Physics 207 WebPages - Fall 1999
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CHAPTER 12: Rolling, Torque, and Angular MomentumCharles Phelps, Shawn Ryan, Liz Mackey
If you forget everything in this chapter, be proud that you at least remember: The angular momentum of any closed system is always constant. It will never
change no matter what you do. This is the big and powerful equation of -> Key concepts: Kinetic Energy of a rolling object- Okay, this chapter really isn’t too bad. Take kinetic energy of a rolling
object for example. If the object is rotating, then the Kinetic energy equality
is So if you have a rotating translational object, also known as a rolling object, then you just add the two equations together to get total Kinetic energy. Remember that I is the rotational inertia of the object through the center of mass and v is the velocity of the object through the center of mass. Torque- Torque is defined as P is a particle in this closed system. It position is at a distance r
from the reference point of O. O can be any reference point and the
distance to the particle from that point will be r. The torque created by
the force F on particle P (in relation to point O) is given
by Angular Momentum- Angular momentum is defined as Given a particle P with mass m, its angular momentum (in
reference to point O) is found by the equation Newton’s Second Law for Angular Momentum- Just like Newton’s Second law when dealing with forces, remember Angular Momentum of a Rigid Body rotating on a fixed axis- If an object is just spinning around its axis, then its angular momentum is
defined by
Major Problem Types- Rolling down ramps- For any object with a rotational inertia of I, the acceleration of the
object down an incline with an angle f is given by
the equation Finding Angular Momentum of an object with respect to a point- Any moving object or particle has an angular momentum with respect to a point
as defined by the equation A change in angular velocity of an object in a closed system- Problems like this are usually characterized by an object rotating on a fixed axis with a certain angular velocity and its rotational inertia changes to give it a new angular velocity. Two ways an object can change its rotational inertia are by redistributing its mass, like growing, shrinking, or by changing its configuration or by changing is mass altogether, such as having an object added to it, or mass taken away. Because angular momentum is always constant you can use the equation Inelastic collisions that spin an object about its axis- These problems are easy to recognize because they involve a moving object hitting a stationary object that will end up rotating because of the collision. The moving object will usually end up sticking to the rotating one (it makes things way easier). The moving object (sometimes a bullet or a rock) has an angular momentum with
respect to any point. The best point to pick is the point of rotation of the
soon to be spinning object. Since the moving object is the only thing moving, it
has all the angular momentum so
ProblemsA uniform sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to be 0.10g? (b) For this angle, what would be the acceleration of a frictionless block sliding down the incline?
Solution: (a). This is a rolling down a ramp problem, so the equation of choice is For this problem use the equation so the final equation is
Starting equation Expand the I g cancels from both sides and the MR2 cancel
(b) This is just an easy kinematics problem 
A solid cylinder of radius 10cm and mass 12 kg starts from rest and rolls without slipping a distance of 6m down a house roof that is inclined at 30° . (a) What is the angular speed of the cylinder about its center as it leaves the house roof? (b) The outside wall of the house is 5m high. How far from the edge of the roof does the cylinder hit the level ground?
Solution: (a) This is rolling down a ramp equation where v= w o = v/R = I = ½ MR2 -Rotational inertia for a cylinder a = g sinq /(1+ Icm /( MR2 )) -Equation 12-9 a = g sinq / (1 + ½ ) -MR2 cancels with the one from the I a = 2/3 g sinq -The acceleration of the object w o = ( w o = w o = 63 rad/s (b) This is just a projectile motion equation h = vyt + ½at2 vy = -v sin q h = -(v sin q )t - ½gt2 -5 = -(6.3 sin 30)t – 4.9t2 (Use quadratic to solve) t = 0.7386 s x = vxt vx = v cos q x = (v cos q )t x = (6.3 cos 30)(0.7386) x = 4. 0 m
What is the net torque about the origin on a flea located at coordinates (0, -4m, 5m) when forces F1 = (3N)k and F2 = (-2N)j acting on the flea?
Solution: This is just a torque about a point problem and you should use (aybz-azby)i +(azbx-axbz)j +(axby-aybx)k where a is the r vector and b is the F vector. The j and k values go to zero because the x component of both vectors are zero. t = r x Fnet = [-4j + 5k] x [3k + -2j] t = (-4(3)-(5(-2))i t = (-2i) N· m
What is the magnitude of the angular momentum, about the Earth’s center, of an 84kg person on the equator due to the rotation of the Earth? (The radius of the Earth is 6.37 x 106 meters.)
