Answer: It will fall short of its goal.
Explanation: I took the quiz.
The complete question is missing, so i have attached the complete question.
Answer:
A) FBD is attached.
B) The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
Explanation:
A) I've attached the image of the free body diagram.
B) The formula for the net force is given as;
F_net = mv²/r
We know that angular velocity;ω = v/r
Thus;
F_net = mω²r
Now, the minimum downward force is the weight and so;
mg = m(ω_min)²r
m will cancel out to give;
g = (ω_min)²r
(ω_min)² = g/r
ω_min = √(g/r)
The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
Tell your instructor or teacher
Answer:
Explanation:
Moment of inertia of the metal rod pivoted in the middle
= M l² / 12
If the spring is compressed by small distance x twisting the rod by angle θ
restoring force by spring
= k x
moment of torque about axis
= k x l /2
= k θ( l /2 )² ( x / .5 l = θ )
=
moment of torque = moment of inertia of rod x angular acceleration
k θ( l /2 )² = M l² / 12 d²θ/dt²
d²θ/dt² = 3 k/M θ
acceleration = ω² θ
ω² = 3 k/M
ω = √ 3 k / M
Answer:
a.) the speed at the bottom is greater for the steeperhill
Explanation:
since the energy at the bottom of the steeper hilis greater

As we can see from above that v is higher when h ishigher.