(5A) Define the center of mass (CM) of an objects which rotates and/or several p
ID: 1466618 • Letter: #
Question
(5A) Define the center of mass (CM) of an objects which rotates and/or several parts of the object move relative to one another.Give an equation expressing the center of mass of an object of two parts,explaining the symbols you use.
(5B) The sun subtends an angles of 0.5 degrees to use on earth,150 million km away. Estimate(i.e. calculate) the radius of the sun.
(5C) (c) and (d)
An automobile engine slows doqn from 3500 rpm to 1200 rpm in 2.5s
(5c) caculate its angular acceleration,assumed constant.
(5d) A 7150kg railroad car travels alone on a level frictionless track with a constant speed of 15.0m/s. A 3350kg load, initiallt at rest, is dropped ontothe car. What will be the car's new speed?
Show work and calculations
please!
Explanation / Answer
5A.explanation
The terms "center of mass" and "center of gravity" are used synonymously in a uniform gravity field to represent the unique point in an object or system which can be used to describe the system's response to external forces and torques. The concept of the center of mass is that of an average of the masses factored by their distances from a reference point. In one plane, that is like the balancing of a seesaw about a pivot point with respect to the torques produced.
for two masses Xcm = (m1 x1 + m2 x2) / m1 + m2
If you are making measurements from the center of mass point for a two-mass system then the center of mass condition can be expressed as
m1*r1 = m2*r2
m1 = m2* (r2/r1)
5C. answer
a)
angular acceleration= change in rpm's/change in time
convert rpm's in to rps
3500 rev/min * 1 min/60 sec= 58.33 rev/sec
1200 rev/min* 1 min/60 sec= 20 rev/sec
angular accel = (20-58.33) / 2.5 rev/sec= -15.33 rev/s/s
5d. answer
this is a conservation of momentum problem
the car that is dropped has no initial horizontal momentum, so the horizontal momentum of the system is conserved, this means that
M v = (M+m) V
where M = 7150 kg, v = 15m/s , m = 3350 kg and V is the speed of the combined system after the car is dropped onto the first car
so we have
V = 7150kg*15m/s / (7150kg + 3350 kg) = 10.21m/s
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