Solution: The Earth makes a rotation every day which is 24 hours long.
The angular momentum equation = 2.5 x 1011 kg· m2/s
A 2 kg object moves in a plane with velocity components vx = 30m/s, vy = 60m/s, and it passes through the point (x,y) = (3, -4)m. (a) What is its angular momentum relative to the origin at this moment? (b) What is its angulare momentum relative to the point (-2, -2)m at this same moment?
Solution: This is just a finding angular momentum of a particle problem and you should
use (aybz-azby)i +(azbx-axbz)j +(axby-aybx)k where a is the r vector and b is the v vector. The i and j values go to zero because the z component of both vectors are zero.
= (6 x 102 kg· m2 /s)k
A projectile of mass m is fired from the ground with an initial speed v0 and an initial angle q above the horizontal. (a) Find an expression for the magnitude of its angular momentum about the firing point as a function of time. (b) Find the rate at which the angular momentum changes with time. (c) Evaluate the magnitude of r x F directly and compare the result with (b). Why should the results be identical?
Solution: (a) This is a kinematics problem with some angular momentum stuff thrown in
to make it a little harder. The moment arm, r, is going to change with
the position of the projectile. Its vector is given as (vo cos q
)i + (vo sin q )j. Use the
equation r = (v0xt)i + (v0yt – ½gt2)j = (v0 cos q t)i + (v0 sin q – ½gt2)j Next the velocity is constant for the x direction but for the y direction the velocity is v = voy – gt = v0 sin q – gt. Set the velocity vector equal to this for both x and y directions. v = vx i + vy j = (v0 cos q )i + (v0 sin q – gt)j Cross the two together using the formula l = m(r x v) = -½mv0 cos q gt2k
(b) Just take the derivative of the above equation knowing that -½mv0 cos
q g is a constant and that dl/dt = -v0 mgt cos q 0 k
t = r x F = [(v0 cos q 0 t)i + ry j] x (-mg j) = -v0 mgt cos q 0 k (this is the same as in part b and this is because dl/dt = t = r x F)
A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 6m long, weighs 10N, and rotates at 240 rev/min clockwise when seen from above. Calculate (a) the rotational inertia of the rod about the axis of rotation and (b) the angular momentum of the rod about that axis.
Solution:
So the mass is W / g. The I of the rod equal to 1/3 ML2. I = 1/3 Ml2 = (1/3)[(10N)/(9.8m/s2 )](62 ) I = 12.2 kg· m2 (b) Angular momentum is defined as Iw . Remember that w is 2p * (rev/sec). L = Iw = (12.2 kg· m2 )[(2p )(240)/(60s)] L = 308 kg· m2/s downward
The figure shows a rigid structure consisting of a circular hoop, of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis with a period of rotation of 2.5s. Assuming R = .5m and m = 2 kg, calculate (a) the structures rotational inertia about the axis of rotation and (b) its angular momentum about that axis.
Solution: (a) The moment of inertia for the Square will the two thin rods perpendicular to the axis of rotation and the thin bar that is not on the axis of rotation (The other bar is entirely at a distance of zero from the axis and has a rotational inertia of zero). Square (moment of inertia): bar on axis of rotation does not contribute Other vertical bar = mR2 Each horizontal bar = 1/3 mR2 Total = mR2 + 2/3 mR2 For the hoop, use the parallel axis theorem once the rotational inertia has been found. Hoop (moment of inertia) = mR2 + ½mR2 = 3/2 mR2 I = 3/2 mR2 + mR2 + (2/3)mR2 = (19/6)mR2 = (19/6)(2kg)(.5m)2 I = 1.6 kg· m2 (b) Now, with the rotational inertia found, the angular momentum is easy.
Just remember to convert the period into angular speed with the equation L = Iw = I(2p /T) = [2p (1.6 kg· m2 )] / 2.5s L= 4 kg· m2
A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a weight in each hand. The rotational inertia of the system of man, weights, and platform about the central axis is 6 kg· m2 . If by moving the weights the man decreases the rotational inertia of the system to 2.0 kg· m2 , (a) what is the resulting angular speed of the platform and (b) what is the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What provided the added kinetic energy?
