Day 10, Physics 131

HW Problem 10-10

• A 2.5-inch-diameter floppy disk in the computer rotates with a period of 0.200 s.

• (a) What is the angular speed of the disk ?• (b) What is the linear speed of a point on the

rim of the disk ?• (c) Does a point near the center of the disk

have an angular speed that is greater than, less than, or the same as the angular speed found in part (a)?

HW Problem 10-20

• A discus thrower starts from rest and begins to rotate with a constant angular acceleration of 2.2 rad/s2.

• ? (a) How many revolutions does it take for the discus thrower’s angular speed to reach 6.3 rad/s ?

• ? (b) How much time does this take ?

HW Problem 10-34

• Jeff of the Jungle swings on a vine that is 7.20 m long. At the bottom of the swing, just before hitting the tree, Jeff’s linear speed is 8.50 m/s.

• ? (a) Find Jeff’s angular speed at this time. ?• ? (b) What centripetal acceleration does Jeff

experience at the bottom of his swing ?• ? (c) What exerts the force that is responsible

for Jeff’s centripetal acceleration ?

HW Problem 10-61

• When a pitcher throws a curve ball, the ball is given a fairly rapid spin.

• ? If a 0.15-kg baseball with a radius of 3.7 cm is thrown with a linear speed of 48 m/s and an angular speed of 42 rad/s, (a) how much of its kinetic energy is translational and (b) how much is rotational ?

• Assume the ball is a uniform, solid sphere.

HW Problem 10-70

• Atwood’s Machine. The two masses (m1 = 5.0 kg and m2 = 3.0 kg) in the Atwood’s machine shown in the figure are released from rest with m2 at a height of 0.75 m above the floor. When m1 hits the ground, its speed is 1.8 m/s.

• Assume the pulley is a uniform disk with a radius of 12 cm.

• ? Find the mass of the pulley ?

HW Problem 11-6• At the local playground, a 16-kg child sits on the

end of a horizontal teeter-totter, 1.5 m from the pivot point. On the other side of the pivot point an adult pushes straight down on the teeter-totter with a force of 95 N.

• Assume the child’s torque would rotate the teeter-totter in the CW direction.

• In which direction (CW or CCW) does the teeter-totter rotate if the adult applies a force at a distance of (a) 3.0 m, (b) 2.5 m, or (c) 2.0 m from the pivot ? Why?

HW Problem 11-13

• A person holds a ladder horizontally at its center.

• Assume the ladder is a uniform rod of length 3.15 m and a mass of 8.42 kg.

• ? Find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 rad/s2 ?

HW Problem 11-23

• To loosen the lid of a jar of jam 8.9 cm in diameter, a torque of 8.5 N m must be applied to the circumference of the lid.

• ? If a jar wrench whose handle extends 15 cm from the center of the jar is attached to the lid, what is the minimum force required to open the jar ?

HW Problem 11-29

• A 0.122–kg remote control 23.0 cm long rests on a table, with a length L overhanging its edge. To operate the power button on this remote requires a force of 0.365 N.

• Assume the mass of the remote is distributed uniformly and that the power button is 1.41 cm from the overhanging end of the remote.

• ? How far can the remote control extend beyond the edge of the table and still not tip over when you press the power button ?

HW Problem 11-43

• A 0.34-kg meter stick balances at its center. If a necklace is suspended from one end of the stick, the balance point moved 9.5 cm toward that end.

• ? (a) Is the mass of the necklace more than, less than, or the same as that of the meter stick ?

• ? (b) Find the mass of the necklace. ?

HW Problem 11-69

• A disk-shaped merry-go-round of radius and mass of 155 kg rotates freely with an angular speed of 0.641 rev/s. A 59.4-kg person running tangentially to the rim of the MGR at 3.41 m/s jumps onto its rim and holds on. Before jumping on the MGR, the person was moving in the same direction as the MGR’s rim.

• ? What is the final angular speed of the MGR?

Rotational Work and Power

• Work = W = , where is torque 11-17• Compare W = Fx

Section 11-9Vector Nature of Rotational Motion

• Not on quiz

Center of Mass of Alligator

• Given: 416-kg alligator• Given: alligator of length 3.5 m• Given: support board of weight 65 N and

length 3.5 m• Scale measures 1880 N• ? What is xcm from the pivot?

Force of Gravity

• Force of gravity, FG = G m1 m2 / r2

• Universal gravitational constant = G• G = 6.67 x 10-11 N m2 /kg2

• r is COM to COM of the two bodies

Pool Table

• A coordinate system (in meters) is constructed on the surface of a pool table, and three objects are placed on the table as follows:

• 2.0-kg object at the origin• 3.0-kg object at (0,2.0)• 4.0-kg object at (4.0,0)• ? Find the resultant gravitational force exerted

by the other two objects on the object at the origin. ?

Kepler’s Laws

• Kepler “acquired” Brahe’s data – 16 years worth of data!

• #1. All planets on elliptical orbits with Sun at a focal point.

• #2. Equal areas in equal times.• #3. T2 = ( 4 2 / G Msun ) r3

Escape Speed

• Gravitational potential Energy, U = -G m1 m2 / r,

equation 12-8 on page 395

Ch 13: Periodic MotionSimple Harmonic Motion

• T = period of one complete cycle of periodic motion• f = 1 / T frequency, in cycles/second, Hz• x = A cos ((2 / T ) t) 13-4• = / T = 2 f• v = - A sin ( t ) 13-6• a = -A 2 cos (t) 13-8

• …. Be sure to use radian mode on your calculator!

Block on Spring

• SHM: visualize block on a spring• = ( k / m )1/2 13-10• T = 2 ( m / k ) ½

• or, rewritten, k = 4 2 m / T2

• Total Energy = ½ k A2 13-15• ½ k A2 = ½ mv2 + ½ k x2

• or, rewritten, v = +_ (( k/m )(A2 –x2))1/2

13-5 Energy Conservation in Oscillatory Motion

• Total Energy • E = ½ k A2 13-15• Potential Energy as a Function of Time• U = ½ k A2 cos2(t) 13-14• Kinetic energy as a Function of Time• K = ½ k A2 sin2 (t) 13-15

Block on Spring Example

• m = 0.40 kg• k = 19.6 N/m• x = 4.0 cm so that A = 4.0 cm• ? (a) max speed ?• ? (b) speed when compressed to 1.5 cm ?• ? (c) speed when stretched to 1.5 cm ?• ? (d) when is speed ½ of max speed ?

One more block on a spring

• A 2.00-kg block on a frictionless horizontal surface is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.00 m to the right from its equilibrium position and then released, initiating simple harmonic motion.

• ? (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released. ?

• ? (b) How many times does the object oscillate in 3.50 s. ?

Simple PendulumChild’s Swing????

• T = 2 ( L / g ) ½

• ? If T = 2s, what is L ?

• ? If T = 1s, what is L ?

Pendula

Tall tower

• A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floor and that its period is 15.5 s.

• ? (a) How tall is the tower ?• ? (b) If this pendulum were taken to the

moon, where the free-fall acceleration is 1.67 m/s2, what is the period there. ?

Oscillations and Resonances

• Damped – shock absorbers• Driven – children’s swings• Dr. Moog in a new elevator• Driven – Front-loaded clothes washing

machine