Problems

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A poorly constructed see-saw has a fulcrum 2/3 of the way along its length. a) If the see-saw weighs 30 kg, where would a 20 kg child have to sit in order to balance the see-saw? b) What is the least mass that a child must have in order to balance the see-saw?

Solution:

a)

The center of gravity of the see-saw is assumed to be in the center, assuming that it is uniform. The force diagram is therefore as shown below. m_ss = 30  kg is the mass of the see-saw, m_c = 20  kg is the mass of the child, and L is the length of the see-saw.

  

Figure 8.3: Problem 8.1

In order to find the distance x from the child to the fulcrum we can do a torque balance about the fulcrum:

$$\sum$$$$\tau$$ = $${\frac{L}{6}}$$m_ssg - xm_cg = 0

so that

x = $${\frac{L m_{ss}}{6 m_c}}$$ = $${\frac{L(30)}{6(20)}}$$ = $${\textstyle\frac{1}{4}}$$L.

Thus, the child must sit L/4 away from the fulcrum.

b)

To find the minimum mass of the child that would work, we assume the child has mass m_c kg and sits as far to the left as possible so that:

$$\sum$$$$\tau$$ = $${L \over 6}$$m_ssg - $${L \over 3}$$m_cg = 0

which gives

m_c = $${1\over 2}$$m_ss = $${1\over 2}$$(30) = 15  kg.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A 10 m ladder weighs 50 N and balances against a smooth wall at 60 ^o to the horizontal. If the ladder is just on the verge of slipping, what is the static coefficient of friction between the floor and the ladder?

Solution:
The force diagram for the ladder is shown in Figure 8.4.

  

Figure 8.4: Problem 8.2

We need to impose both equilibrium conditions: i) $\sum$$\vec{F}$ = 0 and ii) $\sum$$\tau$ = 0.

i)

$\sum$$\vec{F}$ = 0 :

$$\sum$$ = f_s - N₁

$$\Rightarrow \mu_{s}^{}$$ =

$$\sum$$ = N₂ - mg

$$\Rightarrow$$ = mg

where f_s = $\mu_{s}^{}$N₂ is the maximum static frictional force that can exist. Putting these together gives:

0 = $$\mu_{s}^{}$$mg - N₁.

ii)

$\sum$$\vec{\tau}$ = 0 : Choose the axis carefully to have least number of terms to deal with. An axis through the point of contact between the floor and ladder will work well:

$$\sum \tau \theta \theta$$ =

$$\Rightarrow {\frac{mg \cos \theta}{2 \sin \theta}}$$ =

Combining the results of parts (i) and (ii) we have:

$$\mu_{s}^{} {\frac{mg \cos \theta}{2 \sin \theta}}$$ =

$$\Rightarrow \mu_{s}^{} {\frac{\cos \theta}{2 \sin \theta}} {\frac{\cos 60^{\circ}}{2 \sin 60^{\circ}}}$$ =

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Calculate the x and y positions of the center of mass of the following three objects where m₁ = 1.0  kg , m₂ = 2.5  kg , and m₃ = 4.0  kg . See Figure 8.5.

  

Figure 8.5: Problem 8.3

Solution:

$${\frac{\sum m_i x_i}{\sum m_i}}$$ x_cm =

$${\frac{m_1(0) + m_2(0) + m_3(3)}{m_1+m_2+m_3}}$$ =

$${\frac{\sum m_i y_i}{\sum m_i}}$$ [2mm]y_cm =

$${\frac{m_1(2) + m_2(0) + m_3(0)}{m_1+m_2+m_3}}$$ =

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Calculate the moments of inertia of the following 4 masses where m₁ = 2.0  kg and m₂ = 1.0  kg. See Figure 8.6.

  

Figure 8.6: Problem 8.4

a) About the axis through O (center) perpendicular to the page.
b) About an axis joining two of the masses.
Note: The moment of inertia depends strongly on the axis chosen.

Solution:

a)

$$\sum$$ I_O =

= 2m₁(2²) + 2m₂(3²)

= 8m₁ + 18m₂

= 34 kg m ².

b)

We consider the following two axes that pass through two of the masses: one along the x -axis and one along the y -axis.

First consider the x -axis (i.e. the one that passes through the m₂ 's):

$$\sum$$ I_x =

= m₁(2²) + m₂(0²) + m₁(2²) + m₂(0²)

= 2m₁(2²) = 16 kg m ².

For the y -axis (i.e. the one that passes through the two m₁ 's):

$$\sum$$ I_y =

= m₁(0) + m₂(3²) + m₁(0) + m₂(3²)

= 2m₂(3²) = 18 kg m ².

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A round object, mass m , radius r and moment of inertia I_O , rolls down a ramp without slipping as shown in Fig. 8.6 below. a) If the ramp is at an angle $\theta$ to the horizontal, find an expression for the acceleration of the center of mass of the object in terms of m,r,I₀ and $\theta$ . b) What is the velocity of the center of mass of a solid sphere, I_O = ${\frac{2}{5}}$mr ², if it starts from rest and rolls down a 4 m ramp inclined at 30 ^o to the horizontal?

  

Figure 8.7: Problem 8.5

Solution:

a)

First draw and clearly label the diagram as shown in Figure 8.7. Next we apply Newton's laws for translational motion:

$$\sum \theta$$ =

$$\Rightarrow \theta {\frac{f}{m}}$$ =

$$\sum \theta$$ =

where f_s is the magnitude of the static frictional force. We must use the static force because we assume that the object roles without slipping.

