Oscillations and Gravitation questions?
1. Pluto is an average distance of 5.91 x 10^9 km from the Sun. Estimate the length of the Plutonian year given that the Earth is 1.50 x 10^8 km from the Sun on the average.
2.Tarzan plans to cross a gorge by swinging in an arc from a hanging vine. If his arms are capable of exerting a force of 1100 N on the rope, what is the maximum speed he can tolerate at the lowest point of his swing? His mass is 74 kg and the vine is 4.8 m long.
3. A curve of radius 90 m is banked for a design speed of 60 km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve?
4. While fishing, you get bored and start to swing a sinker weight around in a circle below you on a 0.20 m piece of fishing line. The weight makes a complete circle every 0.70 s. What is the angle that the fishing line makes with the vertical?
For the third one the car’s weight isn’t given, so what should I do?
1) Kepler’s law relates the period, T, and the semi-major axis, a.
T^2 is proportional to a^3.
So T2 = T1 * (r2/r1)^(3/2)
This calculation will be flawed, however, if you try to plug in the average distance of pluto from the sun. Pluto is fairly eccentric, so the semi-major axis is significantly less than the average distance.
2) Tarzan will have to hold up his own weight plus the centrifugal force since he’s rotating.
F = mg + mv^2/r
Solve for the maximum speed:
v = sqrt ( (F – mg) * r)
They give you the force and his mass and the length of the vine. You know the acceleration due to gravity at the earth’s surface, g. Plugnchug.
3) The banking cancels the centrifugal force,
F = mv^2/r
Now increase the force by mu (the coefficient of friction) times the car’s weight and recalculate to estimate the max speed. Now decrease the force by mu times the car’s weight to get the min speed. If mu times the car’s weight exceeds the bank’s force (which normally it should), then the car can be stationary on the bank.
4) Find the centrifugal force:
F = m omega^2 r = m (2 pi / T)^2 R
= m (2 pi / T)^2 r cos(theta)
The force of gravity is mg
the tangent of the angle is the vertical (gravity) over horizontal (centrifugal) forces.
tan(theta) = mg / ( m (2 pi / T)^2 r cos(theta))
Solve for the angle:
theta = arcsine (g (T/2pi)^2 / r)
That’s actually the angle with horizontal. It’s complement will be the angle with vertical.
Elmer’s workshop making cod jigs