Two masses (m_A = 4 kg, m_B = 10 kg) are connected via a light kg string around

Question

Two masses (m_A = 4 kg, m_B = 10 kg) are connected via a light kg string around a solid disk frictionless pulley (m_p = 1 kg, r_p = 20 cm). The coefficient of static and kinetic friction between all sur- faces is 0.2 and 0.1 respectively. The angle theta_1 = 30degree while theta _2 = 55 degree. (a) What is the angular acceleration of the disk? (b) What is the tension in the string on both sides of the pulley? If the sys- tem starts from rest, (c) how far does block A travel in 3 s and what is the (d) angular velocity of the disk A space station consists of three modules, to form an equilateral triangle of side length 82.0 m Suppose 100 people, an average mass 75.0 live each capsule and the mass of the modules is negligible compared to the mass of the people. I went to the top left module for a parade what would be the change in position of the center parade Assume the actual station's of mass of the station and top left module, during the artificial mass is negligible)? Suppose now that everyone is back to their respective module and to be gravity is simulated by rotating the station. (c) How fast would one of the modules have an moving to simulate gravity on earth? If the station does not change mass or interact with outside agent, could it actually go from a state of not rotating to a state of rotating? Explain. Suppose that the station is rotating at the artificial Earth gravity speed and the moment of inertia of the station without the people is 2.00 times 10^9 kg m^2. If the distance between each module is reduced to half what will be new (e) tangential velocity the and radial acceleration in units of g?

Explanation / Answer

a) We don't have the mass of the station. We do know that the combined mass of the people is 3 * 100 * 75kg = 22 500 kg
so as far as the people alone are concerned, their center of mass will have moved
sin120 / 82m = sin30 / d d = 47 m

c) radius r = 47 m
a = g = 9.8 m/s² = v² / r = v² / 47 m v = 21.5 m/s

d)

I suppose, if everybody inside had some method of "sticking" to the floor (besides artificial gravity) and if they all started running in the same direction, the station would start spinning in the opposite direction. Of course, when they stopped, the station would stop.

With the people, initial I = 2e9kg·m² + 3 * 100 * 75kg * (47m)² = 2.05e9 kg·m²
and initial L = I * v / r = 9.31e8 kg·m²/s = final L
After retraction, well, who knows? What happens to the MoI of the station? It can't stay constant. Lacking any other data, we might assume that it is 25% of the initial value. Ditto for the occupants (since MoI is proportional to radius²). Then (best guess) I = 5.13e8 kg·m², and

(e)

L = 9.31e8 kg·m²/s = 5.13e8kg·m² * v / 23.7m v = 43 m/s

(f)

and a = (v² / r)/g = (43m/s)² / 23.7m / 9.8m/s² = 5.8 "g"s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.