JPO 152 Additional Physics
Class Test 4A : 28 May 2014
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Student Number:
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Question 1
/20
[9]
a) A person on a spinning office chair extends her arms outwards. Will her angular velocity will
then increase, decrease or remain the same? Explain
(3)
Increase. According to the law of conservation of angular momentum (𝐼𝜔)𝑖 = (𝐼𝜔)𝑓 . By extending
her arms outwards I is increased hence angular velocity must decrease to compensate
b) The diagram alongside shows a uniform metal plate had been square before
25% of it was snipped off. Three lettered points are indicated. Rank them
(greatest first) according to the rotational inertia of the plate around a
perpendicular axis through them. Explain
(3)
c, a, b. Rotational inertia is given by 𝐼 = 𝐼𝑐𝑜𝑚 + 𝑀ℎ2 and since 𝐼𝑐𝑜𝑚 is a constant for
the object regardless of choice of points position then one need only look at h (the distance from the
point to the center of mass since the greater the distance h the greater the rotational inertia. The
center of mass is located between a and b but more towards b.
c) The diagram gives the angular momentum L of a wheel as a function of
time. Rank the four lettered regions according to the magnitude of the
torque acting on the wheel, greatest first. Explain
(3)
D, B, A=C. 𝜏 =
𝑑𝐿
𝑑𝑡
thus the steeper the gradient the grater the magnitude of the
torque acting on the wheel.
Question 2
[11]
A wave 𝑦(𝑥, 𝑡) = 𝐴 sin(𝑘𝑥 − 𝜔𝑡 + 𝜙) is travelling in the positive x direction with a velocity v. It has
an amplitude of 15 𝑐𝑚, a wavelength of 40 𝑐𝑚 and a frequency of 8 𝐻𝑧. The vertical displacement
at 𝑡 = 0 and 𝑥 = 0 is 15 𝑐𝑚.
a) Determine the wave number k and give its units.
𝑘=
(2)
2𝜋 2𝜋
=
= 5𝜋 𝑟𝑎𝑑/𝑚
𝜆
0.4
b) Determine the angular frequency 𝜔 and give its units.
(2)
𝜔 = 2𝜋𝑓 = 2𝜋(8) = 16𝜋 𝑟𝑎𝑑/𝑠
c) Determine the speed of the wave
𝑣=
(2)
𝜔 16𝜋
=
= 3.2 𝑚/𝑠
𝑘
5𝜋
d) Determine the phase constant of the wave
(2)
At 𝑥 = 0 𝑚, 𝑡 = 0 𝑠 substitution into equation yields 15 = 15 sin(𝜙) ⇒ sin(𝜙) = 1
⇒ 𝜙 = sin−1(1) = 𝜋/2
e) Determine the maximum velocity for any fixed point of the medium
(
𝑑𝑦
)
= (−𝜔𝑦𝑚 cos(𝑘𝑥 − 𝜔𝑡))𝑚𝑎𝑥 = 𝜔𝑦𝑚 = (16𝜋)(0.15) = 2.4𝜋 𝑚/𝑠
𝑑𝑡 𝑚𝑎𝑥
(3)
Formula Sheet
𝑣 = 𝜔𝑟
𝑎𝑡 = 𝛼𝑟
𝑎𝑟 = 𝜔2 𝑟
𝐼 = ∑ 𝑚𝑖 𝑟𝑖2
𝐼 = ∫ 𝑟 2 𝑑𝑚
𝐼 = 𝐼𝑐𝑜𝑚 + 𝑀ℎ2
1
𝐸𝐾 = 𝐼𝜔2
2
𝜏 = 𝑟𝐹 sin 𝜃
𝜏𝑛𝑒𝑡 = 𝐼𝛼
𝜏⃗ = 𝑟⃗ × 𝐹⃗
𝑙⃗ = 𝑟⃗ × 𝑝⃗
𝜏⃗𝑛𝑒𝑡 =
𝑑𝐿⃗⃗
𝑑𝑡
𝐿 = 𝐼𝜔
𝑦(𝑥, 𝑡) = 𝑦𝑚 sin(𝑘𝑥 − 𝜔𝑡)
1
1
𝑦(𝑥, 𝑡) = [2𝑦𝑚 cos ( 𝜙)] sin(𝑘𝑥 − 𝜔𝑡 + 𝜙)
2
2
𝑦(𝑥, 𝑡) = 2𝑦𝑚 sin(𝑘𝑥) cos(𝜔𝑡)