King Fahd University of Petroleum and Minerals, Physics Department
PHYS 101 REC, Fall 2014 (141)
SEC # 28, Quiz # 5
Name:
ID #
The angular position of a point on a rotating wheel is given by
𝜃 = 3.0 𝑡 2 − 2.0 𝑡 3 ,
where 𝜃 is in radians and 𝑡 is in seconds.
(a) At 𝑡 = 2 s, what are the point’s angular velocity and angular acceleration?
𝜔=
𝑑𝜃
= 6.0 (𝑡 − 𝑡 2 ).
𝑑𝑡
𝛼=
𝑑𝜔
= 6.0 (1 − 2𝑡).
𝑑𝑡
At 𝑡 = 2 s,
𝜔 = 6.0 (2 − 22 ) = −12
rad
,
s
𝛼 = 6.0 [1 − 2(2)] = −18
rad
.
s2
(b) What are its average angular velocity and average angular acceleration between 𝑡 = 0 and
𝑡 = 4.0 s?
𝜔avg =
𝛼avg =
𝜃(4) − 𝜃(0) −80 rad − 0
rad
=
= −20
.
4s−0
4s−0
s
𝜔(4) − 𝜔(0) −72 rad/s − 0
rad
=
= −18 2 .
4s−0
4s−0
s
King Fahd University of Petroleum and Minerals, Physics Department
PHYS 101 REC, Fall 2014 (141)
SEC # 29, Quiz # 5
Name:
ID #
The figure gives angular speed versus time for a thin
rod that rotates around one end. The scale on the ω
axis is set by ωs = 3.0 rad/s.
(a) What is the magnitude of the rod’s angular acceleration?
The angular acceleration 𝛼 is the slope of the 𝜔(𝑡) curve in the 𝜔 versus 𝑡 graph. Using the
two points (6.0 s, 3.5 rad/s) and (2.0 s, 0.50 rad/s)
𝛼=
𝜔2 − 𝜔1 3.5 rad/s − 0.50 rad/s 3.0 rad/s
rad
=
=
= 0.75 2 .
𝑡2 − 𝑡1
6.0 s − 2.0 s
4.0 s
s
(b) At t = 4.0 s, the rod has a rotational kinetic energy of 10 J. What is its moment of inertia?
At 𝑡 = 4.0 s, 𝜔 = 2.0 rad/s.
𝐼=
2𝐾
2(10 𝐽)
=
= 5.0 kg ∙ m2 .
2
(2.0 rad/s)2
𝜔
King Fahd University of Petroleum and Minerals, Physics Department
PHYS 101 REC, Fall 2014 (141)
SEC # 30, Quiz # 5
Name:
ID #
The figure shows three 1.00 kg particles that have
been glued to a rod of length L = 3.00 m and
negligible mass. The assembly is rotating around a
perpendicular axis through point O at the left end
at constant angular speed.
(a) By what percentage does the rotational kinetic energy of the assembly decrease when the
the innermost particle is removed? The angular speed remains unchanged.
The kientic energies before and after the particle is removed are, respectively
1
1
𝐾1 = 𝐼1 𝜔2 and 𝐾2 = 𝐼2 𝜔2 .
2
2
The percent difference is
1
1
𝐼1 𝜔2 − 2 𝐼2 𝜔2
𝐾1 − 𝐾2
𝐼2
2
× 100 =
× 100 = (1 − ) × 100.
1
𝐾1
𝐼1
2
2 𝐼1 𝜔
Using the figure we evaluate 𝐼1 :
𝐼1 = 𝑚𝑑2 + 𝑚(2𝑑)2 + 𝑚(3𝑑)2 = 14 𝑚𝑑2 .
After the innermost particle is removed we have
𝐼2 = 𝑚(2𝑑)2 + 𝑚(3𝑑)2 = 13 𝑚𝑑2 .
Therefore,
𝐾1 − 𝐾2
13 𝑚𝑑 2
× 100 = (1 −
) × 100 = 7.14 %.
𝐾1
14 𝑚𝑑 2
(b) By what percentage does the rotational kinetic energy of the assembly decrease when the
the outermost particle is removed? The angular speed remains unchanged.
After the outermost particle is removed we have
𝐼2 = 𝑚𝑑2 + 𝑚(2𝑑)2 = 5 𝑚𝑑2 .
Therefore,
𝐾1 − 𝐾2
5 𝑚𝑑2
× 100 = (1 −
) × 100 = 64.3 %.
𝐾1
14 𝑚𝑑 2

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