<span>We put a motion detector at </span>one end of the track<span> and put a cart on the track. ... Next, we put a motorized fan on the cart and let it push the cart down the track. ... This is what I would expect based on the velocity graph, since </span>acceleration<span> equals the slope of the velocity graph, which remains</span>constant<span> in time.</span>
Answer:
Since the spring mass system will execute simple harmonic motion the position as a function of time can be written as
'A' is the amplitude = 6 inches (given)
is the natural frequency of the system
At equilibrium we have
Applying values we get
thus natural frequency equals
Thus the equation of motion becomes
At time t=0 since mass is at it's maximum position thus we have
Thus the position of mass at the given times is as follows
1) at 
2) at 
3) at 
4) at 
5) at 
Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
