PHY2020
Fall 2011, Test 2
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Ignore air friction in all problems.
Please give complete responses to all questions
including units and theoretical justification for
responses.
You must show ALL of your work to receive full credit!
Useful values and formulas:
G = 6.7 10-11 N m2 / kg2
MEarth = 5.98 * 1024 kg
REarth = 6378 km
1 km = 0.62 miles = 3280.8 feet
1 kg = 2.2 lbs
Moment of Inertia
Cylinder of radius r, axis through center: 1/2 m r2
Sphere of radius r, axis through center: 2/5 m r2
Rod of length l,
axis through center: 1/12 m l2
Rod of length l,
axis through end: 1/3 m l2
1. (6 pts) What does it mean for a quantity to be conserved? Give an example of
a conserved quantity in real life.
3 pts: Conserved: for a system of objects, the quantity remains constant
regardless of what happens in the system.
3 pts: Ex: slingshot + rock, cart at top of hill rolling to bottom of hill.
2. (6 pts) If you push against a brick wall are you doing work? Why or why not?
3 pts: No, because you’re not moving it any distance. The definition of work is:
3 pts: W = F * d * cos θ
or
W = change in kinetic energy
3. (8 pts) What is required for an object to be in equilibrium? What are the
different types of equilibrium?
4 pts: An object must have:
Sum of all torques acting on object = 0 (no rotation)
Sum of all forces acting on object = 0 (no linear motion)
4 pts: The types of equilibrium are stable, unstable, and neutral.
4. (20 pts) Ball 1, mass 2 kg, collides with ball 2, 1 kg, at rest, as pictured below.
What is the final velocity of ball 2? Hint: don’t forget to include directional
information.
Use conservation of momentum, but make sure you apply it in both dimensions:
in x and in y!
Momentum in x direction before collision = Momentum in x direction after collision
Momentum in y direction before collision = Momentum in y direction after collision
12 pts: In x: (4 pts) m1 * v1,initial = m1 * v1,final * cos 15 + m2 * v2,final x
2 kg * 3.3 m/s = 2 kg * 1.9 m/s * cos 15 + 1 kg * v2,final x
6.6 kg m/s = 2 kg * 1.9 m/s * 0.97 + 1 kg * v2,final x
6.6 kg m/s = 3.69 kg m/s + 1 kg * v2,final x
(3 pts) 2.91 kg m/s = v2,final x , pointing east.
12 pts: In y: (4 pts) m1 * 0 = m1 * v1,final * sin 15 + m2 * v2,final y
2 kg * 1.9 m/s * sin 15 = - 1 kg * v2,final y
2 kg * 1.9 * 0.26 = - 1 kg * v2,fina l y
(3 pts) - 0.99 kg m/s = v2,final y , pointing south
6 pts: v2, final = 3.07, pointing south east
5. (6 pts) How much work is done lifting a 15 lb cat 4 feet?
3 pts: W = F * d * cos θ
3 pts: W = (15 lb * 1 kg/2.2 lbs) * 9.8 m/s2 * (4 ft * 1 km/3280.8 ft * 1000 m/1 km) *
cos 0
W = 81.46 Joules
6. (8 pts) You need to open the drain plug on your car’s engine, which had been
tightened to 25 Newton-meters. You know you can apply between 100 and 125
N of force to open the plug. What size wrench should you use?
3 pts: Torque = F * l * sin θ
You need a torque of 25 N-m. The angle between the applied force and the
distance (rotating object) is 90°, so the sin θ is 1.
2.5 pts: 25 N-m = 100 N * d
2.5 pts: 25 N-m = 150 N * d
The size of the wrench must be between 0.25 m and 0.2 m.
7. (20 pts) A ball is thrown downward with a velocity of 5 m/s, from a height of
150 km. How fast will it be traveling when it hits the ground?
Use conservation of energy.
8 pts: Energy at the start = KE + PE
= ½ m * (5 m/s)2 + m * g * (150 km * 1000 m/1 km)
Energy when it hits the ground is only KE, because h = 0 (at ground level), thus it
has no PE.
8 pts: Energy when it hits the ground = KE
= ½ m * v2
Apply conservation of energy:
4 pts: Energy at the start = Energy when it hits the ground
½ m * (5 m/s)2 + m * g * (150 km * 1000 m/1 km) = ½ m * v2
(5 m/s)2 + 2 *g * (150 km * 1000 m/1 km) = v2
2,940,025 m2/s2 = v2
v = 1714.65 m/s
8. (6 pts) What is the linear speed of a point 6.0 cm from the center of a 45 rpm
record? What is the angular speed at this same point?
3 pts: Angular speed: w = rad/sec
w = 45 revolutions/min * (1 min/60 s) * (2 pi radians/1 revolution) = 4.71 rad/s
3 pts: linear speed: v = r * w
v = 4.71 rad/s * (6.0 cm * 1m/100 cm) = 0.28 m/s
9. (20 pts) A solid sphere and a solid cylinder roll down an incline. At the bottom
of the incline can the angular velocity of the cylinder ever be smaller than that of
the sphere? Explain thoroughly with the help of equations and math.
10 pts: Rotational KE = ½ I w2, where I is the moment of inertia and w is the
angular velocity.
For the sphere: KE = ½ * (2/5 * m sphere * r2 sphere)* w2sphere = 0.2 * m sphere * r2 sphere*
w2sphere
For the cylinder: KE = ½ * (1/2 * m cylinder * r2cylinder) * w2cylinder = 0.25 * m cylinder *
r2cylinder * w2cylinder
5 pts: If they have the same masses, they have the same starting PE (because
they are both at the same location at the top of the hill). Therefore, they will have
the same total KE at the bottom of the hill.
0.2 * m sphere * r2 sphere* w2sphere = 0.25 * m cylinder * r2cylinder * w2cylinder
but the masses are the same, so we can cancel that from each side of the
equation:
0.2 * r2 sphere* w2sphere = 0.25 * r2cylinder * w2cylinder
from here you can proceed in two ways.
1. assume they also have the same radius. Then
0.2 * r2 sphere* w2sphere = 0.25 * r2cylinder * w2cylinder
becomes
0.2 * w2sphere = 0.25 * w2cylinder, or
w2sphere = 1.25 * w2cylinder
5 pts: therefore, the angular velocity of the cylinder is less than the sphere.
2. the second way to proceed is to NOT assume they have the same radius.
Rearranging,
w2cylinder = 0.8 * (r2 sphere/ r2cylinder) * w2sphere
w2cylinder is less than w2sphere when 0.8 * (r2 sphere/ r2cylinder) is less than 1.
5 pts: This happens if (r2 sphere/ r2cylinder) < 1.25, or (r sphere/ rcylinder) < 1.12