drawing of hand and foot

Raymond A. Serway Chris Vuille Chapter Eight Rota:onal Equilibrium and Rota:onal Dynamics Applica:on of Forces •  The point of applica:on of a force is important –  This was ignored in trea:ng objects as point par:cles •  The concepts of rota:onal equilibrium and rota:onal dynamics are important in many fields of study •  Angular momentum may be conserved •  Angular momentum may be changed by exer:ng a torque Introduc:on Force vs. Torque •  Forces cause accelera:ons •  Torques cause angular accelera:ons •  Force and torque are related Sec:on 8.1 Torque •  The door is free to rotate about an axis through O •  There are three factors that determine the effec:veness of the force in opening the door: –  The magnitude of the force –  The posi-on of the applica:on of the force –  The angle at which the force is applied Sec:on 8.1 Torque, cont •  Torque, τ, is the tendency of a force to rotate an object about some axis –  τ = r F •  τ is the torque –  Symbol is the Greek tau •  r is the length of the posi:on vector •  F is the tangen:al force •  SI unit is Newton . meter (N.m) Sec:on 8.1 Torque
Only the tangential component of force causes a torque:
General Definition of Torque
Torque is defined as the cross product :
  
τ =r×F

A particle is located at position r relative to its axis

of rotation. When a force F is applied to the particle,
only the tangential component F⊥ produces a torque.
 
This torque has a magnitude τ = | r || F | sinθ
and is directed outward from the page. θ is the angle
between the position vector and the force.
Right Hand Rule •  Point the fingers in the direc:on of the posi:on vector •  Curl the fingers toward the force vector •  The thumb points in the direc:on of the torque Sec:on 8.1 Lever Arm •  The lever arm, d, is the perpendicular distance from the axis of rota:on to a line drawn along the direc:on of the force •  d = r sin θ
•  This also gives τ = r F sin θ
Sec:on 8.1 The Swinging Door: (a) A man applies a force of F=3.00 × 102 N at an angle of 60.0° to the door of Figure a, 2.00 m from the hinges. Find the torque on the door, choosing the posi:on of the hinges as the axis of rota:on. (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door? The Swinging Door: (a) A man applies a force of F=3.00 × 102 N at an angle of 60.0° to the door of Figure 8.7a, 2.00 m from the hinges. Find the torque on the door, choosing the posi:on of the hinges as the axis of rota:on. (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door?  
(a) τ F = r1 × F = r1Fsinθ = (2.00m)(3.00 × 10 2 N)sin60  = 5.2 × 10 2 Nm
(b) Mimimum force occurs when the torque from the force
is equal in magnitude to the torque from the wedge.
τ F = τW (2)
τW = r2FWsinθ = r2FW sin90  = r2FW = (1.5m)FW (3)
5.2 × 10 2 Nm
= 347N
(1)^(2)^(3) ⇒ 5.2 × 10 Nm = (1.5m)FW ⇒ FW =
1.5m
2
€
(1)
Mul:ple Torques •  When two or more torques are ac:ng on an object, the torques are added –  As vectors •  If the net torque is zero, the object’s rate of rota:on doesn’t change Sec:on 8.1 Torque and Axis •  The value of the torque depends on the chosen axis of rota:on •  Torques can be computed around any axis –  There doesn’t have to be a physical rota:on axis present •  Once a point is chosen, it must be used consistently throughout a given problem Sec:on 8.1 Net Torque •  The net torque is the sum of all the torques produced by all the forces –  Remember to account for the direc:on of the tendency for rota:on •  Counterclockwise torques are posi:ve •  Clockwise torques are nega:ve Sec:on 8.2 Torque and Equilibrium •  First Condi:on of Equilibrium –  The net external force must be zero –  This is a statement of transla:onal equilibrium •  The Second Condi:on of Equilibrium states –  The net external torque must be zero –  This is a statement of rota:onal equilibrium Sec:on 8.2 Selec:ng an Axis •  It’s usually best to choose an axis that will make at least one torque equal to zero –  This will simplify the torque equa:on Sec:on 8.2 Balancing Act: A woman of mass m = 55.0 kg sits on the lec end of a see-­‐ saw—a plank of length L = 4.00 m, pivoted in the middle as in the Figure. