Unformatted text preview: tential energy. If the level of the hinge is
chosen as the zero level for measuring heights, then: E0 = mgh0 = mgL. Just before the
object hits the floor, the system has only kinetic energy.
Therefore
1
2 mgL = mv 2 Solving for v gives v= 2 gL From Equation 8.9, vT = rω. Solving for ω gives ω = vT/r. As the object rotates downward,
it travels in a circle of radius L. Its speed just before it strikes the floor is its tangential
speed. Therefore, ω= vT
r = 2 gL v
=
L L b. From Equation 8.10: = 2g
=
L 2 2(9.80 m/s )
= 3.61 rad/s
1.50 m aT = rα Solving for α gives α = aT/r. Just before the object hits the floor, its tangential acceleration
is the acceleration due to gravity. Thus,
2 aT g 9.80 m/s
α=
==
=
r
L
1.50 m 6.53 rad/s 2 44. REASONING AND SOLUTION The stone leaves the circular path with a horizontal speed
v0 = vT = rω
so ω = v0/r. We are given that r = x/30 so ω = 30v0/x. Kinematics gives x = v0t. With this
substitution for x the expression for ω becomes ω = 30/t. Kinematics also gives for the
1
2 vertical displacement y that y = v0 y t + a y t 2 (Equation 3.5b). In Equation 3.5b we know
1
2 that v0y = 0 m/s since the stone is launched horizontally, so that y = a y t 2 or t = 2 y / a y .
Using this result for t in the expression for ω and assuming that upward is positive, we find 416 ROTATIONAL KINEMATICS ω = 30 ay
2y −9.80 m/s 2
= 14.8 rad/s
2 ( −20.0 m ) = 30 45. SSM REASONING The magnitude ω of each car’s angular speed can be evaluated from
ac = rω (Equation 8.11), where r is the radius of the turn and ac is the magnitude of the
centripetal acceleration. We are given that the centripetal acceleration of each car is the
same. In addition, the radius of each car’s turn is known. These facts will enable us to
determine the ratio of the angular speeds.
2 SOLUTION Solving Equation 8.11 for the angular speed gives ω = ac / r . Applying this
relation to each car yields:
Car A: ωA = ac, A / rA Car B: ωB = ac, B / rB Taking the ratio of these two angular speeds, and noting that ac, A = ac, B, gives
ac, A ωA
=
ωB rA
ac, B = ac, A rB
ac, B rA = 36 m
= 0.87
48 m rB 46. REASONING Since the car is traveling with a constant speed, its tangential acceleration
must be zero. The radial or centripetal acceleration of the car can be found from Equation
5.2. Since the tangential acceleration is zero, the total acceleration of the car is equal to its
radial acceleration.
SOLUTION
a. Using Equation 5.2, we find that the car’s radial acceleration, and therefore its total
acceleration, is
v 2 (75.0 m/s) 2
a = aR = T =
= 9.00 m/s 2
r
625 m b. The direction of the car’s total acceleration is the same as the direction of its radial
acceleration. That is, the direction is radially inward . Chapter 8 Problems 47. 417 SSM REASONING AND SOLUTION
a. The tangential acceleration of the train is given by Equation 8.10 as
aT = r α = (2.00 × 10 m)(1.50 × 10
2 −3 2 rad/s ) = 0.300 m/s 2 The centripetal acceleration of the train is given by Equation 8.11 as
ac = r ω = (2.00 × 10 m)(0.0500 rad/s) = 0.500 m/s
2 2 2 2 The magnitude of the total acceleration is found from the Pythagorean theorem to be
2 2 a = aT + ac = 0.583 m/s 2 b. The total acceleration vector makes an angle relative to the radial acceleration of
a
T
a
c θ = tan −1 = tan −1 0.300 m/s 2 0.500 = m/s 2 31.0° 48. REASONING
a. According to Equation 8.2, the average angular speed is equal to the magnitude of the
angular displacement divided by the elapsed time. The magnitude of the angular
displacement is one revolution, or 2π rad. The elapsed time is one year, expressed in
seconds.
b. The tangential speed of the earth in its orbit is equal to the product of its orbital radius
and its orbital angular speed (Equation 8.9).
c. Since the earth is moving on a nearly circular orbit, it has a centripetal acceleration that is
directed toward the center of the orbit. The magnitude ac of the centripetal acceleration is
given by Equation 8.11 as ac = rω .
2 SOLUTION
a. The average angular speed is ω= ω= 2π rad
∆θ
=
= 1.99 × 10−7 rad /s
7
∆t 3.16 × 10 s (8.2) b. The tangential speed of the earth in its orbit is ( )( vT = r ω = 1.50 × 10 m 1.99 × 10
11 −7 ) 4 rad/s = 2.98 × 10 m/s (8.9) 418 ROTATIONAL KINEMATICS c. The centripetal acceleration of the earth due to its circular motion around the sun is ( )( a c = r ω 2 = 1.50 × 1011 m 1.99 × 10−7 rad /s ) 2 = 5.94 × 10−3 m /s2 (8.11) The acceleration is directed toward the center of the orbit. 49. REASONING The centripetal acceleration ac at either corner is related to the angular speed ω of the plate by ac = rω 2 (Equation 8.11), where r is the radial distance of the corner from
the rotation axis of the plate. The angular speed ω is the same for all points on the plate,
including both corners. But the radial distance rA of corner A from the rotation axis of the
plate is different from the radial distance rB of corner B. The fact that the centripetal
acceleration at corner A is n times as...
View
Full
Document
This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details