Ap Physics Worksheet With Answers Page 7
ADVERTISEMENT
1
2
3
4
5
6
7
849. A baseball player swings his bat with his arms fully extended. If his arms are pulled in closer to his
body, which of the following choices correctly describes the impact of his motion on his swing’s
angular momentum and kinetic energy?
Angular Momentum
Kinetic Energy
A
B
C
D
A.
increases
increases
B.
increases
remains constant
C.
remains constant
increases
D.
remains constant
remains constant
50. An ice skater performs a fast spin by pulling her outstretched arms down and close to her body.
What happens to her angular momentum with respect to the axis of rotation?
A. increases
A
B
C
D
B. decreases
C. stays the same
D. depends on her initial rotational velocity
51. The system below rotates with an angular velocity, ω. If the mass of
the rod supports is negligible, what is the ratio of the angular momentum
of the two upper spheres to the two lower spheres?
A. 1:4
A
B
C
D
B. 1:2
C. 2:1
D. 4:1
52. A rod rotates about a pivot at its center at 2 rad/s. Its angular velocity increases uniformly to 14 rad/s in 3 s. Find the rod’s
angular acceleration over this time period.
ω
= ω
+ 2αt
f
i
α = (ω
- ω
)/2t
f
i
2
α = (14 rad/s – 2 rad/s)/(2*3 s) = 2 rad/s
2
53. A bicycle wheel of radius 40.0 cm and angular velocity of 10.0 rad/s starts accelerating at 80.0 rad/s
. (a) What is the tangential
acceleration of the wheel at this time point? (b) What is the centripetal acceleration of the wheel at this time point?
2
2
(a) a
= rα = (0.40 m)(80.0 rad/s
) = 32.0 m/s
tan
2
2
2
(b) a
= a
= v
/r = ω
r = (10.0 rad/s)2(0.40 m) = 40 m/s
r
c
54. A record on a turntable is spinning and coming to rest. At t = 0 s, the angular velocity of the record is 20.0 rad/s with a constant
2
acceleration of – 5.0 rad/s
. How long does it take for the record to come to rest?
ω
= ω
+ 2αt
f
i
t = (ω
- ω
)/2 α
f
i
2
t = (0 rad/s – 20.0 rad/s)/2*(-5.0 rad/s
) = 2.0 S
55. As shown to the right, a wooden square of side length 1.0 m is on a horizontal tabletop and is
free to rotate about its center axis. The square is subject to two forces and rotates.
Find the net torque of the system.
The moment arm, r, from the contact point of the force to the center axis is 0.707 m. This is
found by using the Pythagorean Theorem and the fact that the ‘halfway’ distance horizontally
and vertically to the center point is 0.5 m.
τ
= τ
+ τ
net
4N
8N
o
o
τ
= -(4 N)(0.707 m)(sin 135
) + (8 N)(0.707 m)(sin 135
) { τ
is negative since its direction is CW}
net
4N
τ
= 2 Nm
net
56. Two masses of mass 10.0 and 6.0 kg are hung from massless strings at the end of a light rod. The rod is virtually weightless. A
pivot (fulcrum) is placed off center and the system is free to rotate.
If the 6.0 kg mass is 4.0 m away from the pivot, how far away is the 10.0 kg mass if the system is not rotating?
Στ = 0 (if not rotating)
τ
+ τ
= 0
10kg
6kg
τ
=- τ
10kg
6kg
o
o
(10 kg)(9.8 N/kg)(x)(sin 90
) = (6 kg)(9.8 N/kg)(4 m)(sin 90
)
x = (6 kg)(4 m)/(10 kg) = 2.4 m
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education