For the rotational motion we have:

$$\sum \tau \alpha$$ = - f_sr

$$\Rightarrow {\frac{-I_O \alpha}{r}}$$ =

The angular acceleration of the object about its center of mass is the same as the angular acceleration of the center of mass about the point in contact with the ramp. If it rolls without slipping the acceleration of the center of mass is equal to minus the tangential acceleration as it rotates about the stationary point in contact with the ramp:

a_cm = - a_t = - r$$\alpha$$.

Thus we can write

f_s = $${I_O a_{cm}\over r^2}$$.

Subsituting this into the above equations gives:

a_cm = gsin $$\theta$$ - I_O$${a_{cm}\over mr^2}$$

which can easily be solved to yield:

a_cm = $${{g\sin\theta \over ( 1 + {I_O\over mr^2})}.}$$

b)

Use v ² = v₀² + 2a_cmx where v₀ = 0 and x = 4  m :

$$\sqrt{\displaystyle{2 g \sin\theta \; x \over (1 + {2/5 mr^2\over mr^2})}}$$ v_x =

$$\sqrt{2\times 9.8 \times \sin30^{\circ} \times 4 \over (1+2/5)}$$ =

= 5.3  m/s. (17)

Note:

• The larger the moment of inertia, for a given mass, the slower the acceleration of the center of mass. For example, for a hollow cylinder, I = mR ², and a_cm = g/2sin $\theta$ . For a uniform disc, I = 1/2mR ², so a_cm = gsin $\theta$/(1 + 1/2) = 2/3gsin $\theta$ . For a uniform sphere, I = 2/5mR ² and a_cm = gsin $\theta$/(1 + 2/5) = 5/7gsin $\theta$ .

  Note that the mass and radius of the object cancel out of the final expression.

• If the object slips down the hill without rolling the moment of inertia is effectively zero, and the acceleration along the ramp is

  a_cm = gsin $$\theta$$

  as expected.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Do the above problem (part b) using energy conservation.

Solution:

$$\Delta \Delta \Delta$$ W_nc = 0 =

$${1\over 2} {1\over 2} \omega^{2}_{}$$ =

$$\Rightarrow \omega^{2}_{}$$ =

where h is the vertical distance moved by the object. If L is the distance moved along the ramp then:

h = Lsin $$\theta$$        (fromthediagram)

Also we have that:

$$\omega {\frac{v_{t}}{r}} {\frac{v_{cm}}{r}}$$ =

$${2\over 5}$$ I _sphere =

Putting these Equations together yields:

$${2\over 5} {\frac{v_{cm}^2}{r^2}} \theta$$ =

$$\theta$$ =

$$\Rightarrow \sqrt{\frac{10 g L \sin \theta}{7}}$$ =

which agrees with what we had before.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A cylinder with moment of inertia I = 34 kg m ² rotates at $\omega_{1i}^{}$ = 1.4  rads/s . It drops onto a second identical cylinder, and they stick together end to end. a) Calculate the final angular velocity of the two cylinders. b) Calculate the kinetic energy lost in the process.

Solution:

a)

Use conservation of angular momentum:

$$\omega_{1i}^{} \omega_{2i}^{} \omega_{1f}^{} \omega_{2f}^{}$$ =

$$\omega_{1i}^{} \omega_{f}^{}$$ =

$$\Rightarrow \omega_{f}^{} {1\over 2} \omega_{1i}^{}$$ =

b)

$$\Delta$$ = KE_rf - KE_ri

$${1\over 2} \omega_{f}^{2} {1\over 2} \omega_{1i}^{2}$$ =

$$\omega_{f}^{2} {1\over 2} \omega_{1i}^{2}$$ =

= - 16.7  J

where the negative sign indicates that the rotational kinetic energy has decreased.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A student stands on the center of a rotating platform that has frictionless bearings. He has a 2.0 kg object in each hand, held 1.0 m from the axis of rotation of the system. The system is initially rotating at 10 rpm. Determine a) the initial angular velocity in radians per second, b) the angular velocity of the sytem in radians per second after the objects are brought to a distance of 0.20 m from the axis of rotation, c) the change in the rotational kinetic energy of the system as the objects are pulled closer to the center of rotation. d) What causes the increase in rotational kinetic energy? (Assume that the moment of inertia of the platform + student remains constant at 1.0 kg m ².)

Solution:

a)

Use 1 rev = 2 $\pi$ radians and 1 minute = 60 seconds to convert:

$$\pi$$ 10  rev/min =

= 1.05  rad /s.

b)

Use conservation of angular momentum. We therefore need to compute the initial and final moments of inertia of the system:

I_i = I _{(student + platform)} + mr_i² + mr_i²

= 1.0 kg m ² + (2.0  kg )(1.0  m )² + (2.0  kg )(1.0  m )²

= 5.0 kg m ²

I_f = I _{(student + platform)} + mr_f² + mr_f²

= 1.0 kg m ² + (2.0  kg )(0.2  m )² + (2.0  kg )(0.2  m )²

= 1.16 kg m ².

Determine final velocity using conservation of angular momentum:

$$\omega_{i}^{} \omega_{f}^{}$$

So that:

$$\omega_{f}^{} {I_i\over I_f} \omega_{i}^{}$$ =

$${5.0\over 1.16}$$ =

= 4.5  rads/s.

c)

The change in kinetic energy is:

$$\Delta$$ = KE_f - KE_i

$${1\over 2} \omega_{f}^{2} {1\over 2} \omega_{i}^{2}$$ =

$${1\over 2} {1\over 2}$$ =

= 11.8  J - 2.75  J

= 9.05  J.

d)

As the student pulls the object in toward the center of rotation, he is doing work on the system. This work goes into increasing the kinetic energy of the system.