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 75.0 kg sit if the system (seesaw plus man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of mpl = 12.0 kg. (c) Repeat part (a), but this :me compute the torques about an axis through the lec end of the plank. Balancing Act: A woman of mass m = 55.0 kg sits on the lec end of a see-­‐ saw—a plank of length L = 4.00 m, pivoted in the middle as in the Figure. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 75.0 kg sit if the system (seesaw plus man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of mpl = 12.0 kg. (c) Repeat part (a), but this :me compute the torques about an axis through the lec end of the plank. τ boy = −xMg τ girl = (2.00m) mg τ plank = 0 τ n = 0
(a) For rotational equilibrium ∑τ = 0 ⇒
τ boy + τ girl + τ plank + τ n = 0 ⇒ −xMg + (2.00m) mg = 0 ⇒
x=
(2.00m) m (2.00m)55.0kg
M
(
=
75kg
= 1.47m
)
(b) n = mpl + M + m g
(c) τ boy = −( x + 2.00m) Mg τ girl = 0 τ plank = −(2.00m) mpl g τ n = (2.00m)n
τ boy + τ girl + τ plank + τ n = 0 ⇒
(
)
−( x + 2.00m) Mg − (2.00m) mpl g + (2.00m) mpl + M + m g = 0 ⇒
x = 1.47m
Center of Gravity •  In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at a single point Sec:on 8.3 Calcula:ng the Center of Gravity •  The object is divided up into a large number of very small par:cles of weight (mg) •  Each par:cle will have a set of coordinates indica:ng its loca:on (x,y) Sec:on 8.3 Calcula:ng the Center of Gravity, cont. •  We assume the object is free to rotate about its center •  The torque produced by each par:cle about the axis of rota:on is equal to its weight :mes its lever arm –  For example, τ1 = m1 g x1 Sec:on 8.3 Calcula:ng the Center of Gravity, cont. •  We wish to locate the point of applica:on of the single force whose magnitude is equal to the weight of the object, and whose effect on the rota:on is the same as all the individual par:cles. •  This point is called the center of gravity of the object Sec:on 8.3 Coordinates of the Center of Gravity •  The coordinates of the center of gravity can be found from the sum of the torques ac:ng on the individual par:cles being set equal to the torque produced by the weight of the object Sec:on 8.3 Center of Gravity and Center of Mass •  The three equa:ons giving the coordinates of the center of gravity of an object are iden:cal to the equa:ons giving the coordinates of the center of mass of the object •  The center of gravity and the center of mass of the object are the same if the value of g does not vary significantly over the object Sec:on 8.3 Center of Gravity of a Uniform Object •  The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry •  Ocen, the center of gravity of such an object is the geometric center of the object Sec:on 8.3 Experimentally Determining the Center of Gravity •  The wrench is hung freely from two different pivots •  The intersec:on of the lines indicates the center of gravity •  A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is ac:ng upward through the object’s center of gravity Sec:on 8.3 Where Is the Center of Gravity?: (a) Three objects are located in a coordinate system as shown in Figure a. Find the center of gravity. (b) How does the answer change if the object on the lec is displaced upward by 1.00 m and the object on the right is displaced downward by 0.500 m (Figure b)? Treat the objects as point par:cles. Where Is the Center of Gravity?: (a) Three objects are located in a coordinate system as shown in Figure a. Find the center of gravity. (b) How does the answer change if the object on the lec is displaced upward by 1.00 m and the object on the right is displaced downward by 0.500 m (Figure b)? Treat the objects as point par:cles. Rcm =
∑rm
∑m
i
i
i
(a)
Rcm =
(b) Rcm,x =
(−0.500m)5.00kg + 0 + (1.00m) 4.00kg
11.00kg
∑x m ,
∑m
i
i
Rcm,y =
