first block)

0 cos

g m N

0 0

a m 0 sin g m N

T

2 2

2 2

2 k 2

= θ

− +

+

= + θ +

μ

−

−

The z-component of angular acceleration of the pulley, related to the acceleration of the blocks, is related to the net torque exerted on the pulley. With the reference point at the center of the pulley

R T R R T

MR a 2 1

2 1

2

⋅ = −

−

The above set of linear equations has five unknowns (N

₁

, T

₁

, N

₂

, T

₂

and a). The rest is algebra. Solving the equations simultaneously for the unknown component of the acceleration (of the center of mass of each block) and the two tensions we obtain

( )

( )

2 2

2 1

2 1

k 2

s 309 m . s 0

8 m . 9 kg

6 kg 2 kg 2 10 1

30 cos kg 6 kg 2 36 . 0 30 sin kg 6

g m

m 2 M

1

cos m m sin

a m

=

⋅ +

+

⋅

°

⋅ +

⋅

−

°

= ⋅

=

⋅ +

+

θ +

μ

−

= θ

( ) 7 . 67 N

s 8 m . 9 36 . s 0

309 m . 0 kg 2 g a

m

T

1 1 k 2 2

⎟ =

⎠

⎜ ⎞

⎝

⎛ + ⋅

⋅

= μ +

⋅

=

( )

( )

( ) 9 . 22 N

s 309 m . s 0

8 m . 9 30 sin 30

cos 36 . 0 kg

6

a g sin cos

m T

2 2

k 2

2

⎟ =

⎠

⎜ ⎞

⎝

⎛ − ⋅ ° + ° ⋅ −

⋅

=

=

−

⋅ θ + θ μ

−

⋅

=

Problem 10.73

A string is wound around a uniform disk of radius R and mass M. The disk is released from rest with the string vertical and its top end tied to a fixed support (Figure P11.47). As the disk descends, show that (a) the tension in the string is one third the weight of the disk, (b) the magnitude of the acceleration of the center of mass is 2g/3, and (c) the speed of the center of mass is (4gh/3)

^1/2

. Verify your answer to (c) using the energy approach.

a) Using Newton's second law for rotational motion we can relate the net torque about the instantaneous axis with the angular acceleration of the cylinder.

For the direction of the axis of rotation (see the figure)

(1) I

_A

α

_z

= τ

_net

= 0 - MgR I used here the fact that the total gravitational torque is such that the entire gravitational force was applied at the

center of gravity (coinciding with the center of mass).

Using also the parallel axis theorem we can determine the rotational inertial of the disk about the instantaneous axis and from (1) find the angular acceleration

(2)

R 3

g 2 MR

2 MR 1

MgR

2

z 2

= −

+

= − α

Now we can relate the acceleration (of the center of mass) with the above angular acceleration and use Newton's second law for the translational motion of a system of particles. For the vertical components

(3) T - W = Ma

_y

= Mα

_z

R Therefore

R M

x y

z

h

(4) Mg 3 R 1 R 3

g M 2 Mg R

M W

T = + α

_z

= − =

b) (In fact we needed the acceleration in part (a).) The motion is in a vertical direction therefore the horizontal components of the acceleration are zero. Recalling again the relationship between the acceleration and the angular acceleration the magnitude of the acceleration is

(5) ( )

3 g R 2

0 0

a =

²

+

²

+ − α

²

=

(The minus sign indicates that consistent with my drawing the acceleration is down.)

c) Solution 1.

Since we found the tension force in part (b), the easiest approach would be to use the work-energy theorem for the translational kinetic energy.

The change in the translational kinetic energy is directly related to the speed of the center of mass

(6) 0

2 K Mv

2

T

=

cm

−

Δ

(The initial kinetic energy is zero).

Tension and gravity alone produce external forces which are exerted on the cylinder. The work done on the center of mass due to these interactions is

(7) Mgh

3 Mgh 2 3 Mgh

) 1 m

( h

T W

i i

cm

= − ⋅ + ⋅ Δ = − + =

Δ ∑ g r

i

From the work-energy theorem (for translational kinetic energy)

(8) Mgh 3 2 2

Mv

²_cm

=

Solving for the unknown speed we obtain (9)

3 gh v

_cm

= 4

Note. The problem was not properly formulated. You can only verify the formula given in the problem if h represents the displacement of the center of mass. In the included figure h should be marked as the distance from the “ceiling” to the top of the cylinder and not as the distance to its center.

Solution 2. (work-energy theorem for total kinetic energy)

Using the work-energy theorem for the total kinetic energy of the cylinder, we will be able to relate the angular velocity (axial component) of the cylinder (all its particles) with the displacement of the center of mass. In this version of the theorem only gravitational work is performed on the system.

(10) W

_ext

= 0 + W

_g

= ∑ ( m

_i

⋅ Δ ) = Mgh

i

r

i

g

(The gravitational work again depends only on the displacement of the center of mass.)

