Idea Transcript
PHYSICS 218 Honors EXAM 2 Retest Choose 5 of the following 6 problems. Indicate which problem is not to be graded.
1. A rope is affixed at one end to the rim of a pulley, and wrapped five turns around the pulley. The pulley has a mass of 5 kg, a radius of 0.2 m, and its mass is distributed as a uniform disk. A mass of 10 kg is suspended from the loose end of the rope, so that it hangs 2 m from the floor. The mass is then released. What is the velocity of the mass just before it strikes the floor? Solution: The rope will unwind by an angle 2m θ = L/ R = = 10rad 0.2m We require Newton's law for force and for torque: (M=10 kg is mass of weight, m=5 kg is mass of pulley)
Mg − T = Ma TR = Iα a = Rα mR 2 2 mRα Mg − = Ma 2 Mg a= = 8m / s 2 M + m/2 I=
Now solve the equation of motion: 1 x = at 2 2 t = 2x / a v = at = 2ax = 5.6m / s
2. A lawn roller in the form of a thin-walled hollow cylinder of mass M is pulled horizontally with a constant force F applied by a handle attached to its axle. If it rolls without slipping, find the acceleration and the friction force. Solution: There will be a friction force Ff opposing the external force F, so as to increase the angular velocity as the linear velocity increases (no slipping). Require Newton’s law for both forces and torques: F − F f = Ma F f R = Iα a = Rα I = Mr 2 (thin cylindrical shell, all mass on the surface (radius R) F − Iα / R = Ma F − Ma = Ma a = F / 2M F f = Ma = F / 2
3. Consider a uniform solid disk of radius 25 cm, with a mass of 20 kg. Three 10 cm diameter holes are bored through the disk as shown. What is the moment of inertia of the holey disk, about its axis?
Solution: Use the parallel axis theorem: The moment of inertia of the big disk with no holes, about its axis, is 1 I disk = MR 2 = 0.625 kg m 2 2 The moment of inertia of each small disk (radius r, distance from center of big disk d, mass m = Mr2/R2) that is removed, about the central axis, is .008 .009 1 2 2 I hole = mr + md = .001 + .018 = .019 2 .032 .033 3
So overall, I = I disk − ∑ I holen = 0.565 kg m2 n =1
4. Two astronauts must retrieve a satellite that has begun to tumble in orbit. The satellite is in the shape of a long cylinder, whose mass of 1,000 kg is uniformly distributed inside. The satellite is tumbling end over end, with a period of 5 minutes. The astronauts want to stop the tumbling by catching the satellite at its ends. If they each apply constant force over a maximum stroke of 0.3 m, how much force must they apply to stop the satellite?
Solution: Call the overall length of the cylindrical satellite L. The moment of inertia of a rod rotating about its center in a tumbling motion is 1 I = ML2 12 L The torque exerted by the two astronauts is τ = 2( ) F 2 The angular acceleration is then α = τ / I = 12 F / ML 2π The initial angular velocity is ω = = .021rad / s 5 min The angle through which the ship rotates before stopping is 1 θ = ω 0 t − αt 2 2 ω = ω 0 − αt
ω = 0 when θ =
ω 02 = .018 L / F 2α
We require that the angle θ correspond to a stroke s = 0.3 m. L s = ( )θ = .009 L2 / F = 0.3 2 F = .03L2
5. Communications satellites are placed in geosynchronous orbit: the satellite orbit remains over a fixed location on the Earth’s surface. Calculate the radius of a geosynchronous orbit. What is the highest latitude on Earth’s surface that can receive line-of-sight signals from such satellites? The universal constant of gravitation is 6.67 x 10-11 N m2/kg2. The mass of the earth is 6 x 1024 kg. Hint: you don’t need these, only g and the radius of the Earth, 6400 km, and the length of a day!
Solution: For geosynchronous orbit, we must have that the orbit period equals one day: 2π ω= = 7.3 ⋅ 10 −5 rad / s 24(3600s ) Apply Newton’s law to require that the acceleration from gravity equals that needed to keep it in orbit: R2 F = GMm / r 2 = GM / RE2 m 2E 14243 r g
(
)
v2 F =m = mω 2 r r r = 3
gRE2
ω
2
(9.8m / s )⋅ (6.4 ⋅ 10 m) = (7.3 ⋅ 10 / s ) 2
6
−5
2
2
= 7.5 ⋅ 10 24 m 3
r = 1.9 ⋅ 10 8 m The highest latitude that can receive line-of-sight signals is an angle β from the pole, R 6.4 ⋅ 10 6 m = .034rad = 2° where sin β = E = r 1.9 ⋅ 10 8 m So the highest latitude is 90o-2o = 88o.
6. A uniform pulley wheel 8 cm diameter, mass 1 kg, has a 5 m long cord wrapped around its periphery. Starting from rest, the wheel is given an angular acceleration of 1.47 rad/s2. A) through what angle must the wheel turn for the cord to unwind? B) how long does it take? C) what is the final angular momentum of the wheel?
Solution: A) θ = L / R =
5m = 62.5rad .08m
1 B) 2 t = 2θ / α = 9.2 s 1 L = Iω = MR 2 ω 2 C) ω = αt = 111rad / s
θ = αt 2
L = .08kgm 2 / s