Chapter 8: 8-7P (page 198)

(a) A grinding wheel 0.35 m in diameter rotates at 2200 rpm. Calculate its angular velocity in \({{{\bf{rad}}}\mathord{\left/{\vphantom{{{\bf{rad}}} {\bf{s}}}} \right.} {\bf{s}}}\).(b) What are the linear speed and acceleration of a point on the edge of the grinding wheel?

Short Answer

Expert verified

(a)The angular velocity of the wheel in terms of\({{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\)is 230.26 rad/s.

(b) The linear speed and acceleration of the point on the edge of the grinding wheel are 40.3 m/s and \(9.28 \times {10^3}\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\), respectively.

Step by step solution

Calculation of radial acceleration

The value of the radial acceleration of an object can be calculated by multiplying the square of the angular velocity with the radius of the rounded path.

Given information

Given data:

The diameter of the wheel is\(d = 0.35\;{\rm{m}}\).

The angular velocity is \(\omega = 2200\;{\rm{rpm}}\).

Calculate the angular velocity of the wheel in terms of rad/s

(a)

The angular velocity of the wheel in terms of\({{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\)can be calculated as follows:

\(\begin{aligned}{l}\omega &= \left( {2200\;{\rm{rpm}}} \right)\left( {\frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}}} \right)\left( {\frac{{1\;\min }}{{60\;{\rm{s}}}}} \right)\\\omega &= 230.26\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the angular velocity of the wheel in terms of rad/s is 230.26 rad/s.

Calculate the linear speed and acceleration of the point on the edge of the grinding wheel

(b)

The radius of the wheel can be calculated as follows:

\(\begin{aligned}{l}r &= \frac{d}{2}\\r &= \frac{{\left( {0.35\;{\rm{m}}} \right)}}{2}\\r &= 0.175\;{\rm{m}}\end{aligned}\)

The linear speed of the pointon the edge of the grinding wheel can be calculated as follows:

\(\begin{aligned}{l}v &= r\omega \\v &= \left( {0.175\;{\rm{m}}} \right)\left( {230.26\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\v &= 40.3\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

The acceleration of the pointon the edge of the grinding wheel can be calculated as follows:

\(\begin{aligned}{l}{a_R} &= {\omega ^2}r\\{a_R} &= {\left( {230.26\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)^2}\left( {0.175\;{\rm{m}}} \right)\\{a_R} &= 9.28 \times {10^3}\;{{\rm{m}} \mathord{\left/{\vphantom{{\rm{m}}{{{\rm{s}}^{\rm{2}}}}}}\right.}{{{\rm{s}}^{\rm{2}}}}}\end{aligned}\)

Thus, the linear speed and acceleration of the point on the edge of the grinding wheel are 40.3 m/s and \(9.28 \times {10^3}\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\), respectively.