i
∑y m
∑m
i
i
i
Rcm,x = 0.136m
Rcm,y =
(1.00m)5.00kg − (0.50m)4.00kg
11.00kg
2
2
Rcm = Rcm,x
+ Rcm,y
= 0.3 m
€
= 0.27m
= 0.136m
Locating Your Lab Partner’s Center of Gravity: In this example we show how to find the loca:on of a person’s center of gravity. Suppose your lab partner has a height L of 173 cm (5 c, 8 in) and a weight w of 715 N (160 lb). You can determine the posi:on of his center of gravity by having him stretch out on a uniform board supported at one end by a scale, as shown in Figure 8.12. If the board’s weight wb is 49 N and the scale reading F is 3.50 × 102 N, find the distance of your lab partner’s center of gravity from the lec end of the board. Loca=ng Your Lab Partner’s Center of Gravity: In this example we show how to find the loca:on of a person’s center of gravity. Suppose your lab partner has a height L of 173 cm (5 c, 8 in) and a weight w of 715 N (160 lb). You can determine the posi:on of his center of gravity by having him stretch out on a uniform board supported at one end by a scale, as shown in Figure 8.12. If the board’s weight wb is 49 N and the scale reading F is 3.50 × 102 N, find the distance of your lab partner’s center of gravity from the lec end of the board. €
∑F = 0 ⇒ n + F = w + w
b
⇒ n = 715N + 49N - 350N = 414N
⎛ L ⎞
FL − ⎜ ⎟ w b
⎛ L ⎞
⎝ 2 ⎠
= 0.079m
∑τ = 0 ⇒ −x cg w − ⎜⎝ 2 ⎟⎠wb + FL = 0 ⇒ x cg =
w
€
Notes About Equilibrium •  A zero net torque does not mean the absence of rota:onal mo:on –  An object that rotates at uniform angular velocity can be under the influence of a zero net torque •  This is analogous to the transla:onal situa:on where a zero net force does not mean the object is not in mo:on Sec:on 8.4 Solving Equilibrium Problems •  Diagram the system –  Include coordinates and choose a convenient rota:on axis •  Draw a free body diagram showing all the external forces ac:ng on the object –  For systems containing more than one object, draw a separate free body diagram for each object Sec:on 8.4 Problem Solving, cont. •  Apply the Second Condi:on of Equilibrium –  This will yield a single equa:on, ocen with one unknown which can be solved immediately •  Apply the First Condi:on of Equilibrium –  This will give you two more equa:ons •  Solve the resul:ng simultaneous equa:ons for all of the unknowns –  Solving by subs:tu:on is generally easiest Sec:on 8.4 A Weighted Forearm: A 50.0-­‐N (11-­‐lb) bowling ball is held in a person’s hand with the forearm horizontal, as in Figure a. The biceps muscle is akached 0.030 0 m from the joint, and the ball is 0.350 m from the S joint. Find the upward force F exerted by the biceps on the forearm (the ulna) and the downward force R exerted by the humerus on the forearm, ac:ng at the joint. Neglect the weight of the forearm and slight devia:on from the ver:cal of the biceps. A Weighted Forearm: A 50.0-­‐N (11-­‐lb) bowling ball is held in a person’s hand with the forearm horizontal, as in Figure a. The biceps muscle is akached 0.030 0 m from the joint, and the ball is 0.350 m from the joint. Find the upward force F exerted by the biceps on the forearm (the ulna) and the downward force R exerted by the humerus on the forearm, ac:ng at the joint. Neglect the weight of the forearm and slight devia:on from the ver:cal of the biceps. ∑τ = 0 ⇒ F (0.03m) − (50N)(0.35m) = 0 ⇒
F=
(50N)(0.05m)
0.03m
= 583.3N
∑ F = 0 ⇒ R + 50.0N = F ⇒ R = 583.3N - 50.0N = 533N
€
Don’t Climb the Ladder: A uniform ladder 10.0 m long and weighing 50.0 N rests against a smooth ver:cal wall as in Figure a. If the ladder is just on the verge of slipping when it makes a 50.0° angle with the ground, find the coefficient of sta:c fric:on between the ladder and ground. Don’t Climb the Ladder: A uniform ladder 10.0 m long and weighing 50.0 N rests against a smooth ver:cal wall as in Figure a. If the ladder is just on the verge of slipping when it makes a 50.0° angle with the ground, find the coefficient of sta:c fric:on between the ladder and ground. 
