make your answer clear and logic legible if you want a positive rating Two masse
ID: 1458979 • Letter: M
Question
make your answer clear and logic legible if you want a positive rating
Two masses (mA= 1 kg, mB= 5 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively.
a.) Where is the center of mass of this system?
cm, from mass A.
b.) The system is then hung from a string, so that it stays horizontal. Where should the string be placed?
cm, from mass A.
c.) Now, if mass B was removed, how much force would need to be exerted at the 100 cm mark in order to keep the meter stick level*?
size: N, dir: ---Selectanswer, its either---( upwards or downwards)
d.) Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant*.
rad/s2
*For parts c-d, assume the string remains attached at the same location you found in part b.
Explanation / Answer
Let:
T be the tension in the string,
x cm. be the distance of the centre of mass from the 0cm. mark.
(a)
Moments about the centre of mass:
1 * x = 5(75 - x)
375 - 5x = x
x = 375 / 6
= 62.5 cm, from mass A.
(b)
As the masses have no moment about the C of M, that is where the string should be attached.
(c)
Let:
F be the downward force needed at 100cm,
x cm. be the new string position,
9.81 m/s^2 be the acceleration due to gravity.
Moments about the end of the string:
1 * 0.625 * 9.81 = F(1 - 0.625)
F = 16.35 N.
(d)
Let:
a be the acceleration (initially downwards) of the mass.
g be the acceleration due to gravity.
mg = ma
a = g.
The angular acceleration is:
9.81 / 0.625
= 15.7 rad / s.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.