# At the instant shown, the sphere is moving in the circular, and at the same instant ? = 30 ?...

## Question:

At the instant shown, the sphere is moving in the circular, and at the same instant {eq}\beta= 30^{\circ} {/eq} {eq}\dot{\beta }= 3 rad/s {/eq} and {eq}\ddot{\beta }= -5 rad/s^2 {/eq} . Determine the absolute velocity and acceleration of the sphere at this instant.

## Velocity and acceleration:

The first derivative of distance and the time function with respect to time gives the magnitude of velocity, whereas the second derivative of distance and the time function with respect to time gives acceleration.

## Answer and Explanation:

**Given Data**

- The distance of the circular arc from centre is: {eq}r = 15\;{\rm{in}} {/eq}

- The angular velocity is: {eq}\dot \beta = 3\;{\rm{rad/s}} {/eq}

The distance travelled by the sphere is,

{eq}x = r \times \beta {/eq}

The velocity is,

{eq}v = \dfrac{{dx}}{{dt}} {/eq}

Substitute the values in above equation.

{eq}\begin{align*} v &= \dfrac{{d\left( {r\beta } \right)}}{{dt}}\\ & = r\dot \beta \end{align*} {/eq}

Substitute the values in above equation.

{eq}\begin{align*} v &= 15\;{\rm{in}}\left( {\dfrac{{1\;{\rm{ft}}}}{{12\;{\rm{in}}}}} \right) \times {\rm{3}}\;{\rm{rad/s}}\\ & = {\rm{3}}{\rm{.75}}\;{\rm{ft/s}} \end{align*} {/eq}

Thus, the absolute velocity is {eq}3.75\;{\rm{ft/s}}. {/eq}

The normal acceleration is,

{eq}{a_n} = \dfrac{{{v^2}}}{r} {/eq}

Substitute the values in above equation.

{eq}\begin{align*} {a_n}& = \dfrac{{{{\left( {3.75\;{\rm{ft/s}}} \right)}^2}}}{{15\;{\rm{in}}\left( {\dfrac{{1\;{\rm{ft}}}}{{12\;{\rm{in}}}}} \right)}}\\ & = 11.25\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{align*} {/eq}

The tangential acceleration is,

{eq}\begin{align*} {a_t}& = \dfrac{{dv}}{{dt}}\\ & = \dfrac{{d\left( {r\dot \beta } \right)}}{{dt}}\\ & = r\ddot \beta \end{align*} {/eq}

Substitute the values in above equation.

{eq}\begin{align*} {a_t}& = 15\;{\rm{in}}\left( {\dfrac{{1\;{\rm{ft}}}}{{12\;{\rm{in}}}}} \right) \times \left( { - 5\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}} \right)\\ & = - 6.25\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{align*} {/eq}

The absolute acceleration is,

{eq}a = \sqrt {{{\left( {{a_n}} \right)}^2} + {{\left( {{a_t}} \right)}^2}} {/eq}

Substitute the values in above equation.

{eq}\begin{align*} a &= \sqrt {{{\left( {11.25} \right)}^2} + {{\left( { - 6.25} \right)}^2}} \\ & = \sqrt {165.625} \\ & = 12.869\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{align*} {/eq}

Thus, the absolute acceleration is {eq}12.869\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} {/eq}

#### Learn more about this topic:

from Physics 101: Help and Review

Chapter 17 / Lesson 15