∑τ = 0 ⇒ −(5m)(50N) sin140 + (10m)P sin130
(5m)(50N) sin140 
= 21N
P=

10m
sin130
( )
∑F
∑F
x
y
= 0 ⇒ f − P = 0 ⇒ f = P = 21N
= 0 ⇒ n = 50N
f = µsn ⇒ µs =
€
f 21N
=
= 0.42
n 50N

=0
Walking A horizontal Beam: A uniform horizontal beam 5.00 m long and weighing 3.00 × 102 N is akached to a wall by a pin connec:on that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal (Fig. a). If a person weighing 6.00 × 102 N stands 1.50 m from the wall, find the magnitude of the tension T in the cable and the force R exerted by the wall on the beam. Walking A horizontal Beam: A uniform horizontal beam 5.00 m long and weighing 3.00 × 102 N is akached to a wall by a pin connec:on that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal (Fig. a). If a person weighing 6.00 × 102 N stands 1.50 m from the wall, find the magnitude of the tension T in the cable and the force R exerted by the wall on the beam. €
∑τ = 0 ⇒ −(1.50m)(600N) − (2.50m)(300N) + (5.00m)T sin127
T=
(1.50m)(600N) + (2.50m)( 300N)
= 413N
(5.00m) sin127 
x
= 0 ⇒ Rx − Tx = 0 ⇒ Rx = Tx = T cos53 = 249N
y
= 0 ⇒ Ry − 600N − 300N − T sin53 ⇒ Ry = 570N
∑F
∑F

=0
Torque and Angular Accelera:on •  When a rigid object is subject to a net torque (Στ ≠ 0), it undergoes an angular accelera:on •  The angular accelera:on is directly propor:onal to the net torque –  The rela:onship is analogous to ∑F = ma •  Newton’s Second Law Sec:on 8.5 Moment of Iner:a •  The angular accelera:on is inversely propor:onal to the analogy of the mass in a rota:ng system •  This mass analog is called the moment of iner-a, I, of the object –  SI units are kg m2 Sec:on 8.5 Newton’s Second Law for a Rota:ng Object •  The angular accelera:on is directly propor:onal to the net torque •  The angular accelera:on is inversely propor:onal to the moment of iner:a of the object Sec:on 8.5 More About Moment of Iner:a •  There is a major difference between moment of iner:a and mass: the moment of iner:a depends on the quan:ty of maker and its distribu:on in the rigid object •  The moment of iner:a also depends upon the loca:on of the axis of rota:on Sec:on 8.5 Moment of Iner:a of a Uniform Ring •  Imagine the hoop is divided into a number of small segments, m1 … •  These segments are equidistant from the axis Sec:on 8.5 Other Moments of Iner:a Sec:on 8.5 The Baton Twirler: In an effort to be the star of the halcime show, a majoreke twirls an unusual baton made up of four spheres fastened to the ends of very light rods (see Figure). Each rod is 1.0 m long. (a) Find the moment of iner:a of the baton about an axis perpendicular to the page and passing through the point where the rods cross. (b) The majoreke tries spinning her strange baton about the axis OO’, as shown in the Figure. Calculate the moment of iner:a of the baton about this axis. The Baton Twirler: In an effort to be the star of the halcime show, a majoreke twirls an unusual baton made up of four spheres fastened to the ends of very light rods (see Figure). Each rod is 1.0 m long. (a) Find the moment of iner:a of the baton about an axis perpendicular to the page and passing through the point where the rods cross. (b) The majoreke tries spinning her strange baton about the axis OO’, as shown in the Figure. Calculate the moment of iner:a of the baton about this axis. 2
2
(a) I = ∑ mi ri2 = 2(0.2kg)(0.5m) + 2(0.3kg)(0.5m) = 0.25 kg m2
(b) I = ∑ mi ri2 ≅ 2(0.3kg)(0.5m)
€
2
Warming Up: A baseball player loosening up his arm before a game tosses a
0.150-kg baseball, using only the rotation of his forearm to accelerate the ball
(see Figure). The forearm has a mass of 1.50 kg and a length of 0.350 m. The
ball starts at rest and is released with a speed of 30.0 m/s in 0.300 s. (a) Find
the constant angular acceleration of the arm and ball. (b) Calculate the
moment of inertia of the system consisting of the forearm and ball. (c) Find the
torque exerted on the system that results in the angular acceleration found in
part (a).
