April 4, 2008
PHY2053 Discussion
Quiz 9 (Chapter 8.6 -9.5)
Name:
UFID:
**1. (5 pts) A spherical shell with a mass of 2.00 kg and radius of 0.100 m starts from
rest at the top of a 45.0˚ incline and rolls 2.00 m down the incline without slipping. Find
the final velocity of the shell. (The moment of inertia of a spherical shell is (2/3)MR².)
Since the sphere rolls without slipping the translational velocity is related to the
angular velocity as
ω = v/R.
The mechanical energy of the sphere is conserved, thus we have
Mgh = (1/2)Mv²+(1/2)Iω²
Substitute the expression for moment of inertia and angular velocity and solve for
linear velocity:
MgLsinθ = (1/2)Mv²+(1/2)(2/3)MR²(v/R)²
⇒ gLsinθ = (5/6)v² ⇒ v = √((6/5) gLsinθ) = 4.08 m/s
**2. (5 pts) A merry-go-round modeled as a disk with a mass of 80.0 kg and radius of
2.00 m is initially rotating about a frictionless axle with an angular speed of 0.800 rad/s.
A 50.0-kg person jumps onto the rim of the disk and then walks to a point 1.00 m from
the center. Assuming the person does not have angular momentum around the axle
before he/she gets on the merry-go-round, find the final angular speed of the
merry-go-round. (The moment of inertia of a solid disk is (1/2)MR².)
We consider the system that consists of the person and the merry-go-round, then no
external torque is exerted on the system when the person jumps on to the
merry-go-round or walks on it. Applying the angular conservation law to the system, we
have
Li = Lf ⇒ Iω = (I+I’)ω’ ⇒ ω’ = Iω/(I+I’)
where I = (1/2)MR² = (1/2)×80×2² = 160 kgm² and I’ = mr² = 50×1² = 50 kgm².
Thus ω’ = 160×0.8/(160+50) = 0.610 rad/s
*3. (5 pts) Determine the elongation of an aluminum wire with a diameter of 0.5 cm and
a length of 2.00 m if it is under a tension of 8.00×10³N. (The Young’s modulus of
Aluminum is 7.0×10¹⁰Pa.)
Since the tensile stress is proportional to the tensile strain, we have
F/A = Y(ΔL/L) ⇒ ΔL = FL/(Yπr²)
= 8.00×10³×2/(7.0×10¹⁰×3.14(2.5×10⁻³)²) = 1.16×10⁻² m
Note that the radius is half of the diameter.
*4. (5 pts) A tank is filled to a depth of 2.00 m with salt water having a density of 1.05×
10³kg/m³. On top of the water floats a 1.50-m-think layer of oil with a density of 7.00×
10² kg/m³. What is the absolute pressure at the bottom of the container? (Take 1.01×
10⁵ Pa as atmospheric pressure.)
The pressure at the bottom of the oil layer is given by
P₁ = P₀+ρgh = 1.01×10⁵+700×9.8×1.5 = 1.11×10⁵ Pa
Using P₁ as a new starting pressure, the pressure at the bottom of the tank is given by
P₂ = P₁+ρ’gh’ = 1.11×10⁵+1050×9.8×2 = 1.32×10⁵ Pa