Solution: (a) The total angular momentum of the system of man, weights, & platform is conserved Li = Lf Iiw i = Ifw f w f = (Ii / If )w i = [(6 kg· m2 ) / (2 kg· m2 )](1.2 rev/s) w f = 3.6 rev/s (b) Remember that the Kinetic energy of a rotating body is Ki = ½Iiw i2 Kf = ½If w f 2 Kf / Ki = (If w f 2 ) / (Iiw i2 ) Kf / Ki = [(2kg· m2 )(3.6 rev/s)2 ] / [(6 kg· m2)(1.2 rev/s)2 ] Kf / Ki = 3.0 (c) The added kinetic energy came from the man doing work in decreasing the rotational inertia by moving the weights closer to his body.
A phonograph record of mass 0.1kg and radius 0.1m rotates about a vertical axis through its center with an angular speed of 4.7 rad/s. The rotational inertia of the record about its axis of rotation is 5 x 10-4 kg· m2 . A wad of putty of mass 0.02 kg drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately after the putty sticks to it?
Solution: Angular momentum is always conserved, so with the increase in mass to the record, the angular speed should now be slower. The putty is a point mass with a rotational inertia of Mr2. Putty: mass m Record: mass M, and radius r Li = Iiw i = Lf = If w f Initial momentum must be equal to the final momentum. w f = (Ii w i)/If = (Iiw i ) / (Ii + mr2 ) Solve for w f. w f = [5 x 10-4 kg· m2 )(4.7 rad/s)] / [5x10-4 kg· m2 + (.02kg)(.1m) 2 w f = 3.4 rad/s
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Email comments or questions about these WebPages
to jacarlis@vcu.edu This page was last updated on 12/13/99
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Do not forget this!
where I is the rotational
inertia of the object rotating around the axis through its center of mass and w
is the angular speed of the object. And if you have a pure translational moving
object (no rotation) then Kinetic energy found by 
.
Notice the remarkable similarities. The nice thing about uniform objects is that
the center of mass is smack dab in the center.
. This applies
to any particle that moves in any path in relation to a fixed point, not just
rotating systems.
or 
also known as 
. Remember that this
just isn’t only for rotating objects. Every object has a chance to share in
this equation.
.
or 
. You can think of r^
as the closest particle P will pass to point O with the given
velocity. Choosing a good reference point, such as point O, will become
important when dealing with Equilibrium type of equations as seen from Chapter
13.
,
Angular momentum has its own version of this law, which is 
.
This is for a single particle. However for a system of particles the equation is
pretty much the same thing 
where 
.
For a system of particles, the total angular momentum is equal to the total of
all the angular momentum of each particle in the system.
. Notice the similarities
to the linear momentum equation 
.
Freaky, isn’t it?
.
where Io is the object’s starting rotational inertia and If
is its final rotational inertia.
. When
it hits the stationary object, it will start spinning and has an angular
momentum of 
. I of course
depends on several things which are usually supplied, include the mass of the
object, M, and the mass of the first object, m. This will make the
total mass equal to M+m (this is important, don’t forget it). So, now
the final angular momentum of the system is 
.
Because angular momentum never changes in a system, then 
which means 
. The only tricky thing to
put in here is to give the rotating object an intitial angular velocity of w
o which means 
where Io
is the object’s rotational inertia before the small object is added to it, so
its mass is just M. This leads to a final equation of 
.
and you want 
.
.
The sphere has a mass M and radius R.
-Velocity of the objet after
accelerating for a distance d.
/
R -This is the angular speed equation
)/
R = 
Substitute for a into
equation
Solve for w
.
The flea is at (0, -4, 5) which is the moment arm, r. The force vector is
the 3k + -2j which is equal to (-2, 3). To cross the two vectors
use the formula:
. This is
the time it takes for 1 revolution.
. The object is at (3i, -4j)
which is the moment arm, r. The velocity vector is (30i + 60j).
To cross the two vectors use the formula:
for the distance in the x
direction and 
for the y
direction. Set r equal to this.
and
then simplify.
.
Multiply the two together and you get …
.
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