Warming Up: A baseball player loosening up his arm before a game tosses a
0.150-kg baseball, using only the rotation of his forearm to accelerate the ball
(see Figure). The forearm has a mass of 1.50 kg and a length of 0.350 m. The
ball starts at rest and is released with a speed of 30.0 m/s in 0.300 s. (a) Find
the constant angular acceleration of the arm and ball. (b) Calculate the
moment of inertia of the system consisting of the forearm and ball. (c) Find the
torque exerted on the system that results in the angular acceleration found in
part (a).
(a)
v t = ωr ⇒ ω =
v t 30.0m/s
=
= 85.7rad/s
r 0.350m
ω f = α ( Δt ) ⇒ α =
ωf
85.7rad/s
=
= 286rad /s2
(Δt ) 0.300s
2
(b) I = ∑ mi ri2 = IBall + IArm = (0.15kg)(0.350m) +
1
2
1.50kg)(0.350m)
(
2
I = 7.97 × 10 -2 kgm2
(c) τ = Iα ⇒ ( 7.97 × 10 -2 kg m2 )(286 rad/s2 ) = 22.8 N m
€
Rota:onal Kine:c Energy •  An object rota:ng about some axis with an angular speed, ω, has rota:onal kine:c energy KEr = ½Iω2 •  Energy concepts can be useful for simplifying the analysis of rota:onal mo:on Sec:on 8.6 Conserva:on of Energy •  Conserva:on of Mechanical Energy –  Remember, this is for conserva:ve forces, no dissipa:ve forces such as fric:on can be present –  Poten:al energies of any other conserva:ve forces could be added Sec:on 8.6 Work-­‐Energy in a Rota:ng System •  In the case where there are dissipa:ve forces such as fric:on, use the generalized Work-­‐
Energy Theorem instead of Conserva:on of Energy •  Wnc = ΔE = ΔKEt + ΔKEr + ΔPE Sec:on 8.6 Angular Momentum •  Similarly to the rela:onship between force and momentum in a linear system, we can show the rela:onship between torque and angular momentum •  Angular momentum is defined as –  L = I ω –  and Sec:on 8.7 Angular Momentum, cont •  If the net torque is zero, the angular momentum remains constant •  Conserva-on of Angular Momentum states: Let Li and Lf be the angular momenta of a system at two different :mes, and suppose there is no net external torque so Σ τ = 0, then angular momentum is said to be conserved Sec:on 8.7 Conserva:on of Angular Momentum •  Mathema:cally, when –  Applies to macroscopic objects as well as atoms and molecules Sec:on 8.7 Conserva:on of Angular Momentum, Example •  With hands and feet drawn closer to the body, the skater’s angular speed increases –  L is conserved, I decreases, ω increases Sec:on 8.7 Conserva:on of Angular Momentum, Example, cont •  Coming out of the spin, arms and legs are extended and rota:on is slowed –  L is conserved, I increases, ω decreases Sec:on 8.7 Conserva:on of Angular Moment, Astronomy Example •  Crab Nebula, result of supernova •  Center is a neutron star –  As the star’s moment of iner:a