# Problems

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. mss = 30  kg is the mass of the see-saw, mc = 20  kg is the mass of the child, and L is the length of the see-saw.

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

= mssg - xmcg = 0

so that

x = = = 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 mc kg and sits as far to the left as possible so that:

= mssg - mcg = 0

which gives

mc = mss = (30) = 15  kg.

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.

We need to impose both equilibrium conditions: i) = 0 and ii) = 0.

i)
= 0 :
 Fx = 0 = fs - N1 0 = N2 - N1 [2mm]Fy = 0 = N2 - mg N2 = mg
where fs = N2 is the maximum static frictional force that can exist. Putting these together gives:

0 = mg - N1.

ii)
= 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:
 = 0 = - mg(L/2)cos + N1Lsin N1 = .
Combining the results of parts (i) and (ii) we have:
 = mg - = = = 0.29

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.

Solution:

 xcm = = = 1.6  m [2mm]ycm = = = 0.27  m.

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

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)

 IO = miri2 = 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):

 Ix = miri2 = 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):

 Iy = miri2 = m1(0) + m2(32) + m1(0) + m2(32) = 2m2(32) = 18 kg m 2.

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

Solution:

a)
First draw and clearly label the diagram as shown in Figure 8.7. Next we apply Newton's laws for translational motion:
 Fx = macm = mgsin - f acm = gsin - [2mm]Fy = 0 = N - mgcos .
where fs 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:

 = IO = - fsr fs = .
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:

acm = - at = - r.

Thus we can write

fs = .

Subsituting this into the above equations gives:

acm = gsin - IO

which can easily be solved to yield:

acm =

b)
Use v 2 = v02 + 2acmx where v0 = 0 and x = 4  m :
 vx = = = 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 acm = g/2sin . For a uniform disc, I = 1/2mR 2, so acm = gsin /(1 + 1/2) = 2/3gsin . For a uniform sphere, I = 2/5mR 2 and acm = gsin /(1 + 2/5) = 5/7gsin .

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

acm = gsin

as expected.

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

Solution:

 Wnc = 0 = KEt + KEr + PEg = (mvcm2 - 0) + (I - 0) + (0 - mgh) 0 = mvcm2 + I - 2mgh.
where h is the vertical distance moved by the object. If L is the distance moved along the ramp then:

h = Lsin         (fromthediagram)

Also we have that:
 = = I sphere = mr 2.
Putting these Equations together yields:
 = mvcm2 + (mr 2) - 2mgLsin = 5vcm2 + 2vcm2 - 10gLsin vcm = = 5.3  m/s
which agrees with what we had before.

A cylinder with moment of inertia I = 34 kg m 2 rotates at = 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:
 I + I = I + I I = 2I = = 0.7  rads/s.
b)

 KEr = KErf - KEri = (I + I) - I = I - I = - 16.7  J
where the negative sign indicates that the rotational kinetic energy has decreased.

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 radians and 1 minute = 60 seconds to convert:
 10  rev/min = (10  rev/min )(2  rads/rev )(1  min /60  s ) = 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:
 Ii = I (student + platform) + mri2 + mri2 = 1.0 kg m 2 + (2.0  kg )(1.0  m )2 + (2.0  kg )(1.0  m )2 = 5.0 kg m 2

 If = I (student + platform) + mrf2 + mrf2 = 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:

 Ii = If.

So that: