Chapter 11; Rotational Dynamics and Static Equilibrium
 ToC, Chapter 11  Course Calendar 
D11.1 A solid, uniform cylinder of 12 cm radius with mass of 5.0 kg is free to rotate about its symmetry axis. A cord is wound around the drum and a 1.2 kg mass is attached to the end of the cord. Find the acceleration of the hanging mass, the angular acceleration of the cylinder, and the tension in the cord.
For the 1.2 kg mass, take down as positive and apply Newton's Second Law,
For the rotating cylinder, apply the rotational form of Newton's Second Law,
since = 90°
For a solid cylinder,
Therefore,
At the moment, we have three unknowns, T, a, and , in two equations. The angular acceleration of the cylinder and the linear acceleration of the cord or of the hanging mass are connected by
Now we have two equations in two unknowns
T = 8.1 N
D11.2 A 10 kg block sits on a horizontal
surface with coefficient of friction of 0.2 between the block and the surface.
A string runs from this block over a wheel of radius 10 cm and moment of inertia
of 2.0 kg m^{2} and is attached to a hanging 5 kg mass. Find the acceleration
of the masses, the angular acceleration of the wheel, and the tension in the
string on each side.
Notice that the tension is different on
the two sides of the pulley because the pulley is (very) massive.
For the 10kg block on the horizontal plane, we can
write
and
F_{N} = 100 N
We can use this information to determine the friction force,
Then the F_{net,x} equation becomes
For the massive pulley, we have
We now have four unknowns  T_{L}, T_{R}, a, and  in two equations. The angular acceleration of the cylinder and the linear acceleration of the cord or of the hanging mass are connected by
T_{R}  T_{L} = (200 kg) a
We now have three unknowns T_{L}, T_{R}, and a  in two equations. We still must apply Newton's Second Law to the hanging mass. There we may as well take down as positive,
Now we have three equations for three unknowns,
T_{R} = 20 N + (190 kg) (0.15 _{}) = 49.2 N
D11.3 Assume a playground merrygoround to be a uniform cylinder or disk of 150 kg and 1.8 m radius. What is its moment of inertia?
It is initially at rest when a 50 kg child, running at 4 m/s, in a direction tangential to the edge of the merrygo round, jumps on. What is its angular velocity after the child sits down on the edge?
For a solid disk or cylinder,
Therefore, the moment of inertia of the merrygoround is
The initial angular momentum of the child is
The initial angular momentum of
the merrygoround is zero.
So the total initial angular momentum of the entire system
is
If the bearings of the merrygoround are well oiled this should be the total final angular momentum,
but now the moment of inertia is the moment of inertia of the merrygoround and the child
D11.4 A basket of tomatoes of mass 20.0 kg is being hoisted by a windlass. The rope is wrapped around an axle that is a solid cylinder of wood having a radius of 0.1 m and a mass of 10 kg. The mass of the crank handles is neglibible. The operator lets go of the handle when the basket is 6 m above the ground. With what linear speed does the basket strike the ground?
The moment of inertia for a solid cylinder (or a disk) is
The velocity of the load and the angular velocity of the windlass are connected by
As always, good diagrams make life much simpler.
Initially, before the windlass and load are released, we have
In this problem, as in the
previous one, the Kinetic Energy of the wheel or drum
or cylinder is far greater than the Kinetic Energy of the
masses and dominates the entire motion.
Finally, after the windlass has increased its angular speed and the
load has increased its linear speed, and just before the load hits
the floor, we have
D11.5 To demonstrate conservation of angular momentum a Physics professor stands on a frictionless turntable with a 2 kg mass in each outstretched hand. An assistant gives her a small initial angular velocity of 2 rad/s. She then drops her hands to her sides and her angular velocity increases dramatically. As a rough estimate, consider her arms to have mass of 5 kg each and to be 1 m long rods hinged at the axis of rotation. The rest of her body has an approximate moment of inertia of 0.55 kg m^{2}. Find her final angular velocity when the masses are 0.25 m from the axis of rotation. Calculate the initial and final values of the rotational kinetic energy and explain the cause of the difference in these values.
The initial angular momentum is
(It is only a coincident
that L and KE happen to have the same numerical value!).
As she pulls in her hands, the total moment of inertia
changes,
Total angular momentum is conserved,
Work must be done on the weights to pull them in toward the axis of rotation. This work that is done by the dizzy professor shows up as increased kinetic energy.
