Question

a wheel with a tire mounted on it rotates at a constant rate of 2.89 times a second. A tack is stuck in the tire at a distance 39.1cm from the rotation axis. noting that for every rotation the tack travels one circumference. find the tacks tangential speed.
__________m/s
what is the tacks radial acceleration?
___________m/s^2

Answer 1

Concepts and reason

The concepts required to solve the given problem tangential speed, relation of tangential speed with angular speed and centripetal acceleration.

Initially, determine the angular speed of the wheel when it rotates 2.89 times in one second. Then, use relation of angular speed with tangential speed to find the tangential speed of the tack by knowing its distance from the axis of rotation of the wheel. Finally, determine the radial acceleration of the tack by using expression for centripetal acceleration.

Fundamentals

The tangential speed v of an object rotating at an angular speed $\omega$ is given by following expression.

$v = r\omega$

Here, r is the distance to the object from the axis of rotation.

The centripetal acceleration ${a_C}$ of a rotation object with tangential speed v is given by following expression.

${a_C} = \frac{{{v^2}}}{r}$

An object needs to cover $2\pi$ radians to complete one revolution.

$1\,{\rm{rev = 2}}\pi \,{\rm{rad}}$

The wheel rotates 2.89 times a second that means wheel complete 2.89 revolutions in a second. Therefore, the angular speed $\omega$ of the wheel is given as follows:

$\begin{array}{c}\\\omega = 2.89\,{\rm{rev/s}}\left( {\frac{{2\pi \,{\rm{rad}}}}{{1\,{\rm{s}}}}} \right)\\\\ = 18.158\,{\rm{rad/s}}\\\end{array}$

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression.

$v = r\omega$

Here, r is the distance to the tack from axis of rotation.

Substitute 18.158 rad/s for $\omega ,$ and 39.1 cm for r in the above equation to solve for v.

$\begin{array}{c}\\v = \left( {39.1\,{\rm{cm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)\left( {18.158\,{\rm{rad/s}}} \right)\\\\ = 7.10\,{\rm{m/s}}\\\end{array}$

Use expression for centripetal acceleration to find the radial acceleration.

The expression for centripetal acceleration ${a_C}$ is,

${a_C} = \frac{{{v^2}}}{r}$

Substitute 7.10 m/s for v, and 39.1 cm for r in the above equation to solve for centripetal acceleration.

$\begin{array}{c}\\{a_C} = \frac{{{{\left( {7.10\,{\rm{m/s}}} \right)}^2}}}{{\left( {39.1\,{\rm{cm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)}}\\\\ = 129\,{\rm{m/}}{{\rm{s}}^2}\\\end{array}$

Ans:

The tangential speed of the tack is 7.10 m/s.

The radial acceleration of the tack is $129\,{\rm{m/}}{{\rm{s}}^2}.$