Q BgQuestion:

      
Mentor
Karma Points: 485
Respect (100%):
posted by  nba on 4/23/2009 1:13:36 AM  |  status: Live  |  Earned Karma: 485

Center Of Mass - IMMEDIATE LIFESAVER - (100% Respect)

Course Textbook Chapter Problem Needs by
N/A N/A N/A N/A N/A
Question Details:
The vector position of a 3.50g particle moving in the xy plane varies in the time according to r1 = (3i + 3j)t + 2jt2. At the same time, the vector position of a 5.50g particle varies as r2 = 3i - 2it2 - 6jt, where t is in seconds and r is in cm. At t = 2.50s, determine
a.)the vector position of the center of mass
b.)the linear momentum of the system
c.) the velocity of the center of mass
d.)the acceleration of the center of mass
e.) the net force exerted on the two particle system.
Please show work
Tags: Physics
Bonus Point Alert! Earn +8 additional karma points for helping this gold member.

AAnswers:

Answer Question Ask for clarification
Mentor
Karma Points: 485
posted by nba on 4/23/2009 8:43:01 AM  |  status: Live
Asker's Rating: N/A-Posted by Person Asking Question   
Response Details:
?
Mentor
Karma Points: 485
posted by nba on 4/23/2009 1:37:54 PM  |  status: Live
Asker's Rating: N/A-Posted by Person Asking Question   
Response Details:
?
Oracle
Karma Points: 73,637
posted by zsm28 on 4/23/2009 2:04:16 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
m1 = 3.50 g, r1 = 3t i + (3t + 2t2) j
r1' = 3 i + (3 + 4t) j
r1'' = 4 j
t = 2.50s, r1 = 7.50 i + 20.0 j, r1' = 3 i + 13.0 j, r1'' = 4 j
m2 = 5.50 g, r2 = (3 - 2t2) i - 6t j,
r2' = -4t i - 6 j,
r2'' = -4 i,
t = 2.50s, r2 = -9.5 i - 15.0 j,  r2' = -10 i - 6 j, r2'' = -4 i,
a) rc = (m1r1 + m2r2)/(m1 + m2) = (-2.89 i - 1.39 j) cm
b) P = m1r1' + m2r2' = (-44.5 i + 12.5 j) g-cm/s
c) rc' = P/(m1 + m2) = (-4.94 i + 1.39 j) cm/s
d) rc'' = (m1r1'' + m2r2'')/(m1 + m2) = (0.44 i - 0.44 j) cm/s2
e) (m1 + m2) rc'' = (4 i - 4 j) g-cm/s2 = (4 i - 4 j) * 10-5 N

Oracle
Karma Points: 41,557
posted by richmondMike on 4/23/2009 5:19:02 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:
(Note: I did not notice that r is in cm until the last second. For standard unit, all the numerical values below should be divided by 100.)
Given m1 = 0.0035kg, and m2 = 0.0055kg.
Given t = 2.50s.
(a)
The displacement function of particle 1 with respect to time t is .
The displacement function of particle 2 with respect to time t is .
The vector position of the center of mass is .
(note: the unit is in cm)
(b)
The velocity function of particle 1 with respect to time t is .
The velocity function of particle 2 with respect to time t is .
The momentum is P(t) = , and .
(note: the unit is in kg cm/s)
(c)
The velocity of the center of mass is  .
(note: the unit is in cm/s)
(d)
The acceleration function of particle 1 with respect to time t is .
The acceleration function of particle 2 with respect to time t is .
The acceleration of the center of mass is ...(1).
(note: the unit is in cm/s2)
(e)
The net force exerted on the two particle system is (1) * (m1 + m2) = .
(note: the unit is in newton/100)
Answer Question Ask for clarificarion

Join Cramster's Community

Cramster.com brings together students, educators and subject enthusiasts in an online study community. With around-the-clock expert help and a community of over 100,000 knowledgeable members, you can find the help you need, whenever you need it. Join for free today » How Cramster is different from tutoring »