Transcript Hints and Examples in Chapter 10

Hints and Examples in
Chapter 10
Hints
10-10: t=r*F and t=Ia
 10-32: Part a) See previous hint. b)
w2=w20+2*a*q c) W=tq d) P=tw
 10-40: I1=Ibody + Ithin rod at r=1.2 m and I2=Ibody
+ Ithin rod at r=25 cm and then use
conservation of L
 10-54: q= ½ at2 where s=rq
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10:15
A solid uniform cylinder with mass of 8.25 kg and
diameter of 15 cm is spinning at 220 rpm on a
thin frictionless axle that passes along the
cylinder axis. You design a simple friction
brake to stop the cylinder by pressing the
brake against the outer rim with a normal
force. The coefficient of friction between the
brake and rim is 0.333. What must the normal
force be to bring the cylinder to rest after it has
turned through 5.25 revolutions?
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The total angular distance is 5.25 revolutions
or 33 radians.
w0=220 rpm=23 rad/s
tq=W=-mk*n*R
Iaq=W=-mk*n*R
I= ½ *m*R2 = ½ (8.25)(0.075 m)2=0.023 kg*m2
w2=w20+2*a*q so –w02/2q=a=-8.04 rad/s2
n= .023*8.04*33/0.333/.075=7.47 N
10:39
A small block on a frictionless horizontal surface has a
mass 0.025 kg. It is attached to a massless cord
passing through a hole in the surface. The block is
originally revolving at a distance of 0.3 m from the
hole with an angular speed of 1.75 rad/s . The cord
is then pulled from below shortening the distance the
radius in which the block revolves to 0.15 m. Treat
the block as a particle
a) Is angular momentum conserved?
b) What is the new angular speed?
c) Find the change in kinetic energy of the block.
d) How much work was done by pulling the cord?
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Part a) Since there is no torque acting on the
particle (its angular velocity is not changing),
the L is conserved.
Part b) Let I=mr2 so I1w1=I2w2 so w2=w1*(r1/r2)2
w2=4*w1=7 rad/s
Part c) DK= ½ I2w22-½ I1w21=1.03 x 10-2 J
Part d) no other force does work so just
change in KE
10:43
A large wooden turntable in the shape of a flat
disk has a radius of 2 m and total mass of
120 kg. The turntable is initially rotating at 3
rad/s about a vertical axis through its center.
Suddenly a 70 kg parachutist makes a soft
landing on the edge.
Find the angular speed after the parachutist
lands
 I1w1=I2w2
 I1= ½ mr2=0.5
(120 kg)*(2)2= 240 kg*m2
w1=3 rad/s
 I2=I1+ m2r2=240+ 70*22=520 kg*m2
 3*240/520=w2=1.38 rad/s
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