1.An object is attached to a 1.60 meter rope and twirled in a horizontal circle on a frictionless table. The tension on the string is 585 Newtons and the ball takes 1.46 seconds to complete one revolution. Determine the mass of the object (in kg to two decimal places).
1a. if the mass of the object were doubled (keeping a constant radius on the string), what would have to happen to the period in order to maintain the same tension on the string? Please explain fully.
2. A 2.4 gram penny is on a turntable. It starts from rest at a radius of 0.14 meters and the turntable speeds up until the penny flies off with a tangential speed of .88 m/s. What coefficient of static friction (to three decimal places) exists between the penny and turntable if the penny doesn't fly off until it reaches this speed?
2a. Suppose a 5.7-gram quarter has the same coefficient of static friction between it and the turntable as does the penny? In what way will the speed at which it flies off differ from that of the penny? Please explain fully.
3. Planet Zwork orbits a massive sun much as the planets of our solar system orbit our sun. It takes planet Zwork 32.8 earth years to orbit the sun as planet Zwork is a distance of 89.7 x 10^12 meters from its center to the center of its sun. Determine the period for planet Kep to orbit if the distance from its center to the center of the planets' sun is 64 x 10^12 meters.
3a. If the sun's mass suddenly increased, would Kep's period need to increase or decrease in order to stay in orbit? Please explain fully.
4. In a popular amusement park ride, a rotating cylinder of radius 3 meters is set into rotation with a period of 2.87 seconds. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction (to three decimal places) is needed between a rider's clothing and the wall to keep the rider from slipping?
4a. Suppose the period of rotation of the cylinder is doubled. By what factor would the minimum coefficient of friction change need to change in order to keep the rider from slipping? Please explain fully.
Please see the attached file.
Solution to problem 1
As the object tied to the rope goes around the circular path, the tension in the rope provides the centripetal force required for the circular motion. Therefore, we can write the following equation :
F = mv2/r = mω2r = m(2Π/T)2r ..............(1)
Where, F = Tension in the rope = 585N
m = mass of the object(to determine)
T = Time period of rotation = 1.46 sec.
r = Radius of the circular path = 1.6 m
Putting values we get, 585 = m(2Π/1.46)21.6
Solving for m we get, m = 19.73 kg
1a) Rearranging (1) for T we get : T = √4mΠ2r/F ........(2)
Putting in (2) m = 2 x 19.73 = 39.46 kg, F = 585 N and r = 1.6 m
we get : T = √4x39.46xΠ2 x 1.6/585 = 2.065 sec
Solution to problem 2
As the turn table rotates the penny, ...
These solutions on circular and planetary motion give an insight into the concepts of friction, centripetal force, Kepler's laws etc.