Question

A rotating merry-go-round makes one complete revolution in 4.0 s (Fig. 8–41). (a) What is the linear speed of a child seated 1.2 m from the center? (b) What is her acceleration (give components)?

FIGURE 8-41 Problem 10

Short answer

(a) The linear speed of the child is 1.9 rad/s.

(b) The acceleration of the child is 3.0m/s².

Step-by-step solution

 Meaning of acceleration

The acceleration of a particle may be defined as the capacity of a particle to increase the object's velocity. Its value varies linearly to the value of the applied force.

 Given information

Given data:

The time is\({\rm{T}} = 4.0\;{\rm{s}}\).

The distance of the child from the center is \({\rm{r}} = 1.2\;{\rm{m}}\).

 Calculate the linear speed of the child

(a)

The angular velocity can be calculated as follow:

\(\begin{aligned}{c}\omega &= \frac{{2\pi }}{T}\\\omega &= \frac{{2\pi }}{{\left({4.0\;{\rm{s}}} \right)}}\\\omega &= \frac{\pi }{2}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

The linear speed of the child can be calculated as follow:

\(\begin{aligned}{c}v &= r\omega \\v &= \left( {1.2\;{\rm{m}}} \right)\left( {\frac{\pi }{2}\;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\v &= 1.9\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the linear speed of the child is 1.9 rad/s.

Calculate the acceleration of the child

(b)

For a uniform circular motion, \(\frac{{dv}}{{dt}} = 0\). Thus, the tangential component of acceleration will be as follows:

\(\begin{aligned}{c}{{\vec a}_t} = \frac{{dv}}{{dt}}\\{{\vec a}_t} = 0\;{{\rm{m}} \mathord{\left/{\vphantom{{\rm{m}}{{{\rm{s}}^{\rm{2}}}}}}\right.}{{{\rm{s}}^{\rm{2}}}}}\end{aligned}\)

The radial component of the acceleration can be calculated as follows:

\(\begin{aligned}{c}{{\vec a}_r} &= \frac{{{v^2}}}{r}\\{{\vec a}_r} &= \frac{{{{\left( {1.9\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)}^2}}}{{\left( {1.2\;{\rm{m}}} \right)}}\\{{\vec a}_r} &= 3.0\;{{\rm{m}} \mathord{\left/{\vphantom{{\rm{m}}{{{\rm{s}}^{\rm{2}}}}}}\right.}{{{\rm{s}}^{\rm{2}}}}}\end{aligned}\)

The magnitude of the acceleration of the child can be calculated as follows:

\(\begin{aligned}{c}a = \sqrt {a_r^2 + a_t^2} \\a = \sqrt {{{\left( {3.0\;{{\rm{m}} \mathord{\left/{\vphantom{{\rm{m}}{{{\rm{s}}^{\rm{2}}}}}}\right.} {{{\rm{s}}^{\rm{2}}}}}} \right)}^2} + 0} \\a = 3.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\end{aligned}\)

Thus, the acceleration of the child is 3.0 m/s².