decreases, its rota:onal speed increases Sec:on 8.7 The Falling Bucket: A solid, uniform, fric:onless cylindrical reel of mass M = 3.00 kg and radius R = 0.400 m is used to draw water from a well. A bucket of mass m = 2.00 kg is akached to a cord that is wrapped around the cylinder. (a) Find the tension T in the cord and accelera:on a of the bucket. (b) If the bucket starts from rest at the top of the well and falls for 3.00 s before hivng the water, how far does it fall? The Falling Bucket: A solid, uniform, fric:onless cylindrical reel of mass M = 3.00 kg and radius R = 0.400 m is used to draw water from a well. A bucket of mass m = 2.00 kg is akached to a cord that is wrapped around the cylinder. (a) Find the tension T in the cord and accelera:on a of the bucket. (b) If the bucket starts from rest at the top of the well and falls for 3.00 s before hivng the water, how far does it fall? (a)
∑ F = ma ∑τ = Iα
∑ F = ma
⇒ T - mg = ma ⇒ T = m( a + g)
1
2T
∑τ = Iα ⇒ −Tr = 2 m r α ⇒ α = − m r
2
d
d
a = αr = −
2T
md
⎛ 2m ⎞
⎛ 2T
⎞
(2.00kg)(9.8m/s2 )
mg
T = m⎜ −
+ g⎟ ⇒ T⎜1+
=
= 8.4 N
⎟ = mg ⇒ T =
2m
2(2.00kg)
⎝ md ⎠
⎝ md
⎠
1+
1+
md
3.00kg
2T 2(8.4 N)
=
= −5.6 m/s2
md
3.00kg
1
(b) Δy = at 2 = −25.2m
2
a=−
€
A Ball Rolling Down an Incline: A ball of mass M and radius R starts from rest at a height of 2.00 m and rolls down a 30.0° slope, as in the Figure. What is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping. A Ball Rolling Down an Incline: A ball of mass M and radius R starts from rest at a height of 2.00 m and rolls down a 30.0° slope, as in Figure 8.27. What is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping. Ei = E f
Mgh =
I=
1
1
Mv 2 + Iω 2
2
2
2
MR 2
5
For no slipping v = v t = ωR
2
1
1 2 1
12
1
1
2
2
2 v
Mgh = Mv + Iω = Mv +
MR 2 = Mv 2 + Mv 2 ⇒
2
2
2
25
2
5
R
7
10gh
Mgh = Mv 2 ⇒ v 2 =
10
7
€
Blocks and Pulley: Two blocks with masses m1 = 5.00 kg and m2 = 7.00 kg
are attached by a string as in the Figure, over a pulley with mass M = 2.00 kg.
The pulley, which turns on a frictionless axle, is a hollow cylinder with radius
0.0500 m over which the string moves without slipping. The horizontal surface
has coefficient of kinetic friction 0.350. Find the speed of the system when the
block of mass m2 has dropped 2.00 m.
Blocks and Pulley: Two blocks with masses m1 = 5.00 kg and m2 = 7.00 kg
are attached by a string as in the Figure, over a pulley with mass M = 2.00 kg.
The pulley, which turns on a frictionless axle, is a hollow cylinder with radius
0.0500 m over which the string moves without slipping. The horizontal surface
has coefficient of kinetic friction 0.350. Find the speed of the system when the
block of mass m2 has dropped 2.00 m.
