Answer:
[tex]v_{m}=2.9 m/s[/tex]
Explanation:
At the top of the circular motion, we will have these forces:
[tex]N+W=F_{c}[/tex]
Where:
N is the normal force
W is the weight
Fc is the centripretal force
Now, if we assume the normal force equal to zero, we could find the minimum value of the speed water to spill out, It means at any value lower than this value, water fallout of the pail.
So we have:
[tex]mg=ma_{c}=m\frac{v^{2}}{R}[/tex]
[tex]g=\frac{v^{2}}{R}[/tex]
[tex]v_{m}=\sqrt{gR}[/tex]
v(m) means the minimum value of speed.
[tex]v_{m}=\sqrt{9.81*0.86}[/tex]
[tex]v_{m}=2.9 m/s[/tex]
I hope it helps you!
The weight of the bucket with water must be equal to the centripetal force. The answer is V = 2.9 m/s
Given that a bucket full of water is rotated in a vertical circle of radius 0.86 m. To calculate the minimum speed of the pail at the top of the circle if no water is to spill out, the weight of the bucket with water must be equal to the centripetal force. That is,
[tex]mV^{2}/r[/tex] = mg
mass m will cancel out.
[tex]V^{2}/r[/tex] = g
[tex]V^{2}[/tex] = rg
V = [tex]\sqrt{rg}[/tex]
V = [tex]\sqrt{0.86 * 9.8}[/tex]
V = [tex]\sqrt{8.428}[/tex]
V = 2.9 m/s
Therefore, the minimum speed of the pail at the top of the circle if no water is to spill out is 2.9 m/s approximately.
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