suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the str
ing when the ball is at the highest point
1 answer:
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).
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Answer:
31.25 meters or ~31 meters approximately
Explanation:
Let's see which of the 5 variables we are given since this is a constant acceleration problem.
We want to find the height of the bridge, aka the vertical displacement of the rock. Let's set the upwards direction to be positive and the downwards direction to be negative.
We are told that the acceleration is 10 m/s² downward, so we have a = -10 m/s².
We are also told that the time it takes the rock to hit the water is 2.5 seconds. Time is the same regardless of the x- or y- direction, so we can say that t = 2.5 seconds.
Now, we aren't told this directly, but we can figure out that the velocity in the y-direction is 0 m/s, since the rock is dropped from rest off the bridge. Therefore, .
We want to find the vertical displacement, the height of the bridge, so we can say that
We have 4 out of 5 variables:
Look through the constant acceleration equations to see which equation has all 4 of these variables. You should come up with this one (no final velocity):
Subtract from both sides of the equation to get:
Substitute in our known variables and solve for delta x.
0 m/s multiplied by 2.5 s is 0, so we have:
Evaluate the exponent first and multiply the terms together.
The vertical displacement is -31.25 meters from the rock's starting position, so we can say that the height of the bridge is 31.25 meters, which is approximately 31 meters tall.