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Solved: A circular saw blade with radius 0.120 m starts
Chapter 9, Problem 63P(choose chapter or problem)
A circular saw blade with radius 0.120 m starts from rest and turns in a vertical plane with a constant angular acceleration of \(3.00 \ rev/s^2\). After the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor. How far does the piece travel horizontally from where it broke off the blade until it strikes the floor?
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QUESTION:
A circular saw blade with radius 0.120 m starts from rest and turns in a vertical plane with a constant angular acceleration of \(3.00 \ rev/s^2\). After the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor. How far does the piece travel horizontally from where it broke off the blade until it strikes the floor?
ANSWER:
Step 1 of 5
We know that, the angular displacement, \(\boldsymbol{\theta}=\boldsymbol{\omega}_{0} \mathrm{t}+1 / 2 \boldsymbol{\alpha} \mathrm{t}^{2}\)
Where,
\(\boldsymbol{\omega}_{0}\) - Initial angular velocity
\(\alpha\) - angular acceleration
\(t\) - time for angular displacement
Provided, the blade starts moving from the rest. Therefore, \(\boldsymbol{\omega}_{0}=0 \ \mathrm{rad} / \mathrm{s}\)
Therefore, we can write, \(\boldsymbol{\theta}=1 / 2 \boldsymbol{\alpha} \mathrm{t}^{2}\)
Provided, \(\boldsymbol{\theta}=155 \ \mathrm{rev}\) and \(\boldsymbol{\alpha}=3 \ \mathrm{rev} / \mathrm{s}^{2}\)
Substituting these values in the equation, we get, \(155\mathrm{\ rev}=1/2\times3\mathrm{\ rev}/\mathrm{s}^2\times\mathrm{t}^2\)
Rearranging, \(\mathrm{t}^{2}=(155 \mathrm{rev} \times 2) / 3 \mathrm{rev} / \mathrm{s}^{2}=310 \mathrm{rev} / 3 \mathrm{rev} / \mathrm{s}^{2}\)
\(\mathrm{t}^{2}=103.33 \mathrm{~s}^{2}\)
Taking square root on both sides, \(\mathrm{t}=10.16 \mathrm{~s}\)
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