Every suggest in a body rotating around a addressed axis moves in acircle whose facility is on the axis has actually a radius, r. Any allude on theradius will move out the exact same angle in the same time.

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One radian, rad, is defined as the edge subtended by an arcwhose size is same to the radius. In the illustration to the left, thelength, l, is same to the radius, r. = 1 rad.

lBy meaning = --- = Angular Displacementr

360o = 2rad

*

= --- = Angular Velocity t

Omega, , isthe typical when measured over lengthy periods the time and instantaneous once thetime philosophies zero. Units are rad/s.

As presented in the drawing to the left, the instantaneousvelocity is bigger as you relocate farther from the axis, however ,is constant. Therefore the is elevation of the radius.

= ------ = AngularAcceleration t

is the averageangular acceleration when measured over lengthy periods that time and also instantaneouswhen the time viewpoints zero.

There is a basic relationship between Angular and also Linearmeasurements.

x = rv = r

aT = r

Observe the the relationship only provides the tangential acceleration. Inthe section on circular motion, we saw that one object additionally has centripetalacceleration, aC. The complete linear acceleration of a particleis

a = at +aC

Remember that aT and also aC areperpendicular to every other.

In the section on circular motion, we associated centripetalacceleration come the straight velocity and also radius. Us can additionally relate them toangular velocity.

v2( r)2C = -----= ---------- = 2rrr

At times us relate the angular velocity to the frequency ofrotation, f. The frequency is the variety of complete rotations in oneseconds. One transformation is same to an angular displacement the 2rad.

1 rev 2rad------- = ---------- for this reason = 2f 1 s 1s

Period, T, is the time for 1 revolution.

11T = ---- and also f= -----fT

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Problems

1. The sun subtends an edge of about 0.5o when viewed from Earth. The sunlight is 150 million km from Earth. What is the radius of the Sun?
2. A 0.25 m grinding wheel rotates at 3500 rpm. Calculate its angular velocity in rad/s.
3. A laser beam is directed at the Moon, 380,000 km from Earth. The beam diverges at a the 1.8 x 10 -5 rad. What is the diameter that the clues on the Moon?
4. A bicycle through 68.0 cm diameter tires travels 8.50 km. (a) How numerous revolutions carry out the tires do in this trip? (b) If it renders the expedition in 55.0 minutes, what is the angular velocity in rad/s? (c) What is the period of rotation of a tire?
5. (a) calculate the angular velocity that the planet as that revolves around the Sun. (b) calculation the angular velocity of the planet as it rotates.
6. A 70.0 cm diameter wheel speeds up uniformly indigenous 160. Rpm come 350. Rpm in 3.50 s. Calculation (a) angular acceleration, (b) tangential acceleration, and also (c) centripetal acceleration.

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Answers

1. The sunlight subtends an angle of about 0.5o when viewed from Earth. The sunlight is 150 million km from Earth. What is the radius the the Sun?
2rad 0.5o x ----------- = 0.009 rad 360o

x = diameter of sun = r = 1.50 x 10 8 km x 0.009 rad = 1 x 10 6 km

diameter 1 x 10 6 kilometres radius = ------------- = ------------------ = 5 x 10 5 km 2 2

Rad is not a true unit and does not have to be canceled.

2. A 0.25 m grind wheel rotates in ~ 3500 rpm. Calculation its angular velocity in rad/s.
3500 rev 60 s 2rad ------------ x ---------- x ------------ = 3700 rad/s 1 min 1 min 360o
3. A laser beam is command at the Moon, 380,000 km from Earth. The beam diverges at a that 1.8 x 10 -5 rad. What is the diameter of the clues on the Moon?
x = diameter of clues = r = 380,000 kilometres x 1.8 x 10 -5 rad = 6.8 meters
4. A bicycle v 68.0 centimeter diameter tires travel 8.50 km. (a) How plenty of revolutions carry out the tires make in this trip? (b) If it provides the trip in 55.0 minutes, what is the angular velocity in rad/s? (c) What is the duration of rotation that a tire?
(a) one of tire = d = x 0.340 m = 1.07 m pilgrimage 8.50 x 103 m ------------------- = --------------------- = 7940 revolutions circumference 1.07 m

(b) 7940 rev 2rad 1 min -------------- x ----------- x ----------- = 15.1 rad/s 55.0 min 1 rev 60 s

(c) 55.0 min 60 s ------------- x --------- = 0.416 s 7940 rev 1 min

5.

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(a) calculation the angular velocity of the planet as that revolves roughly the Sun. (b) calculate the angular velocity the the planet as it rotates.
(a) 1 rev 1 d 1 h 2rad ------------ x -------- x ----------- x ----------- = 1.996 x 10 -7 rad/s 364.25 d 24 h 3600 s 1 rev (b) 1 rev 1 h 2rad -------- x ----------- x ------------ = 7.27 x 10 -5 rad/s 24 h 3600 s 1 rev
6. A 70.0 centimeter diameter wheel speeds up uniformly native 160. Rpm to 350. Rpm in 3.50 s. Calculation (a) angular acceleration, (b) tangential acceleration, and also (c) centripetal acceleration.
(a) /\ = 350. Rpm - 160. Rpm = 190. Rpm 190. Rev 2rad 1 min ----------- x ----------- x ----------- = 0.332 rad/s = 1 min 1 rev 3600 s