D11.6 A piece
of pipe may sometimes be used as a "cheater" to effectively lengthen the moment
arm of a wrench. If a pipe is used as shown in the figure to allow a 150 N force
to be exerted 30 cm from the bolt, what torque is exerted on the bolt?
= r F sin
= (0.30 m) (150 N) (1)
= 45 mN
As always, a free body diagram or a force diagram is essential.
From the first condition of equilibrium, we have
But that really means








More information is needed to solve for T and F_{x}. We can get that information from the second condition of equilibrium. Before we start to calculate torques, we must decide on the reference point about which we will calculate those torques. If we choose the lower, left end of the pole, there are two forces, F_{x} and F_{y}, that will have zero torque. That will reduce the number of terms in all of our equations. Therefore, that is a good choice for the origin or reference point or axis of rotation.
F_{x}:
= 0 (since r = 0 in
= r F sin
)
F_{y}:
= 0 (since r = 0 in
= r F sin
)
T:
_{ccw}
= (4 m)(T)(sin 30°) = (4 m) T (0.50)
= (2 m) T
600 N:
_{cw}
= (6 m)(600 N)(sin 60°) = (6 m)(600 N)(0.866) = 3118 mN
_{ccw}
=
_{cw}_{
}(2 m) T = 3118 mN
T = 1559 N
Therefore,
F_{x} = 1559 N
And we already knew
D11.8 A uniform pole 6 m long weighs 300 N and is attached by a pivot at one end to a wall. The pole is held at an angle of 30° above the horizontal by a horizontal guy wire attached to the pole 4.0 m from the end attached to the wall. A load of 600 N hangs from the upper end of the pole. Find the tension in the guy wire and the components of the force exerted on the pole by the wall.
(Compare this to the previous question, D11.7).
As always, a good free body diagram is essential!
From the first condition of equilibrium, we have
But that is a vector equation so it really means








More information is needed to solve for T and F_{x}. We can get that information from the second condition of equilibrium. Before we start to calculate torques, we must decide on the reference point about which we will calculate those torques. If we choose the lower, left end of the pole, there are two forces, F_{x} and F_{y}, that will have zero torque. That will reduce the number of terms in all of our equations. Therefore, that is a good choice for the origin or reference point or axis of rotation.
F_{x}:
= 0 (since r = 0 in
= r F sin
)
F_{y}:
= 0 (since r = 0 in
= r F sin
)
T: _{
ccw} = (4 m)(T)(sin 30°) = (4 m) T
(0.50) = (2 m) T
600 N:
_{cw}
= (6 m)(600 N)(sin 60°) = (6 m)(600 N)(0.866) = 3118 mN
300 N: _{cw} = (3 m)(300 N)(sin 60°) = (3 m)(300 N)(0.866) = 779 mN
_{
ccw} =
_{cw}_{
}(2 m) T = 3118 mN + 779
mN
(2 m) T = 3897 N
T = (3897/2) N
T = 1948.5 N
Therefore,
F_{x} = 1948.5 N
And we already knew
F_{y } = 900 N
D11.9 A carpenter's square has the dimensions shown in the figure. It is made of uniformly thick metal. Locate its center of gravity or center of mass.
Think of the square as being made of two rectangles.
The center of gravity of each rectangle is at the geometric center of that rectangle and the mass (or weight) located at that position is proportional to the area of that rectangle.
Notice that the positions are all measured from the lower left corner of the carpenter's square. Now we can use the defintions of center of gravity, Equations 8.10 and 8.11, to locate the center of gravity of the entire object,
Using the first condition of equilibrium, we have
F_{wall} = F_{f}
and
F_{N} = 1200 N
To solve for F_{wall} and, thus, for F_{f}, we must use the second condition of equilibrium. In calculating the torques, let us calculate torques about the foot of the ladder. That choice means two of the forces, F_{N} and F_{f }, will provide zero torque and that reduces the number of terms in our equations. Be very careful of the angles!
F_{wall}: _{cw} = (6.0 m)(F_{wall})(sin 60°) = (6.0 m)(F_{wall})(0.866)_{cw} = (5.2 m) F_{wall}800 N: _{ccw} = (6.0 m)(800 N)(sin 30°) = 2400 mN
400 N: _{ccw} = (2.0 m)(400 N)(sin 30°) = 400 mNF_{wall} : = 0
F_{f }: = 0
(C) 2005; Doug Davis, all rights reserved.
(c) 2005, Doug Davis; all rights reserved .