We can relate the change in the total kinetic energy with the speed of the cylinder

(11)

²

2 2

2 2

A 2

tot

Mv

4 3 R

MR v 2 1 2 Mv 1 2 I 1

2 Mv 1 2

K 1 ⎟ =

⎠

⎜ ⎞

⎝

⋅ ⎛ +

= ω +

=

Δ

From the work-energy theorem for the total kinetic energy

(12) Mv Mgh

4 3

₂

=

we find the same value for the speed

(13)

3 gh v = 4

Solution 3 (from Newton’s second law)

The center of mass is moving with constant acceleration. Consistent with the figure, the distance traveled by the center of mass of the cylinder is related with time as follows

(14)

2 h at

=

2

Therefore, the center of mass is displaced by distance h at instant (15)

a h t = 2

We can find the vertical component of the velocity of the center of mass from the inverse relationship between acceleration and velocity

(16)

3 hg 4 a

h a 2 at

v = − = − = −

Problem 11.55

Two astronauts, each having mass m, are connected by a rope of length d having negligible mass. They are isolated in space, orbiting their center of mass at speed v.

Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum of the two-astronaut system and (b) the rotational energy of the system.

By pulling on the rope, one of the astronauts shortens the distance between them to d/2. (c) What are the astronauts’ new speeds? (d) How much chemical potential energy in the body of the astronauts was converted to mechanical energy in the system when he shortened the rope?

a,b) Directly from the definition, the magnitude of astronaut’s angular momentum is

2 mvd 90 1

sin 2 mv

m d

l = r × v = ⋅ ⋅ ° = ⋅ and his or her kinetic energy

2 K mv

2 i

=

Since the angular momentum of the second astronaut is in the same direction, the magnitude of the total angular momentum of the two astronauts is

mvd L

_i

= L

_i

=

The total kinetic energy of the two astronauts is

2 i

, 2 i

, 1 i

,

tot

K K mv

K = + =

r

2

p

2

l

2

m r

1

p

1

l

1

m

L

c) Since the system is isolated, the total angular momentum is conserved.

However, when the astronauts are closer, their speed will double to compensate for the change in the radius of their paths

v md 2

d mv 2 md

L 2 md

L

v

_f

= 2

^f

=

ⁱ

=

ⁱ

=

d) In the absence of external influence, the (internal) work performed by the astronauts is equal to the change in the total kinetic energy of the system

2 i

, tot f

, tot

int

K K 3 mv

W = − =

Δ

Solution 2. Calculation of the internal work directly from its definition is possible but a bit more complicated. In the process, each astronaut gets closer to the center of mass while the tension force, holding the

astronaut in his or her circular motion, must be equal to the required centripetal force.

[ ] ( ) _{( )}

2 2

2 2 4 2

d

2 2 d 2 4 2

d

2 d 3 2 2

4 d

2 d

2 4

d

2 d

2 int

mv d 3

4 d

d 16 4 mv 1 r

d 1 4 mv dr 1

r d 1 2 mv 1

r dr mr

2 / m mvd 2

r dr r v 2 m

W

⎟ =

⎠

⎜ ⎞

⎝

⎛ −

⋅

=

⋅

=

⋅

⋅

−

=

=

⋅

⎟ ⎠

⎜ ⎞

⎝

⎛

⋅

−

=

−

⋅ ⋅

⋅

= Δ

∫

∫

∫

(0.3)

Problem 12.43

A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. The beam is uniform, weights 200 N and 6 m long; the basket weights 80 N. (a) Draw a free-body diagram to the beam. (b) When the bear is at x = 1 m, find the tension in the wire and component of the force exerted by the wall on the left end of the beam. (c) If the wire can withstand a maximum tension of 900 N, what is the maximum distance the bear can walk before the wire breaks?

a) The following forces are exerted on the beam F – normal force exerted by the wall

N – normal force exerted by the bear (equal to the weight of the bear)

W – gravitational force exerted by the earth

T

1

– the tension force exerted by the wire

T

2

– the tension force exerted by the hanging basket (equal to the weight of the basket)

b) Since the beam is in (static) equilibrium, the net external torque about any point and the net external force exerted on the beam are zero vectors. If we select the reference point at the wall (the origin of the indicated reference frame), the only non-zero components of all the torques are along the z-direction. (The analysis of the x- and y-components leads to trivial equations.) For the z-components

x

F

N W

T

1

T

2

x

y

z

xN

,

N

= −

τ ;

2 W L

z ,

W

= −

τ ;

2 z

,

T

LT

2

= −

τ ;

°

=

τ

_{T ,z}

LT

₁

sin 120

2

The balance of the torques therefore requires 0

xN 2 W

LT L 120

sin

LT

₁

° −

₂

− − =

From which, the tension in the wire must be N 120 343

sin

N m 700

6 m N 1

2 200 N 1

80

T

₁

=

°

⋅ +

⋅

= +

c) The same equation leads to the maximum distance the bear can walk

m 14 . N 5

700

N 2 200 N 1

80 120

sin N 900 m

6

N

2 W LT L

120 sin LT x

2 1

=

⎟ ⎠

⎜ ⎞

⎝

⎛ ⋅ ° − − ⋅

⋅

=

− =

−

= °

Note that the problem could be solved only from the balance of torques.

Balance of forces allows one to determine the force F exerted by the wall (not asked).