**Chapter
10; Rotational Kinematics and Energy **

**| ToC,
Chapter 10 | Course Calendar
|**

**D10.1** It requires 6 seconds for a phonograph
turntable initially rotating at 33 rpm to reach 45 rpm. Assume the angular acceleration
is uniform. How many revolutions are made during this time?

=

**
= _{i}
+ _{i}
t + **

= 0 + (3.46

= 20.76 radians + 3.75 radians = 24.51 radians

= 24.51 radians (

**D10.2** A ventilator fan is turning at
600 rpm when the power is cut off and it turns through 1000 revolutions while
coasting to a stop. Find the angular acceleration and the time required to stop.

_{i}
= 600 **
****
****
= 62.8 **

w_{f} = 0

( _{f}
- _{i}
) = 1000 rev **
= 6 280 rad
**

(0)

(12 560 rad) = - 3 948

= - 0.31

=

0 = 62.8

(0.31

t = 203 s

**D10.3** 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**

F

12 N - T = (1.2 kg) a

**For the rotating cylinder,
apply the rotational form of Newton's Second Law,**

**since
= 90°**

**For a solid
cylinder,**

I = (0.5) (5.0 kg) (0.12 m)

I = 0.036 kg m

**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**

a = (0.12 m)

= a / 0.12 m

(0.12 m) T = (0.036 kg m

T = (0.036 kg m

T = (2.5 kg) a

**Now we have two
equations in two unknowns**

12 N - (2.5 kg) a = (1.2 kg) a

12 N - (2.5 kg) a = (1.2 kg) a

12 N = (3.7 kg) a

a =

a = 3.24

**T = 8.1
N**

**D10.4** 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 10-kg 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**

(0.10 m) T

**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**

a = (0.10 m)

= a / 0.10 m

(0.10 m) T

(0.10 m) T

T

**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,**

(5 kg)(10 m/s

50 kg m/s

**Now we have three
equations for three unknowns,**

T

50 N - T

T

T

50 N - [20 N + (190 kg) a] = (5 kg) a

30 N = (195 kg) a

a =

a = 0.15

= a / 0.10 m = 1.5

T

**T _{R} = 20 N + (190 kg) (0.15
**

**D10.5** Assume a playground merry-go-round
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 merry-go- 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 merry-go-round is**

**The initial angular momentum of
the child is**

**The initial angular momentum of
the merry-go-round is zero.
So the total initial angular momentum of the entire system
is**

**If the bearings of the
merry-go-round are well oiled this should be the total final
angular momentum,**

L

**but now the moment of inertia
is the moment of inertia of the merry-go-round and the
child**

I

I

L

(405 kg m

= 0.89

**D10.6** What is the kinetic energy of
tire with moment of inertia of 60 kg m^{2} that rotates at 150 rpm?

= 150

KE =

**D10.7** A disk rolls without slipping
down a hill of height 10.0 m. If the disk starts from rest at the top of the
hill, what is its speed at the bottom?

**Initially, the disk has gravitational potential energy**

**At the bottom all of this
energy has become kinetic energy-but we have both kinetic
energy of translation and kinetic energy of
rotation,**

KE

**If, as we assume, the disk
rolls without slipping, then**

**or**

**Notice that we are not
given the radius or the mass of the disk. This may be somewhat
disconcerting, but we will continue anyway and expect that these will
"disappear" in the end.**

KE

KE

For a disk,

I =

**Therefore,**

KE

E

v

v = 11.5 m/s

**D10.8** An Atwood's machine is composed
of a 2-kg mass and a 2.5-kg mass attahed to a string hanging over a wheel that
has a radius of 10 cm and a moment of inertia of 12.5 kg-m^{2}. The
2.5-kg mass is initially 1.0 m above the floor. Use conservation of energy to
find the speed of this mass just before it hits the floor.

**Initially,**

PE

PE

E

**Finally (just before the 2.5 kg
block hits the floor),**

KE

KE

= (

PE

PE

E

E

E

v

v = 0.088 m/s

v = 8.8 cm/s

**D10.9** 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**

I = (

The velocity of the load and the angular velocity of the windlass are connected by

v = (0.1 m)

=

As always, good diagrams make life much simpler.

**Initially, before the windlass
and load are released, we have**

PE

E

**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

KE

KE

PE

E

(12.5 kg) v

v

v

v = 9.7 m/s

v = 9.7 m/s

**D10.10** 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.

I

I

I

I

**The initial angular momentum
is**

L

KE

KE

**(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,

I

I

I

**Total angular momentum is
conserved,**

L

(1.01 kg m

KE

KE

__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.

(c) 2005, Doug Davis; all rights reserved** .**