Q57P - University Physics with Modern Physics

Question

A circular saw blade with a radius of 0.120m starts from rest and turns in a vertical plane with a constant angular acceleration of     2.00 rev/s sq. After the blade has turned through 155 rev, a small piece of the blade 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 0.820m to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor.

Step-by-Step Solution

Answer

The horizontal distance from where it broke off the blade, until it strikes the floor, is \(7.70\;{\rm{m}}\).

Step 1: Centrifugal force

On a body moving on a circular path, a force acts on a body radially outward, whose magnitude is equal to the centripetal force. This force is known as centrifugal force. It is a pseudo force and not a reaction to centripetal force.

Step 2: Given Data

Radius of saw blade: \(r = 0.120\;{\rm{m}}\) 

Angular acceleration: \(\alpha  = 2\;{{{\rm{rev}}{\rm{.}}} \mathord{\left/{\vphantom {{{\rm{rev}}{\rm{.}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\) 

Number of revolutions of the blade: \(n = 150\;{\rm{rev}}{\rm{.}}\) 

Distance traveled vertically: \({d_v} = 0.820\;{\rm{m}}\) 

Step 3: Calculate the value of \({{\bf{\alpha }}_{\bf{x}}}\,\,\,{\bf{and}}\,\,\,{\bf{\theta }}\)

Let\(y - z\)   would be the vertical place along which the blade is rotating.

The blade rotates about the x-axis.

For \(R = 0.120\;{\rm{m, }}{\alpha _x} = 2.00\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\;{\rm{and}}\;n = 155\;{\rm{rev}}\), 

The angular acceleration along the x-axis, can be written as-

\(\begin{aligned}{\underline{\phantom{xx}}}{\alpha _x} &= \left( {2.00\;\frac{{{\rm{rev}}}}{{{{\rm{s}}^{\rm{2}}}}}} \right)\left( {\frac{{2\pi \;{\rm{rad}}}}{{1\,\,{\rm{rev}}}}} \right)\\ &= 4.00\pi \;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\end{aligned}\)

The angular displacement for \(155\;{\rm{revolutions}}\) 

\(\begin{aligned}{\underline{\phantom{xx}}}\theta  &= 155\;{\rm{rev}}\left( {\frac{{2\pi \,\,\,{\rm{rad}}}}{{1\,\,{\rm{rev}}}}} \right)\\ &= 310\pi \;{\rm{rad}}\end{aligned}\) 

Since \({\alpha _x}\) is constant and the body started from rest, the third equation of motion, gives- 

\(\begin{aligned}{c}\omega _x^2 &= \omega _{0x}^2 + 2{\alpha _x}\left( {\theta  - {\theta _0}} \right)\\\omega _x^2 &= {\left( 0 \right)^2} + 2\left( {4.00\pi \;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}} \right)\left( {310\pi \;{\rm{rad}} - 0} \right)\\{\omega _x} &= 49.8\pi \;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.}{\rm{s}}}\end{aligned}\)

Here \({\omega _x}\) is the angular velocity and \({\alpha _x}\) is the angular acceleration along x-axis.

Step 2: Determine how far the piece travels horizontally, from where it broke off the blade until it strikes the floor

The velocity of P at that instant is as follows:

\(\begin{aligned}{\underline{\phantom{xx}}}{v_p} &= {\omega _x}R\\ &= \left( {49.8\pi \;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}\right)\left( {0.120\;{\rm{m}}} \right)\\ &=

                 

5.976\pi \;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

 

As the pieces undergo projectile motion in the y-z plane. Breaking the velocity, acceleration and displacement of the piece into two components along y and z- axis.

At the instant when P begins to move along a projectile path with a horizontal initial velocity, let time be taken as \(t = 0\).

So, along the z-axis (vertical motion),

\({u_z} = 0\) 

\({a_z} = 9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\) 

\({s_z} = z\) 

 \({t_z} = t\) 

Applying second equation of motion and inserting the given values-

\(\begin{aligned}{\underline{\phantom{xx}}}z &= {u_z}t - \frac{1}{2}{a_z}{t^2}\\0.820\;{\rm{m}} &= \left( 0 \right)t - \frac{1}{2}\left( {9.8\;{{\rm{m}}\mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right){t^2}\\ - 0.820 &=  - \left( {4.9\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right){t^2}\\{t^2} &= \frac{{0.820\;{\rm{m}}}}{{4.9\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}}}\end{aligned}\) 

\(t = 0.41\;{\rm{s}}\) 

 

For motion along y-axis (horizontal motion)-

 \(\begin{aligned}{l}{u_y} &= {v_p} &= 5.98\pi \;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\\{a_y} &= 0\\{s_y} &= y\\{t_y} &= t &= 0.41\;{\rm{s}}\\{v_y} &= 0\end{aligned}\) 

Applying second equation of motion, we get-

\(\begin{aligned}{l}{s_y} &= {u_y}t + \frac{1}{2}{a_y}{t^2}\\y &= {v_P}t + \frac{1}{2}\left( 0 \right){t^2}\\{v_p} &= \frac{d}{t}\\y &= {v_p}t\end{aligned}\)

For the given values,

\(\begin{aligned}{\underline{\phantom{xx}}}y &= \left( {5.98\pi \;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\left( {0.41} \right)\\ &= 7.70\;{\rm{m}}\end{aligned}\)

The distance, the piece covers horizontally is \(7.70\;{\rm{m}}\).