Mechanics II Flashcards
Center of Mass, Uniform Circular Motion, Torque, Equilibrium, Rotational Inertia
Equation for finding center of mass
(x1m1) + (x2m2) + (x3m3) … + (Xnmn)
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m1 + m2 + m3 + mn
Is the center of mass to the R or L of the origin?
Depends on where you place your zero mark. Usually you place your zero mark all the way to the L end of the origin. If this is the case, the center of mass would be to the R of that origin (as you would always have a positive number). However if you chose to use the middle of a pole/stick for an origin, a positive number would mean that the center of mass is to the R of that origin, while a negative number would mean that the center of mass is to the L of that origin.
An object moving in a circular path is said to execute uniform circular motion if its ____ is constant
speed
Which way does velocity, acceleration, and centripetal force point when an object undergoes uniform circular motion?
the velocity vector always points tangent to the objects path, even though the magnitudes of these vectors are always the same (because speed is constant)
centripetal acceleration and centripetal force always point towards the center of the circle
Equation for centripetal acceleration (ac)
ac = v2/r
Equation for centripetal force (Fc)
Fc = mac = mv2 / r
Equation for translation velocity (for an object experiencing uniform centripetal motion)
Will centripetal acceleration increase or decrease for an object if it is moved further from the axis of rotation?
ac will increases for an object when placed further from the axis of rotation
translational velocity: v = rw
ac = v2 / r
therefore, ac = rw2 → ac is proportional to r, so as r increases, there is a proportional increase in ac
If an object undergoing uniform circular motion is being acted upon by a constant force toward the center, why doesn’t the object fall into the center?
Actually, it is falling toward the center, but because of its speed, the object remains in a circular orbit around the center. Remember the direction of v is not necessarily the same as the direction of Fnet. So, just because Fnet points toward the center doesn’t mean that v must point toward the center. It’s the direction of the acceleration, no the velocity that always matches the direction of Fnet. Furthermore, if it weren’t for Fnet point toward the center (that is, if the Fc were suddenly removed), then the object’s velocity wouldn’t change (it would fly off in a straight line tangent to the circle at the point where the force was removed)
How would the net force on an object undergoing uniform circular motion have to change if the object’s speed doubled?
Fc would have to increase by a factor of 4
Fc = mv2/r → Fc is proportional to the square of the speed
If speed increases by a factor of 2, Fc would increase by a factor of 22 = 4
Equation for force of gravity (Fgrav)
Fgrav = GMm / r2
The moon orbits the eart in a nearly circular path at nearly constant speed. If M is the mass of the earth, m is the mass of the moon, and r is the radius of the moon’s orbit, find an expression for the speed of the moon’s orbit.
First, determine what is providing the centripetal force? The answer is the gravitational pull by the earth. Thus, set Fgrav equal to Fc:
GMm/r2=mv2/r → GM/r=v2 → v=(GM/r)0.5
Note that the mass of the moon cancels out; thus the mass of any object orbiting at the same distance from the earth as the moon must move at the same speed as the moon.
A rope is tied to the handle of a bucket, and the bucket is then whirled in a vertical circle. At the bottom of its path, the tension of the rope is 50N. How would you find the net force of this bucket?
The net force of the bucket (and therefore the centripetal force of the bucket) would be the tension force minus the weight force:
Fnet = FT - Fw = 50N - mg
m = mass of bucket
What’s the difference between centripetal acceleration and tangential acceleration in regards to speed?
centripetal acceleration only makes an object turn so that it moves in a circular path; therefore it doesn’t change the speed.
tangential acceleration changes the speed of the object, as it is the force component (vector) that opposes the direction of the object’s velocity (therefore is the opposite tangent to the velocity tangent; both tangents are perpendicular to the center of the circle, just in opposite directions). This is the reason that an object’s speed decreases when traveling from the bottom to the top of a circle, and increases when traveling from the top to the bottom of a circle
Equation for torque
T = rFsin(theta)
or
T = lF (l is the lever arm, which is always perpendicular to the line of action of the Force applied, which means theta is 90º, and sin90º=1)
(T is really the Greek letter tau)