Since friction is present apply the work - energy theorem : W nc = ΔE
2
1
1
2 v
2
m
+
m
v
+
MR
− m2 g( h + 2.00m)
(
2) f
2 1
2
R2
E i = KE i + PE i = −m2 gh (assumed table level has PE = 0 and h is the
initial distance of m2 from the table)
E f = KE f + PE f =
2
1
1
2 v
2
ΔE = ( m1 + m2 )v f + MR 2 − m2 g(2.00m)
2
2
R
W nc = − f k (2.00m) = −µk n (2.00m) = − µk m1g(2.00m)
−µk m1g(2.00m) =
1
1
m1 + m2 )v 2f + Mv 2f − m2 g(2.00m) ⇒
(
2
2
1
(m + m2 + M ) = m2g(2.00m) − µk m1g(2.00m) ⇒
2 1
g( 4.00m)( m2 − µk m1 )
v 2f =
⇒ v f = 3.83 m/s
m1 + m2 + M
v 2f
€
The Spinning Stool: A student sits on a pivoted stool while holding a pair of weights. (See Figure) The stool is free to rotate about a ver:cal axis with negligible fric:on. The moment of iner:a of student, weights, and stool is 2.25 kg m2 . The student is set in rota:on with arms outstretched, making one complete turn every 1.26 s, arms outstretched. (a) What is the ini:al angular speed of the system? (b) As he rotates, he pulls the weights inward so that the new moment of iner:a of the system (student, objects, and stool) becomes 1.80 kg m2 . What is the new angular speed of the system? (c) Find the work done by the student on the system while pulling in the weights. (Ignore energy lost through dissipa:on in his muscles.) The Spinning Stool: A student sits on a pivoted stool while holding a pair of weights. (See Figure) The stool is free to rotate about a ver:cal axis with negligible fric:on. The moment of iner:a of student, weights, and stool is 2.25 kg m2 . The student is set in rota:on with arms outstretched, making one complete turn every 1.26 s, arms outstretched. (a) What is the ini:al angular speed of the system? (b) As he rotates, he pulls the weights inward so that the new moment of iner:a of the system (student, objects, and stool) becomes 1.80 kg m2 . What is the new angular speed of the system? (c) Find the work done by the student on the system while pulling in the weights. (Ignore energy lost through dissipa:on in his muscles.) Since the net external torque on the student is zero we can
apply conservation of angular momentum.
L i = L f ⇒ I1ω1 = I2ω 2
(a) ω1 =
1 rev 2π rad
rad
=
= 4.99
1.26 s
s
1.26 s
2
(b) ω 2 =
I1ω1
=
I2
⎛
(2.25 kg m )⎜⎝ 4.99
1.80 kg m2
rad ⎞
⎟
s ⎠
= 6.24
rad
s
1
1
(c) Wnet = ΔKE = KE f − KE i = I2ω 22 − I1ω12 = 7.03 J
2
2
€
The Merry-Go-Round: A merry-go-round modeled as a disk of mass M = 1.00 ×
102 kg and radius R = 2.00 m is rotating in a horizontal plane about a frictionless
vertical axle. (a) After a student with mass m = 60.0 kg jumps on the rim of the
merry-go-round, the system’s angular speed decreases to 2.00 rad/s. If the
student walks slowly from the edge toward the center, find the angular speed of
the system when she reaches a point 0.500 m from the center. (b) Find the
change in the system’s rotational kinetic energy caused by her movement to r =
0.500 m. (c) Find the work done on the student as she walks to r = 0.500 m.
The Merry-­‐Go-­‐Round: A merry-­‐go-­‐round modeled as a disk of mass M=1.00×102 kg and radius R = 2.00 m is rota:ng in a horizontal plane about a fric:onless ver:cal axle. (a) Acer a student with mass m = 60.0 kg jumps on the rim of the merry-­‐go-­‐
round, the system’s angular speed decreases to 2.00 rad/s. If the student walks slowly from the edge toward the center, find the angular speed of the system when she reaches a point 0.500 m from the center. (b) Find the change in the system’s rota:onal kine:c energy caused by her movement to r = 0.500 m. (c) Find the work done on the student as she walks to r = 0.500 m. The moment of inertia of the disk + student initially is :
1
2
MR 2 + mR 2 = 200 kg m2 + (60kg)(2.00m) = 440 kg m2
2
The final moment of inertia of the disk + student is :
1
2
I2 = MR 2 + m(0.500m) 2 = 200 kg m2 + (60kg)(0.500m) = 215 kg m2
2
I1 =
We apply conservation of angular momentum:
L i = L f ⇒ I1ω1 = I2ω 2
2
I1ω1 ( 440 kg m )(2.00rad/s)
(a) ω 2 =
=
= 4.09rad/s
I2
215 kg m2
1
1
(b) ΔKE system = KE f - KE i = I2ω 22 − I1ω12 = 920 J
2
2
1
1
(c) Wnet,st = ΔKE st = KE f ,st - KE i,st = I2,stω 22 − I1,stω12 ⇒
2
2
2
2
2
2
Wnet,st = 0.5(60kg)(0.500m) ( 4.09rad/s) − 0.5(60kg)(2.00m) (2.00rad/s) ⇒
Wnet,st = −355 J
€