(a)

(b) The car travels a distance equal to the circumference of the tire for every revolution the tire makes if there is no slipping of the tire on the roadway. Thus, the number of revolutions made during the warranty period is

  

7.3) The tires on a new compact car have a diameter of 2.0 ftand are warranted for 60 000 miles. (a) Determine theangle (in radians) through which one of these tires willrotate during the warranty period. (b) How many revolutionsof the tire are equivalent to your answer in part(a)?

radianxmiles

ft

ft

milesradian

ft

milesradian

d

s

r

s 8102.31

528060000

0.1

60000

2/

srevolutionxmiles

ft

ft

milesrev

d

S

r

Sn 7100.5

1

5280

)0.1(2

60000

2/

(a)

(b)

7.5. A dentist’s drill starts from rest. After 3.20 s of constantangular acceleration, it turns at a rate of 2.51x104 rev/min. (a) Find the drill’s angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period.

stst

revxrev

fi

fi

2.3,0

min/1051.2min,/0 4

(a)From , the angular displacement is

(b) From the equation given above for

, observe that when the angular acceleration is constant, the displacement is proportional to the difference in the squares of the final and initial angular speeds. Thus, the angular displacement would

if both of these speeds were doubled.

7.7) A machine part rotates at an angular speed of 0.06 rad/s; its speed is then increased to 2.2 rad/s at an angular acceleration of 0.70 rad/s2. (a) Find the angle through which the part rotates before reaching this final speed. (b) In general, if both the initial and final angular speed are doubled at the same angular acceleration, by what factor is the angular displacement changed?

yieldsforSolving

if

).......(.

)(222

Main Rotor:

Tail Rotor:

7.9) The diameters of the main rotor and tail rotor of asingle-engine helicopter are 7.60 m and 1.02 m, respectively.The respective rotational speeds are 450 rev/minand 4 138 rev/min. Calculate the speeds of the tips ofboth rotors. Compare these speeds with the speed ofsound, 343 m/s.

7.17 a) What is the tangential acceleration of a bug on the rim of a 10.0 in-

diameter disk if the disk accelerates uniformly from rest to an angular

speed of 78.0 rev/min in 3.00s? b) When the disk is at its final speed, what

is the tangential velocity of the bug? One second after the bug starts from

rest, what are its c) tangential acceleration, d) centripetal acceleration, and

c) total accel eration?

(a) The tangential acceleration of the bug as the disk speeds up is

(b) The final tangential speed of the bug is

(c)

Since the bug has constant angular acceleration, and hence constant tangential acceleration

tangential acceleration at t = 1.00 is as above.

(d) At t = 1.00 s, the tangential velocity of the bug is

tangential acceleration at t = 1.00 is as above.

(d) At t = 1.00 s, the tangential velocity of the bug is

and the radial or centripetal acceleration is

(e) The total acceleration is , and the angle this acceleration makes with the

direction of

and the radial or centripetal acceleration is

The total acceleration is , and the angle this acceleration makes with the

direction of is

7.19) One end of a cord is fixed and a small 0.500 kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.00 m, as shown in the nearby figure. When Ө = 20.00, the speed of the object is 8.00 m/s. At this instant, find a) the tension in the string, b) the tangential and radial components of acceleration, and c) the total acceleration

D) If your answer changed if the object is swinging down toward its lowest

point instead of swinging up? E) Explain your answer to part (d)

(a) The tension in the string must counteract the radial component of the object’s weight, and also supply the needed centripetal acceleration.

   

(b) The net tangential force acting on the object is , so the tangential acceleration has magnitude

and is directed

The radial component of the acceleration is

(c) The total acceleration has magnitude

or

at

(d) in answers if the object is swinging toward the equilibrium point instead of away from it.

(e) If the object is swinging toward the equilibrium position, it is gaining speed, whereas it is losing speed if

it is swinging away from the equilibrium position. In both cases, when the cord is 20.0° from the vertical,

the tangential, centripetal, and total accelerations have the magnitudes and directions calculated in parts

(a) through (c).

7.21 (a) From , we have

(b) The tension is by a factor of

21. A 55.0-kg ice skater is moving at 4.00 m/s when shegrabs the loose end of a rope, the opposite end ofwhich is tied to a pole. She then moves in a circle ofradius 0.800 m around the pole. (a) Determine theforce exerted by the horizontal rope on her arms.(b) Compare this force with her weight

26. A space habitat for a long space voyage consists oftwo cabins each connected by a cable to a central hubas shown in Figure P7.26. The cabins are set spinningaround the hub axis, which is connected to the rest ofthe spacecraft to generate artificial gravity. (a) Whatforces are acting on an astronaut in one of the cabins?(b) Write Newton’s second law for an astronaut lyingon the “floor” of one of the habitats, relating the astronaut’smass m, his velocity v, his radial distance fromthe hub r, and the normal force n. (c) What would nhave to equal if the 60.0-kg astronaut is to experiencehalf his normal Earth weight? (d) Calculate the necessarytangential speed of the habitat from Newton’ssecond law. (e) Calculate the angular speed from thetangential speed. (f) Calculate the period of rotationfrom the angular speed. (g) If the astronaut stands up,will his head be moving faster, slower, or at the samespeed as his feet? Why? Calculate the tangential speedat the top of his head if he is 1.80 m tall.10.0 m

7.26 (a) The only force acting on the astronaut is exerted on him by the “floor” of the cabin.

(b)

(c) If , then

(d) From the equation in Part (b),

(e) Since , we have

(f) The period of rotation is

(g) Upon standing, the astronaut’s head is moving slower than his feet because his head is closer to the axis

of rotation. When standing, the radius of the circular path followed by the head is

, and the tangential speed of the head is

7.37 (a) At the midpoint between the two masses, the forces exerted by the 200-kg and 500-kg masses are

oppositely directed, so from and , we have

or

(b) At a point between the two masses and distance d from the 500-kg mass, the net force will be zero when

or

Note that the above equation yields a second solution . At that point, the two gravitational

forces do have equal magnitudes, but are in the same direction and cannot add to zero.